1GyaanSankalp

Electric charges and fields

ELECTRIC CHARGES AND FIELDSC H A P T E R

1 6 LEARNING OBJECTIVES

(i) Electric and magnetic forces determine the properties of atoms, molecules and bulk matter. Coulomb force and gravitationalforce follow the same inverse-square law. But gravitational force has only one sign (always attractive), while Coulomb force canbe of both signs (attractive and repulsive), allowing possibility of cancellation of electric forces. This is how gravity, despitebeing a much weaker force, can be a dominating and more pervasive force in nature.(ii) Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This isnot always true for every scalar. For example, kinetic energy is a scalar under rotation, but is not invariant for frames of referencein relative motion. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge.Conservation refers to invariance in time in a given frame of reference. A quantity may be scalar but not conserved (like kineticenergy in an inelastic collision). On the other hand, one can have conserved vector quantity (e.g., angular momentum of anisolated system). Quantisation of electric charge is a basic (unexplained) law of nature; interestingly, there is no analogous law onquantisation of mass.(iii) Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 andinversely proportional to the square of the distance r21 separating them.

Mathematically, 21F

= force on q2 due to q1 = 1 2212

21

k (q q )r

r where 21r is a unit vector in the direction from q1 to q2 and

0

1k4

is the constant of proportionality. In SI units, the unit of charge is coulomb. The experimental value of the constant 0 is

12 2 1 20 8.854 10 C N m . The approximate value of k is k = 9 × 109 Nm2C–2.

(iv) Superposition principle should not be regarded as ‘obvious’, or equated with the law of addition of vectors. It says twothings: force on one charge due to another charge is unaffected by the presence of other charges, and there are no additionalthree-body, four-body, etc., forces which arise only when there are more than two charges.(v) The electric field E at a point due to a charge configuration is the force on a small positive test charge q placed at the pointdivided by the magnitude of the charge. Electric field due to a point charge q has a magnitude | q |/40r2 it is radially outwardsfrom q, if q is positive, and radially inwards if q is negative. Like Coulomb force, electric field also satisfies superposition principle.The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges. For continuousvolume charge distribution, it is defined at any point in the distribution. For a surface charge distribution, electric field isdiscontinuous across the surface.(vi) An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electricfield at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowdnear each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electricfield, the field lines are uniformly spaced parallel straight lines. Some of the important properties of field lines are: (i) Field linesare continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positivecharges and end at negative charges —they cannot form closed loops.(vii) An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vectorp has magnitude 2qa and is in the direction of the dipole axis from –q to q.Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a

distance r from the centre: 2 2 3/20

p 1E4 (a r )

3

0

p4 r

, for r >> a

Dipole electric field on the axis at a distance r from the centre: 2 2 2 30 0

2pr 2pE4 (r a ) 4 r

for r >> a

The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point

charge. In a uniform electric field E , a dipole experiences a torque

given by p E but experiences no net force.

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Electric charges and fields

(viii) The flux of electric field E through a small area element S

is given by E. S

The vector area element S is ˆS S n

where S is the magnitude of the area element and n is normal to the area element, which can be considered planar for sufficientlysmall S. For an area element of a closed surface, n is taken to be the direction of outward normal, by convention.(ix) Gauss’s law: The flux of electric field through any closed surface S is 1/0 times the total charge enclosed by S. The law isespecially useful in determining electric field E

, when the source distribution has simple symmetry:

(a) Thin infinitely long straight wire of uniform linear charge density 0

ˆE n2 r

where r is the perpendicular distance of the point from the wire and n is the radial unit vector in the plane normal to the wirepassing through the point.

(b) Infinite thin plane sheet of uniform surface charge density 0

ˆE n2

(c) Thin spherical shell of uniform surface charge density 20

ˆE r4 r

(r R) E 0

(r < R)

INTRODUCTIONMany students feel that in electrostatic charge remains at rest & hence formulae derive in electrostatic are applicable when chargeis at rest but fact is formulae are applicable when charge is in motion only the difference is when charge is in motion we willconsider additional effect called magnetic effect. Electrostatics deals with the study of forces, fields and potentials arising fromstatic charges. Like mass, electric charge is an intrinsic property of protons and electrons.In nature, atoms are normally found with equal numbers of protons and electrons, i.e. atom is electrically neutral.The charge on an electron or a proton is the smallest amount of free charge that has been discovered. Charges of larger magnitudeare built up on an object by adding or removing electrons. If in a body there is excess of electrons over its neutral configuration,conventionally the body is said to be negatively charged and if there is deficiency of electron it is said to be positively charged.

– ve charged body Body has gained electrons+ ve charged boy Body has lost some electrons+ ve & – ve charge named by benjamin Franklin.

INTERESTING EXPERIMENTTake any two materials from the following list and then rubbed with each other. We can always find that the former one ispositively charged and the later one is negatively charged.

Fur glass paper metal silk plastic amber rubber sulfurWhen a charged body is close enough to a neutral body, they attract each other. One of the applications of this effect is to use tinypaint droplets to paint the automobiles uniformly.

CONDUCTOR AND INSULATORSSuppose you charge a rubber rod and then touch it to a neutral object. Some charge, repelled by the negative charge on the rod,will be transferred to the originally-neutral object. What happens to that charge then depends on the material of which theoriginally-neutral object consists. In the case of some materials, the charge will stay on the spot where the originally neutral objectis touched by the charged rod. Such materials are referred to as insulators, materials through which charge cannot move, or,through which the movement of charge is very limited. Examples of good insulators are quartz, glass, and air. In the case of othermaterials, the charge, almost instantly spreads out all over the material, in response to the force of repulsion (recalling that forcecauses acceleration which leads to the movement) that each elementary particle of the charge exerts on every other elementaryparticle of charge. Materials in which the charge is free to move about are referred to as conductors. Examples of good conductorsare metals and saltwater.When you put some charge on a conductor, it immediately spreads out all over the conductor. The larger the conductor, the moreit spreads out. In the case of a very large object, the charge can spread out so much that any chunk of the object has a negligibleamount of charge and hence, behaves as if were neutral. Near the surface of the earth, the earth itself is large enough to play sucha role. If we bury a good conductor such as a long copper rod or pipe, in the earth, and connect to it another good conductor such

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Electric charges and fieldsas a copper wire, which we might connect to another metal object, such as a cover plate for an electrical socket, above but near thesurface of the earth, we can take advantage of the earth’s nature as a huge object made largely of conducting material. If we toucha charged rubber rod to the metal cover plate just mentioned, and then withdraw the rod, the charge that is transferred to the metalplate spreads out over the earth to the extent that the cover plate is neutral. We use the expression “the charge that wastransferred to the cover plate has flowed into the earth.” A conductor that is connected to the earth in the manner that the coverplate just discussed is connected is called “ground.” The act of touching a charged object to ground is referred to as groundingthe object. If the object itself is a conductor, grounding it (in the absence of other charged objects) causes it to become neutral.

CHARGING BY INDUCTIONWhen a charged particle is taken near to neutral metallic object then the electrons move to one side and there is excess of electronson that side making it negatively charged and deficiency on the other side making that side positively charged. Hence chargesappear on two sides of the body (although total charge of the body is still zero). This phenomenon is called induction and thecharge produced by it is called induced charge.

Step 1 :

+++++

+++

+++

+

Charged body isbrought nearuncharged body

Charged body V'= +ve

q'=0

Step 2 :

+++

+++

+

Uncharged body isconnected to theearth

q' = –veV' = 0

e

Step 3 : Uncharged body isdisconnected from theearth

q' = –veV' = 0

+++

+++

+

Step 4 : Charging bodyis removed

q' = –veV' = –ve

A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermoionic ionisation, (e) photoelectric effectand (f) field emission.

BASIC PROPERTIES OF ELECTRIC CHARGE(1) Charge is a scalar and can be of two types (i.e. + ve or – ve). It adds algebraically.(2) Charge is conserved. During any process (chemical, nuclear, decay etc.) the net electric charge of an isolated system remains

constant.In the process one body gains some amount of – ve charge while the other gains an equal amount of + ve charge.Pair - production, Annihilation are processes understand on basis of charge conservation.

(3) Charge is Quantized (exists as discrete "Packets") : Robert Millikan discovered that electric charge always occurs as some integralmultiple of fundamental unit of charge (e).

q = Ne [N is some integer]

Charge on a body can never be 13

e,

23

e etc. as it is due to transfer of electron.

(4) Through large number of experiments it is also well established that similar charges repel each other while dissimilar attract.Here it is worth noting that true test of electrification is repulsion and not attraction as attraction may also take place between acharged bodies.

(5) Charge is always associated with mass i.e. charge can not exist without mass though mass can exist without charge.(6) Charge is transferable. Process of charge transfer is called conduction.(7) Charge is invariant i.e. it is independent on frame of reference.(8) Accelerated charge radiates energy.

Charge at rest produces Electric & magnetic effect.Accelerate charge particle Electric & magnetic effect + radiate energy (According to electromagnetic theory)

+produces E

v = 0 + v = const.

with no radiation Band produces E

+ v const.

radiates energyB

and produces E and

(9) Charge resides on the outer surface of a conductor(10) How to express charge.

The SI unit for measuring the magnitude of an electric charge is the coulomb (C).Current drift of charge per unit time

I = q/t q = It

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Electric charges and fields1 coulomb 1 ampere × 1 sec

If a charge of 1 coulomb drift per second through cross-section of conductor current flowing is called 1 ampere.Charge on electron = – 1.6 × 10–19 C, Charge on proton = + 1.6 × 10–19 CThe coulomb is related to CGS units of charge through the basic relation.

1 Coulomb = 3 × 109 esu of charge (static coulomb or frankline) = 1

10 emu of charge

Practical units of charge are amp × hr (= 3600 coulomb), Faraday ( = 96500 coulomb)Example 1 :

How many electrons are there in one coulomb of negative charge.Sol. Number N of electrons is

N = qe

= 191.00C

1.6 10 C

= 6.25 × 1018

Charge on 6.25 × 1018 electrons = – 1 CExample 2 :

A copper penny (Z = 29) has a mass of 3g. What is the total charge of all the electrons in the penny ?Sol. The electrons have a total charge given by the number of electrons in the penny, Ne, times the charge of an electron, –e. The

number of electrons is 29 times the number of copper atoms N. To find N, we use the fact that one mole of any substance hasAvogadro's number (NA = 6.02 × 1023) of molecules, and the number of grams in a mole is the molecular mass M, which is 63.5 forcopper. Since each molecule of copper is just one copper atom, we find the number of atoms per gram by dividing NA atoms/moleby M grams/mole.1. The total charge is the number of electrons times the electronic charge : Q = Ne (– e)2. The number of electrons is Z time the number of copper atoms Na : Ne = ZNa

3. Compute the number of copper atoms in 3g of copper : Na = (3g) 236.02 10 atoms / mol

63.5g / mol

= 2.84 × 1022 atoms

4. Compute the number of electrons Ne ,Ne = ZNa = (29 electrons/atom) (2.84 × 1022 atoms) = 8.24 × 1023 electrons.5. Use this value of Ne to find the total charge : Q = Ne (– e) = (8.24 × 1023 electrons) (– 1.6 × 10–19 C/electron) = – 1.32 × 105 C

Example 3 :A glass rod is rubbed with a silk cloth. The glass rod acquires a charge of + 19.2 × 10–19 C.(i) Find the number of electrons lost by glass rod. (ii) Find the negative charge acquired by silk.(iii) Is there transfer of mass from glass to silk ?

Sol. (i) Number of electrons lost by glass rod is n = qe

= 19

1919.2 191.6 10

= 12

(ii) Charge on silk = – 19.2 × 10–19 C(iii) Since an electron has a finite mass (me = 9 × 10–31 kg), there will be transfer of mass from glass rod to silk cloth.Mass transferred = 12 × (9 × 10–31) = 1.08 × 10–29 kg.The mass transferred is negligibly small. This is expected because the mass of an electron is extremely small.

DETECTING CHARGECharge can be detected and measured with the help of gold-leaf electroscope, voltameter, ballistic galvanometer. Gold leafelectroscope consist of two gold leaves attached to a conducting post that has a conducting disc ball on top. The leaves areotherwise insulated from the container. Gold leaf electroscope can be used in 2 ways.Uncharged electroscope when uncharged, the leaves hang together vertically.

(a) If a charged body is brought near to it, charge on the ball of electroscope will beopposite to that of body & on leaves similar to that of body and leaves will diverge.

++

++

++++

++

(b) If a charged body is touched : Ball & leaves both acquire similar charge and leaveswill diverge.From above method you will not be able to tell nature of charge (it may be +ve or – ve)in both case leaves will diverge.

++

++

+

+

++

+++++

++

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Electric charges and fieldsCharged electroscope :If a charged body is brought near a charged electroscope, the leaveswill further diverge if the charge on the body is similar to that on theelectroscope and will usually converge if opposite. Thus we will beable to determine nature of charge on a body.

++

++

+

++

+++

++

++

++

+

++

+++

++

Example 4 :What will happen if x-rays are incident on a charged electroscope.

Sol. Due to ionisation of air by x-rays the electroscope will get discharged and hence its leaves will collapse.Example 5 :

What will happen if x-rays are incident on a charged evaluated electroscope.Sol. X-rays will cause photoelectric effect with gold and so the leaves will further diverge if it is positively charged (or uncharged) and

will converge if it is negatively charged.

TRY IT YOURSELFQ.1 Does the attraction between the comb and the piece of papers last for longer period of time ?Q.2 Is repulsion a true test of electrification ?Q.3 What is the total charge, in coulombs, of all the electrons in three mole of hydrogen atom ?Q.4 The existence of a negative charge on a body implies that it has –

(A) Lost some of its electrons (B) Lost some of its protons(C) Acquired some electrons from outside (D) Acquired some protons from outside

Q.5 Lighting rods are made of –(A) Porcelain (B) Bakelite (C) Plastic (D) Metal

Q.6 A positively charged body is brought near an uncharged gold leaf electroscope, then –(A) No charge is induced in the leaves (B) Positive charge is induced in both the leaves(C) Negative charge is induced in both the leaves (D) Positive charge is induced in one leaf and negative in the other.

Q.7 Static electricity is produced by –(A) Friction only (B) Induction only (C) Friction & induction both (D) Chemical reaction only

Q.8 Five balls respectively from 1 to 5 are suspended from different threads. If pair of balls (1,2), (2,4) and (4,1) represents attractionwhile pair (2,3) and (4,5) represents repulsion then on ball 1.(A) Positive charge (B) Negative (C) Neutral (D) Made of metal

Q.9 On charging two metallic spheres of same mass-(A) Mass of positively charged sphere will be more (B) Mass of positively charged sphere will be less(C) Mass of negatively charged sphere will be more (D) Mass of negatively charged will be less(A) 1, 2 (B) 2, 3 (C) 3, 4 (D) 1, 4

Q.10 The current produced in wire when 107 electron/sec. are flowing in it-(A) 1.6 x 10–26 A (B) 1.6 x 1012 A (C) 1.6 x 1026 A (D) 1.6 x 10–12 A

ANSWERS(4) (C) (5) (D) (6) (B) (7) (C)(8) (C) (9) (B) (10) (D)

COULOMB'S LAWForce between two point charges (interaction force) is directly proportional to the product of magnitude of charges (q1 and q2)and is inversely proportional to the square of the distance between them i.e., (1/r2). This force is conservative in nature. This isalso called inverse square law. The direction of force is always along the line joining the point charges.

F 1 22

q qr

; F = K 1 22

q qr

where K is a constant

K = 0 r

14

[K = 9 × 109 C2/N-m2 ] ;

0 = permittivity of free space = 8.85 × 10–12 N-m2/C2, r = relative permittivity (dielectric constant of medium)

Coulomb’s Law in Vector Form

Suppose the position vectors of two charges q1 and q2 are 21 randr , then, electric force on charge q1 due to charge q2 is,

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Electric charges and fields

1 212 1 23

0 1 2

q q1F r r4 | r r |

Similarly, electric force on q2 due to charge q1 is

1 221 2 13

0 2 1

q q1F r r4 | r r |

Here q1 and q2 are to be substituted with sign. Position vector of

charges q1 and q2 are 1 1 1 1ˆ ˆ ˆr x i y j z k

and 2 2 2 2ˆ ˆ ˆr x i y j z k

respectively. Where (x1, y1, z1) and (x2, y2, z2) are the co-ordinates ofcharges q1 and q2.Superposition TheoremThe interaction between any two charges is independent of the presence of all othercharges. Electrical force is a vector quantity therefore, the net force on any onecharge is the vector sum of the all the forces exerted on it due to each of the othercharges interacting with it independently i.e.

Net force on charge q, F

= 1F

+ 2F

+ 3F

+ ...........

Example 6 :In a hydrogen atom, the electron is separated from the proton by an average distance of about 5.3 × 10–11 m. Calculate themagnitude of the electrostatic force of attraction exerted by the proton on the electron.

Sol. Substitute the given value into Coulomb's law :

F = 1 22

k | q q |r

= 2

2ker

= 9 2 2 19 2

11 2(8.99 10 N.m / C ) (1.6 10 C)

(5.3 10 m)

= 8.19 × 10–9 N

Example 7 :Compute the ratio of the electric force to the gravitational force exerted by a proton on an electron in a hydrogen atom.

Sol. We use Coulomb's law with q1 = e and q2 = – e to find the electric force, and Newton's law of gravity with the mass of the proton,mp = 1.67 × 10–27 kg, and the mass of the electron, me = 9.11 × 10–31 kg.1. Express the magnitude of the electric force Fe and the gravitational force Fg in terms of the charges, masses, separation distancer, and electrical and gravitational constant s:

Fe = 2

2ker

, Fg = p e2

Gm m

r2. Take the ratio. Note that the separation distance r cancels :

e

g

FF =

2

p e

keGm m

3. Substitute numerical values :

e

g

FF =

9 2 2 19 2

11 2 2 27 31(8.99 10 N.m / C ) (1.6 10 C)

(6.67 10 N.m / kg ) (1.67 10 kg) (9.11 10 kg)

= 2.27 × 1039

Example 8 :What is the smallest electric force between two charges placed at a distance of 1.0 m.

Sol. Fe = 0

14

. 1 22

q qr

.............. (i)

For Fe to be minimum q1 q2 should be minimum.We know that (q1)min = (q2)min = e = 1.6 × 10–19 CSubstituting in Eq. (i), we have

(Fe)min = 9 19 19

2(9.0 10 ) (1.6 10 ) (1.6 10 )

(1.0)

= 2.304 × 10–28 N.

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Electric charges and fieldsExample 9 :

Electric force between two point charges q and Q at rest is F. Now if a charge – q is placed next to q what will be the (a) force onQ due to q (b) total force on Q ?

Sol. (a) As electric force between two body interaction, i.e., force between two particles, is independent of presence or absence ofother particles, the force between Q and q will remain unchanged, i.e., F.(b) An electric force is proportional to the magnitude of charges, total force on Q will be given by :

F Qq q 00

F Qq q q [as q' = q + (– q) = 0]

i.e., The resultant force on Q will be zero.Example 10 :

Four identical point charges each of magnitude q are placed at the corners of a square of side a. Find the net electrostatic force onany of the charge.

Sol. Let the concerned charge be at C then charge at C will experience the force due to charges at A, B and D. Let these forces

respectively be AF

, BF

and DF

thus forces are given as

2

2

0A

ACq

41F

along AC =

2j

2i

a24q

20

2

2

2

0B

BCq

41F

along BC = )j(aπε4

q2

0

2

FDFAFB

A B

CD

q

qq

q

y

x

2

2

0D DC

q4

1F

along DC = )i(aπε4

q2

0

2

DBAnet FFFF

1

221j1

221i

aπε4q

20

2

20

2

neta4

q122

12F

2

0

2

a4q2

21

Example 11 :Five point charges, each of value + q are placed on five vertices of a regular hexagon of side L m. What is the magnitude of theforce on a point charge of value – q coulomb placed at the centre of the hexagon ?

Sol. If there had been a sixth charge + q at the remaining vertex of hexagon force due to all the six charges on – q at O will be zero (as

the forces due to individual charges will balance each other), i.e., RF 0

Now if f is the force due to sixth charge and F

due to remaining five charges,

F f 0 i.e., F f

or F = f = 0

14 2

q qL

= 0

14

2qL

Example 12 :Two charged spheres of radius 'R' are kept at a distance 'd' (d > 2R).One has a charge +q and the other –q. The force between them will be-

+ + ++

++

+

++

R R

d(A)

2

20

1 q4 d (B)

2

20

1 q4 d

(C) 2

20

1 q4 d

(D) None of these

Sol. (B). Redistribution of charge will take place due to mutual attraction and hence effective distance will be less than d.

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Electric charges and fieldsTRY IT YOURSELF

Q.1 Two identical balls each having a density are suspended from a common point by two insulating strings of equal length. Boththe balls have equal mass and charge. In equilibrium each string makes an angle with vertical. Now, both the balls are immersedin a liquid. As a result the angle does not change. The density of the liquid is . Find the dielectric constant of the liquid.

Q.2 The ratio of gravity force to the electric force between two electrons-(A) 10–36 (B) 10–42 (C) 1042 (D) 10–47

Q.3 A charge Q1 exerts some force on a second charge Q2. If a 3rd charge Q3 is brought near, the force of Q1 exerted on Q2 –(A) Will increase (B) Will decrease (C) Will remain unchanged(D) Will increase if Q3 is of the same sign as Q1 and will decrease if Q3 is of opposite sign.

Q.4 Two identical pendulums A and B, are suspended from the same point. The bobs are given positive charges, A having more chargethan B. They diverge and reach equilibrium with A and B making angles 1 and 2 with the vertical respectively. Then–(A) 1 > 2 (B) 1 < 2 (C) 1= 2(D) The tension in A is greater than tension in B

Q.5 Force of attraction between two point charges placed at a distance d is F. What distance apart should they be kept in the samemedium so that the force between them is F/3 ?

Q.6 A particle of mass m carrying charge +q1 is revolving around a fixed charge –q2 in a circular path of radius r. Calculate the periodof revolution.

Q.7 Two pieces of copper, each weighing 0.01 kg, are placed at a distance of 0.1m from each other. One electron from per 1000 atomsof one piece is transferred to other piece of copper. What will be the Coulomb force between two pieces after the transfer ofelectrons ? The atomic weight of copper is 63.5 g/mole. Avogadro number = 6 × 1023.

Q.8 Two point charges of +2µC and +6µC repel each other with a force of 12 N. If each is given an additional charge –4µC, then forcewill become-(A) 4 N (attractive) (B) 60 N (attractive) (C) 4 N (Repulsive) (D) 12 N (attractive)

Q.9 Three equal charges (q) are placed at corners of a equilateral triangle. The force on any charge is -

(A) zero (B) 3 2

2Kqa

(C) 2

2Kq3a

(D) 3 32

2Kqa

Q.10 Five point charges, each of value – q coulomb, are placed on five vertices of a regular hexagon of side L meter. The magnitude ofthe force on a point charge of value –q coul. placed at the center of the hexagon is -

(A) 2

2kqL

(B) 52

2kqL

(C) 3 2

2kqL

(D) zero

ANSWERS(1) K = / – d (2) (B) (3) (C) (4) (C) (5) 1.732 d

(6) 0

1 2

mrT 4 r

q q

(7) F = 2.06 × 1014 N (8) (C) (9) (B) (10) (A)

ELECTRIC FIELDThe physical field where a charged particle, irrespective of the fact whether it is in motion or at rest, experiences force is called anelectric field. The concept of electric field was given by michael Faraday. Characteristics of electric field :(1) Electric field intensity (shortly we will call electric field). (2) Electric potential. (3) Electric lines of forces.Electric field intensity E

:Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitudeand directionIf a test charge q0 is placed at a point in an electric field and experiences a force F

due to some charges (called source charges),

the electric field intensity at that point due to source charges is given by 0

FEq

.

The presence of the charge q0 will generally change the original distribution of the other charges, particularly if the charges areon conductors. However, we may choose q0 to be small enough so that its effect on the original charge distribution is negligible.

q 0 00

EE limq

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Electric charges and fieldsElectric field due to a point chargeThe electric field produced by a point charge q can be obtained in general terms from Coulomb's law. First, note that the magnitudeof the force exerted by the charge q on a test charge q0 is F = kqq0/r2. Then, divide this value by q0 to obtained the magnitude ofthe field. Since q0 is eliminated algebraically from the result, the electric field does not depend on the test charge:

Point charge q : E = 2kqr

If (x, y, z) are the co-ordinates of the observation point P, thenˆ ˆ ˆr xi yj zk

Also, r = (x2 + y2 + z2)1/2 and r3 = (x2 + y2 + z2)3/2

Now, E(r) =

0

14 2 2 2 3/2

q(x y z )

ˆ ˆ ˆ(xi yj zk)

Source ChargeqO

r

Pq0

The three rectangular components of E (r) are as follows :

Ex(r ) =

0

14 2 2 2 3/2

q x(x y z )

,

Ey( r ) = 0

14 2 2 2 3/2

q y(x y z )

and Ez(r ) =

0

14 2 2 2 3/2

q z(x y z )

Electric field due to Discrete distribution of charge :Point charges placed at different position, use vector approach (Better term : super position rule)

n

1 2 ii 1

E E E ....... E

with ii i3

0 i

q1E r4 r

Corona Discharge : Dielectric strength of medium mean minimum field required for ionisation of a medium. If value of E increasesabove dielectric strength of medium, medium gets ionised and charge leak out into the medium from body generally it happen atthe corner where E is high. This leakage process is called corona discharge.For air dielectric strength = 3 × 106 v/mThe electric field near a high-voltage power line can be large enough to strip the electrons from air molecules, thus ionizing themand making the air a conductor. The glow resulting from the recombination of free electrons with the ions is an example of coronadischarge. Break-down in air is witnessed during atmospheric lighting.Motion of a charged particle in a uniform electric fieldIf force of gravity does not exist or is balanced by some other force say reaction or neglected then

F qEa constant

m m

[as F qE

]

Here equations of motion are valid.

(i) If the particle is initially at rest then from v = u + at, we get v = at = qE tm

And from Eqn. 21s ut at2

we get 2 21 1 qEs at t2 2 m

+++++++ d

PD=V

E

F+q

+++++++

Y-q

L D

v0

10 GyaanSankalp

Electric charges and fieldsThe motion is accelerated translatory with a t°; v t and s t²

Here 2

21 1 qEW KE mv m t2 2 m

also W = qEd = qV

(ii) If the particle is projected perpendicular to the field with an initial velocity v0

From Eqn. v = u + at and s = ut + 12

at² , ux = v0 and ax = 0,

ux = v0 = constant and x = v0t

for motion along y-axis as uy = 0 and ay = qEm

, yqE

v tm

and

21 qEy t

2 m

So eliminating t between equation for x and y, we have 2

22

0 0

qE x qEy x2m v 2mv

If particle is projected perpendicular to field the path is a parabola.

Example 13 :When a 5nC test charge is placed at a certain point, it experiences a force of 2 × 10–4 N in the x-direction. What is the electric field

E at the point ?

Sol. E = 0F / q

= [(2 × 10–4 N) i ] / (5 × 10–9 C) = (4 × 104 N/C) i )

Example 14 :Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from thecentre is a. Find the electric field at the centre of the pentangon.

Sol. Let the charges be placed at the vertices A, B, C and D of the pentagon ABCDE. If weput a charge q at the corner E also, the field at O will be zero by symmetry. Thus, thefield at the centre due to the charges at A, B, C and D is equal and opposite to the fielddue to the charge q at E alone.

The field at O due to the charge q at E is 20

q4 a

along EO. E

A B

C

D

O

Thus, the field at O due to the given system of charges is 20

q4 a

along OE.

Example 15 :Two positive charges Q1 and Q2 are placed on a line as shown in figure. Determine the position of point O, where the net electricfield is zero.

Sol. Let position of P is at a distance x from Q1. Then the fields at P due to Q1 and Q2 are in opposite directions. They will add up togive zero, only if their (electric field's) magnitude are equal. That is

12

kQx

= 2

2kQ

(R x)

R xx

= 2

1

QQ or x =

2 1

R1 Q / Q

Q P1 2

x R

Q

The distance of point P from charge Q is d = R – x = 1 2

R1 Q / Q

If two negative charges are placed on a line (instead of positive charges), then the position of point P where the net electric fieldis zero, is again

x = R / 2 11 Q / Q , d = R / 1 21 Q / Q .

11GyaanSankalp

Electric charges and fieldsExample 16 :

Calculate the electric field strength required to just support a water drop of mass 10–7 kg and having charge 1.6 × 10–19 C.Sol. Here, m = 10–7 kg, q = 1.6 × 10–19 C

Step 1 : Let E be the electric field strength required to support the water drop.Force acting on the water drop due to electric field E is

F = qE = 1.6 × 10–19 EWeight of drop acting downward, W = mg = 10–7 × 9.8 newton.Step 2 : Drop will be supported if F and W are equal and opposite.

i.e., 1.6 × 10–19 E = 9.8 × 10–7 or E = 7

199.8 101.6 10

= 6.125 × 1012 N C–1.

Example 17 :Two charges of + 10 C and + 40 C respectively are placed 12cm apart. Find the position of the point where electric field is zero.

Sol. Let P be the point at a distance x from the charge + 10 C where electric field due to two charges + 10 C and + 40 C is zero.

Electric field intensity due to q1 at P, E1 = 0

14

12

qx

; along PB

Electric field intensity due to q2 at P, E2 = 0

14

22

q(r x) ; along PA.A.

Clearly, field at P will be zero if E1 = E2

i.e.0

14

12

qx

= 0

14

22

q(r x) or

qx

12 =

22

q(r x)

Here, q1 = 10 C = 10 × 10–6 C ; q2 = 40 C = 40 × 10–6 C

6

210 10

x

=

6

240 10(r x)

(r – x)2 = 4x2 or (r – x) = 2x 3x = r or x = r3

But r = 12cm (given) x = 123

= 4.0 cm.Thus electric field will be zero at a distance of 4.0 cm from the charge + 10µC.

Example 18 :Can a metal sphere of radius 1cm hold a charge of 1 coulomb.

Sol. Electric field at the surface of the sphere.

E = 2KQR

= 9

2 29 10 1(1 10 )

= 9 × 1013 Vm

This field is much greater than the dielectric strength of air (3 × 106 v/m), the air near the sphere will get ionised and charge will leakout. Thus a sphere of radius 1 cm cannot hold a charge of 1 coulomb in air.

TRY IT YOURSELFQ.1 Two charges of opposite nature having magnitude 10 µC are 20 cm apart. The electric field at the centre of line joining these

charges will be-(A) 9 x 106 N/C in the direction of positive charge (B) 18 x 106 N/C in the direction of negative charge(C) 18 x 106 N/C in the direction of positive charge (D) 9 x 106 N/C in the direction of negative charge

Q.2 A point charge A of charge +4 µC and another point charge B of charge –1 µC are placed in air at a distance 1 meter apart. Thenthe distance of the point on the line joining the charges and from the charge B, where the resultant electric field is zero, is- (inmetre)(A) 2 (B) 1 (C) 0·5 (D) 1·5

Q.3 Four charges each +q, are placed at the four corners of a regular pentagon as shownin the fig. The distance of each corner from the centre O is r. Then the electric field atthe center will be-

(A) 0

q4 r² towards OA (B)

0

q2 2 r² towards OA

A

Oq q

q q

(C) Zero (D) 0

qr² towards OA

12 GyaanSankalp

Electric charges and fieldsQ.4 A charge Q is divided into two parts such that charge on each part is q and Q/q = 2. The Coulombic force between the two charges

q and q when placed some distance apart –(A) is maximum irrespective of the medium in which they are placed(B) is minimum(C) is equal in magnitude for both opposite in direction(D) is depenent on the medium in which charges are placed

Q.5 Two free point charges +4Q and +Q are placed at a distance r. A third charge q is so placed such that all the three are in equilibrium–(A) q is placed at a distance r/3 from 4Q (B) q is placed at a distance r/3 from Q(C) q = 4Q/9 (D) q = – 4Q/9

Q.6 A charge q = 1 µC is placed at point (1m, 2m, 4m). Find the electric field at point P (0m, – 4m, 3m).Q.7 A copper ball of diameter d is immersed in an oil of density 0. There is a homogeneous electric field E directed vertically upwards

such that the copper ball is suspended in the oil. Density of copper is C. The charge on the ball is –

(A) 3

c 0d ( )g3E

(B)

3c 0d ( )g2E

(C)

3c 0d ( )g6E

(D)

3c 0d ( )g

12E

Q.8 A charge 10–9 coulomb is located at origin is free space and another charge Q at (2, 0, 0). If the x-component of the electric fieldat (3, 1, 1) is zero, calculate the value of Q. Is the y-component zero at (3, 1, 1) ?

Q.9 A charged particle of mass m = 2 kg and charge 1 µC is thrown from a horizontal ground at an angle = 45° with speed 10 m/s. Inspace a horizontal electric field E = 2 × 107 N/C exists. The range of the projectile is –(A)20m (B) 60m (C) 200m (D) 180m

Q.10 In the fig. distance of the point from A where the electric field is zero is-(A) 20 cm (B) 10 cm(C) 33 cm (D) None of these

80cm10 C 20 C

A B

ANSWERS(1) (B) (2) (B) (3) (A) (4) (A) (5) (BD)

(6) Nˆ ˆ ˆE ( 38.42i 250.52 j 38.42k)C

(7) (C) (8) 3/2

93 3 10 C11

(9) (A) (10) (C)

ELECTRIC FIELD DUE TO CONTINUOUS CHARGE DISTRIBUTIONThe evaluate the electric field of a continuous charge distribution, the following procedure is used. First, we divide the chargedistribution into small elements each of which contains a small charge q. Next, we use Coulomb's law to calculate the electric fielddue to one of these elements at a desired point. Finally, we evaluate the total field at a point due to the charge distribution bysumming the contributions of all the charge elements (that is, by applying the superposition principle.)Let us consider some casesCase 1 : The electric field of a uniform ring of charge.A ring of radius a has a uniform positive charge per unit length, with a total charge Q. To find the electric field along the axis of thering at a point P lying a distance x from the center of the ring follow the procedure.

The magnitude of the electric field at P due to the segment of charge q is E = k 2q

r

This field has an x component Ex = E cos along the axis ofthe ring and a component E perpendicular to the axis. But aswe see in figure, the resultant field at P must lie along the x axissince the perpendicular components sum up to zero. That is, theperpendicular component of any element is canceled by the per-pendicular component of an element on the opposite side of thering. Since r = (x2 + a2)1/2 and cos = x/r, we find that

Ex = E cos = 2 2 2 3/2q x kxk q

rr (x a )

E1

E2

1

2

In this case, all segments of the ring give the same contributionto the field at P since they are all equidistant from this point.Thus, we can easily sum over all segments to get the total field

13GyaanSankalp

Electric charges and fields

at P : Ex = 2 2 3/2kx

(x a ) q = 2 2 3/2

kx(x a )

Q

This result shows that the field is zero at x = 0.At large distances from the ring (x > > a) the electric field along the axis approaches that of a point charge of magnitude Q.Case 2 : The electric field of a uniformly charged disk :A disk of radius R has a uniform charge per unit area . To find the electric field along the axis of the disk, a distance x from its centerfollow the procedure.Consider the disk as a set of concentric rings. We can then makeuse result of case 1,which gives the field of a given ring of radiusr, and sum up contributions of all rings making up the disk. Bysymmetry, the field on an axial point must be parallel to this axis.The ring of radius r and width dr has an area equal to 2r dr. Thecharge dq on this ring is equal to the area of the ring multipliedby the charge per unit area, or dq = 2r dr. Using this result inequation (with a replaced by r) gives for the field due to the ringthe expression.

dE = 2 2 3/2kx

(x a ) (2r dr)

r

dq

x

To get the total field at P, we integrate this expression over the limitsr = 0 to r = R, noting that x is a constant, which gives

E = kx R

2 2 3/20

2r dr(x r ) = kx

0

Rz (x2 + r2)–3/2 d(r2) = kx R2 2 1/2

0

(x r )1/ 2

= kx 2 2 1/2

x x| x | (x R )

The result is valid for all values of x.The field close to the disc along an axial point can also be obtained from equation by letting x 0

This gives E = 2 k = 02

where 0 is the permittivity of free space, the same result is obtained for the field of a uniformly charged infinite sheet.

Case 3 : E

due to an infinite plane of chargeThe field of an infinite plane of charge can be obtained from field by disc by either letting R go to infinity or letting x go to zero.Then Ex = 2k, x > 0Thus, the field due to an infinite-plane charge distribution is uniform; that is, the field does not depend on x. On the other side ofthe infinite plane, for negative values of x, the field points in the negative x direction, so Ex = – 2 k , x < 0Case 4 : Spherical distribution of charge(a) Conducting sphere (Hollow, solid) (b) Non-conducting sphere (Hollow, solid)Case (a) Charge on surface. Case (b) Volume distribution of charge.(a) : Hollow/solid conductor or hollow non-conductorImagine a sphere passing through desired point (point where E isto be calculated), calculate charge inside it and assume itto be concentrated at centre and use point charge formula.Inside sphere r < R

Q

r R ++

+++++

++

+

++ + + +

++

E = 0 (No charge inside imagined sphere)

outside r > R Qr

R +++

+++++

+++ + + + +

+E = 2

KQr

surface r = R , E = 2KQR

14 GyaanSankalp

Electric charges and fields

Graphically

E

KQ/R2

r = Rr

E 1/r2

(b) Inside r < RUniform volume distribution : charge inside volume

43r3

3

Q4 R3

43 r3 = Q' or Q' =

3

3QrR

, E = 1

2KQr

= 3KQR

r

++++

+

+

+ +++

++ + +

++

+

Q

Rr

0ˆE r

3

Outside, E = 3KQr

r > R

++ ++

++

++ +++

+

+

++

+

+Q

Rr

Surface E = 3KQR

,

Graphically

E

r = R r

E1/r 2E

r

Electric field intensities due to various charge distributions are given in table.

Name/TypePoint charge

Infinitely longline charge

Semi-infinite

Finite change ofcharge

+++++++++

xP

E

E

Formula

2 3Kq Kqˆ.r r

| r | r

0

ˆ2K rr2 r r

2kr , x y

K kE , E

r r

xKE [sin sin ]r

yKE [cos cos ]r

If =

||0

E 0, E2 r

Particularq is source charge.r is vector drawn from source chargeto the test point.Electric field is nonuniform, radiallyoutwards due to + charges & inwardsdue to – charges.

is a linear charge density(assumed uniform)r is perpendicular distance of pointfrom line charge.r is radial unit vector drawn from thecharge to test point.

At a point above the end of wire at anangle 45°

Where is the linear charge density

r

E

r

E

Graph

15GyaanSankalp

Electric charges and fieldsInfinite non-conductingthin sheet

Uniformly chargedring

Infinitely largecharged conductingsheet

Uniformly charged hollowconducting/nonconducting/solid conductingsphere

Uniformly chargedsolid nonconductingsphere (insulatingmaterial)

Uniformly charged cylin-der with a charge density (R = radius of cylinder)

Uniformly charged cylin-drical shell with surfacecharge density is

is surface charge density(assumed uniform)n is unit normal vector..Electric field intensity isindependent of distance.

Q is total charge of the ring.x = distance of point on the axisfrom centre of the ring.Electric field is always along the axis.

Maximum at x R / 2

is the surface charge.n is unit normal vectorperpendicular is the surface.Electric field intensity isindependent of distance.

R is radius of the sphere.r is a vector drawn from centreof sphere to the point.Sphere acts like a point charge,placed at centre for pointsoutside the sphere.

E is always along radial direction.

Q is total charge ( 24 R )( = surface charge density)

r is a vector drawn from centreof sphere to the point.Sphere acts like a point charge, placedat centre for points outside the sphere.

E is always along radial direction.

Q is total charge 34( . 4 R )3

( = volume charge density)Inside the sphere E rOutside the sphere E 1/r²

0n

2

2 2 3/ 2KQxE

(R x )

centreE 0

0n

(i) for r R

2kQ ˆE r

| r |

(ii) for r < R

E 0

(i) for r R

2kQ ˆE r

| r |

(ii) for r R

30

KQr rE3R

for r < R, in0

rE2

for r > R, 2

0

RE2 r

for r < R, Ein = 0,

for r > R, 0

rEr

r

E

02

E 1/r

16 GyaanSankalp

Electric charges and fieldsExample 19 :

Electric charge is uniformly distributed around a semicircle of radius a, with total charge Q. What is the electric field at the centreof curvature ?

Sol. Consider a small segment of angular width d, located at an angle from the x-axis.

The length of the segment is ds = ad. The charge on the segment is ad QdQ Q d

a

The magnitude of the electric field at P is given by 2 20 0

1 dQ 1 QdE d4 a 4 a

This electric field has y-component y 2 20

QdE dEsin sin d4 a

The x-component of the field from the right hand half of the ring cancels with that of the left-hand half of the ring. The resultantelectric field is thus in the y-direction, and is given by adding up the dEy from each segment in the ring. This is done by integrating from to rad :

y y 2 200

QE dE sin d4 a

y 2 2 2 20 00

Q QE sin d cos4 a 4 a

; y 02 2

0

QE [( 1) (1)]4 a

; y 2 2

0

QE2 a

Example 20 :A thin non-conducting ring of radius R has linear charge density 0 cos , where 0 is a constant, is the azimuthal angle.Find the magnitude of the electric field strength at the centre of the ring.

Sol. The situation is shown in figure.The half ring on the right hand side will be positive while on the half left sidewill be negative. The reason being that cos for first and fourth quadrants ispositive while for 2nd and 3rd quadrants is negative.Consider a small element dx of the ring. Here dx = R cos Charge on small element dq = = dx = 0 cos (R d)

0

20

R cos d1dE4 R

Electric field along x-axis due to this element

xdE dE cos

= 02

0

R cos d1 cos4 R

= 20

0(cos d )

4 R

= 0

0

1 cos 2 d4 R 2

Electric field due to positive part along x-axis,/2

01

0 /2

1 cos 2E d4 R 2

17GyaanSankalp

Electric charges and fields

or 0 01

0 0E

4 R 2 8 R

Similarly, the electric field due to negative charge along x-axis

02

0E

8 R

Enet = E1 + E2 = 0 0 0

0 0 08 R 8 R 4 R

TRY IT YOURSELF

Q.1 A spherical volume contains a uniformly distributed charge of density . The electric field inside the sphere at a distance r fromcenter is –

(A) 0

r3 (B)

0r

4 (C)

0r

(d) 0

1 r4

Q.2 A point charge 50µC is located in the XY plane at the point of position vector 0ˆ ˆr 2i 3j

. What is the electric field at the point

of position vector ˆ ˆr 8i 5j –

(A) 1200 V/m (B) 0.04 V/m (C) 900 V/m (D) 4500 V/mQ.3 A solid metallic sphere has a charge +3Q concentric with this sphere is a conducting spherical shell having charge –Q. The radius

of the sphere is a and that of the spherical shell is b (>a). What is the electric field at a distance r (a < r < b) from the centre–

(A) 0

1 Q4 r (B)

0

1 3Q4 r (C) 2

0

1 3Q4 r (D) 2

0

1 Q4 r

Q.4 The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be –

(A) 20

1 q4 3 3R (B) 2

0

1 2q4 3R (C) 2

0

1 2q4 3 3R (D) 2

0

1 3q4 2 2R

Q.5 Two conducting plates X and Y, each having large surface area A (on one side) are placed parallel to each other. The plate X isgiven a charge Q where the other is neutral. The electric field at a point in between the plates is given by –

(A) Q

2A(B)

0

Q2A towards left (C)

0

Q2A

towards right (D) 0

Q2 towards right

Q.6 Two infinitely long parallel wires having linear charge densities 1 and 2 respectively are placed at a distance R meter. The force

per length on either wire will be 0

1K4

(A) 1 22

2K

R

(B) 1 22K

R

(C) 1 22K

R

(D) 1 2KR

Q.7 A wheel having mass m has charges + q and – q on diametrically oppositepoints. It remains in equilibrium on a rough inclined plane in the presence ofuniform vertical electric field E =

(A) mgq (B)

mg2q

+q

–qE

(C) mg tan

2q

(D) none

ANSWERS(1) (A) (2) (D) (3) (C) (4) (C)(5) (C) (6) (B) (7) (B)

18 GyaanSankalp

Electric charges and fieldsELECTRIC FIELD LINES

Electric lines of forces :(i) The concept of electric field was introduced by Michael Faraday.

The magnitude of electric field strength at any point is measured by the number of electric line of force passing per unit small areaaround that point normally and the direction of field at any point is given by the tangent to the line of force at the point.

(ii) An electric line of force is that imaginary smooth curve drawn in an electric field along which a free isolated unit positive (initiallyat rest) charge moves.

E

at any point on a line of force.

Properties :(1) The lines of force diverge out radially from a +ve charge

and converge at a – ve charge. More correctly the linesof force are always directed from higher to lower potential. + –

(2) The tangent drawn at any point on line of force gives the direction of force acting on a positive charge placed at that point.(3) Two lines of force never intersect. If they are assumed to intersect. There will be two directions of electric field at the point of

intersection : which is impossible.(4) These lines have a tendency to contract in tension like a stretched

elastic strong. This actually explains attraction between oppositecharges. –

Attraction

+

(5) These lines have a tendency to separate from each other in the directionperpendicular to their length. This explains repulsion between likecharges.

Repulsion

+ +

(6) The no. of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system

0(1/ ) electric lines are associated with unit charge. So if a body encloses a charge q. Total line of force associated with it (called

flux) will be 0

q .

(7) Total lines of force may be fractional as lines of force are imaginary.q >qA B

+ –A B(8) Lines of force ends or strarts normally on the surface of a conductor.(9) If there is no electric field there will be no lines of force.(10) Lines of force per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field

while distant lines represent weak field.(11) Electric lines of force differ from magnetic lines of force.

(a) Electric lines of force never form closed loop while magnetic lines are always closed or extended to infinity.

Electric line of force (A)

Magnetic line of force (B)

19GyaanSankalp

Electric charges and fields(b) Electric lines of force always emerge or terminate normally on the surface of charged conductor, while magnetic lines emerge or

terminate on the surface of a magnetic material at any angle.(c) Electric lines of force do not exist inside a conductor but magnetic

lines of force may exist inside magnetic material.Lines of force do not exist inside a conductor (as field inside aconductor is zero) the plates is as shown. (Electrostatic shielding)

12. Neutral point : Where electric field intensity is zero (test charge does not experience any force)13. The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. This

means, for example that if 100 lines are drawn leaving a + 4µC charge then 75 lines would have to end on a –3µC charge.ELECTRIC FLUX ( ) :

If the lines of force pass through a surface then the surface is said to have flux linked with it. Mathematically it can be formulatedas follows :

The flux linked with small area element on the surface of the body : d = sd.E

Where sd is the area vector of the small area element. The area vector of a closed surface is always in the direction of outwarddrawn normal. The total flux linked with whole of the body,

= E.ds total flux linked with closed surface, where is referred to closed integral done for a closed surface.

(i) Electric flux is a scalar quantity(ii) Units (V - m) or (N - m2/Cb),Dimensions : [M1 L3 T–3 A–1](iii) The value of '' does not depend upon the distribution of charges and the distance between them inside the closed surface.(iv) The value of is zero in the following circumstances :

(a) If a dipole is enclosed by a closed surface.(b) Magnitude of +ve and –ve charges are equal inside a closed surface.(c) If no charge is enclosed by a closed surface.(d) In coming flux (–ve) = out going flux (+ ve).

GAUSS'S LAW

The total flux linked with a closed surface is 0

1 times the charge enclosed by the closed surface (Gaussian surface).

i.e. E ds.z =

q0

Law is valid for symmetrical charge distribution and for all vector fields obeying inverse square law.Gaussian surface :(a) Imaginary surface(b) Is spherical for a point charge, conducting and non-conducting spheres.(c) Is cylindrical for infinite sheet of charge conducting charge surfaces, infinite line of charges, charged cylindrical conductors,etc. For finite charge distribution use Coulomb's law. For infinite charge distribution use Gauss theorem

E ds.z = qnet

0

Application : (1) To Calculate flux (2) To calculate Electric field IntensityStudy following cases to learn application properlyObserve flux through common geometrical figures

(i) out = in = R2E (ii) in = out = Ea2 ; total = 0.

20 GyaanSankalp

Electric charges and fields

(iii)T = 0 (iv) T = q0

, hemisphere = q

2 0

(dotted part shows imaginary part to enclosed the charge completely)

(v) T = q0

, cyl. = q

2 0. (vi) T =

q0

cube = q

2 0

Charge position TCube centre = q/.Face centre = q/2.At corner = q/8.At centre of edge = q/40.

T = 0

18 i

iq

; i = 1, 2, ............ 8.

Electric field due to a line charge :Consider an infinite line which has a linear charge density . Using Gauss’sslaw, let us find the electric field at a distance ‘r’ from the line charge.The cylindrical symmetry tells us that the field strength will be the same at allpoints at a fixed distance r from the line. Thus, the field lines are directedradially outwards, perpendicular to the line charge.The appropriate choice of Gaussian surface is a cylinder of radius r and

S3

S2

S1

E

Sd

E

EESd

Sd

++++++++++++++++++++

r

length L. On the flat end faces, 2S and 3S , E

is perpendicular Sd

, which means

flux is zero on them. On the curved surface 1S , E

is parallel Sd

, so that EdSSd,E

.The charge enclosed by the cylinder is Q = L.Applying Gauss’s law to the curved surface, we have

0εLλ)rLπ2(EdSE or r2

E0

Electric field due to an infinite plane thin sheet of charge :To find electric field due to the plane sheet of charge at any point P distant rfrom it, choose a cylinder of area of cross-section A through the point P as theGaussian surface. The flux due to the electric field of the plane sheet of chargepasses only through the two circular caps of the cylinder. Let surface chargedensity =

According to gauss law in 0E .dS q /

+ +++++

+ ++++++++

EP

E Q

rPlane sheet of charge

Gaussian Surface

0I circular II circular cylindricalsurface surface surface

AE dscos E ds cos E dscos

or EA + EA + 0 = 0

A2

or E = 02

21GyaanSankalp

Electric charges and fieldsElectric field intensity due to uniformly charged spherical shell :

We consider a thin shell of radius R carrying a charge Q on its surface(i) at a point P0 outside the shell (r > R)

According to gauss law , 0S1

E .ds

= 0

Q

or E0 (4r2) = 0

Q

S

Pin

ROS2

Ps S1

P0

E

ds = 0º

E0 = 20

Q4 r

= 0

2

2Rr

where the surface charge density = total chargesurface area = 2

Q4 R

The electric field at any point outside the shell is same as if the entire charge is concentrated at centre of shell.

(ii) at a point Ps on surface of shell (r = R) ES = 20

Q4 r

= 0

(iii) at a point Pin inside the shell (r < R)

According to gauss law S2

E .ds

= in

0

q

As enclosed charge qin = 0 , So Ein = 0

E=0E=0

E

Er2E

r21 1

E Q/4 0R2

O r < R r = R r > Rdistance from centre (r)

The electric field inside the spherical shell is always zero. Electric field intensity due to a spherical uniformly charge distribution :We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the

volume. The charge density = total chargetotal volume

= 3

Q4 R3

= 33Q

4 R

(i) at a point P0 outside the sphere (r > R)

according to gauss law 0E .ds

= 0

Q

or E0 (4r2) = 0

Q

or E0 = 20

Q4 r

= 03

3

2Rr

+ +

+

+

+++

+

+

+

++

+

+

+++

+

+

++ +

P0Pin PsE

dsrO

R

(ii) at a point Ps on surface of sphere (r = R)

Es = 20

Q4 R

= 03

R

(iii) at a point Pin inside the sphere (r < R)According to gauss law

inE .ds

= in

0

q

= 0

1

.43

r3 = 3

30

QrR

O r = R r > R

2r1E

2r1E E

rE

r

E

r < REin(4r2) =

3

30

QrR

or Ein = 30

Qr4 R

= 03

r (Ein r)

ELECTROSTATIC PRESSURETo find force on a charged conductor (due to repulsion of like charge) imaginea small part PR to be cut and just separated from the rest of the conductorMLN. The field in the cavity due to the rest of the conductor is E2, while fielddue to small part is E1. Then

M

N

PR

L

22 GyaanSankalp

Electric charges and fieldsInside the conductor :

E = E1 – E2 = 0 or E1 = E2Outside the conductor :

1 2 0E E E /

Thus 1 20

E E2

To find force, imagine charged part PR (having charge dA placed in the cavity MN having field E2. Thus force

2dF ( dA) E

or2

0dF dA

2

The force per unit area or electric pressure is2

0

dFdA 2

The force is always outwards as 2( ) is positive i.e. whether charged positively or negatively, this force will try to expand thecharged body.A soab bubble or rubber balloon expands on given charge to it. (charge of any kind + or –)Energy associated per unit volume of electric field of intensity E is defined as energy density.

u = dwdv =

02

2E

=

2 0 J/m3

U = z u . dv = 0

2

vz E2 dv ; v is the volume of electric field.

Equilibrium of charged liquid surfaces :Soap Bubble :Pressures (forces) acts on a charged soap bubble, due to(i) Surface tension of a soap bubble PT (inward)(ii) Air out side the bubble p0 (inward)(iii) Electric charges (electrostatic pressure) Pe (outward)(iv) Air inside the soap bubble Pi (outward)Hence, in state of equilibrium

inward pressure = outward pressurePT + P0 = Pi + Pe

Excess pressure(Pex.) = Pi – P0 = PT – Pe

But PT = 4Tr , Pe =

2 0

Pex. = 4Tr

–

2 0

If Pi = P0, then 4Tr =

2 0

Example 21 :

There is a solid sphere of radius R having volume charge density 0r

1R

, where 0 is any constant and r is the distance

from the centre of sphere. Find electric intensity E inside and outside the sphere.

23GyaanSankalp

Electric charges and fieldsSol. (i) Inside : When r < R, electric flux through small area,

d E.dS E dS

2E dS E dS E 4 r

But according to Gauss's law enclosed

0

q

Charge q contained between radius x to x + dx.

20

xdq 4 x 1

R dx

Charge enclosed inside the gaussian surface,rr 3 3 4

2enclosed 0 0

0 0

x x xq 4 x dx 4

R 3 R

=

3 4

0r r43 4R

But3 4

2 0

0

4 r rE 4 r3 4R

0

0

r 1 rE3 4R

0

0

r 3rE 13 4R , when r R

(ii) Outside : When r R, thenR R3 4 3 4

enclosed 0 00 0

x x r rq 4 4

3 4R 3 4R

3

02

0

RE

12 r

R

r

Example 22 :A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1cm. It is covered by a concentric, hollowconducting sphere of radius 5 cm.(a) Find the electric field at a point 2 cm. away from the center(b) A charge of 6 × 10–8 C is placed on the hollow sphere.Find the charge on the outer surface of the hollow sphere.

Sol. (a) Let us consider figure (a). Suppose, we have to findfield at the point P. Draw a concentric spherical throughP. All the points on this surface are equivalent and bysymmetry, the field at all these points will be equal inmagnitude and radial in direction.The flux through this surface

P

(a) (b)= 2E.dS E.dS E dS 4 x E

where x = 2 cm = 2 × 10–2 mFrom Gauss’s law, this flux is equal to the charge q contained inside the surface divided by 0.

Thus, 2

0

q4 x E

or8

9 2 2 52 4 2

0

q 4 10 CE (9 10 Nm / C ) 9 10 N / C4 x 4 10 m

(b) See figure (b). Take a Gaussian surface through the material of the hollow sphere. As the electric field in a conducting material

24 GyaanSankalp

Electric charges and fields

is zero, the flux E.dS through this enclosed must be zero. Hence, the charge on the inner surface of the hollow sphere is

– 4 × 10–8 C. But the total charge given this hollow sphere is 6 × 10–8 C. Hence, the charge on the outer surface will be10 × 10–8 C.

Example 23 :A gaussian surface encloses an object with a net charge of +2.0 C and there are 6 lines leaving the surface. Some charge is addedto the object and now there are 18 lines entering the surface. How much charge was added ?

Sol. Since there are 6 lines when there is +2.0 C, therefore a charge of +1.0 C is equivalent to 3 lines. After charge is added, there are 18lines entering.

So the net charge is now 18 lines 6.0 C

3 lines/coulomb

Therefore, the charge added was f iQ Q Q 6.0 C 2.0 C 8.0C Example 24 :

The length of each side of a cubical closed surface is . If charge q is situated on one of the vertices of the cube, then find the fluxpassing through each face of the cube.

Sol. The charge contributes only (q/8)th to this cube. Further, three faces of the cube meet at one corner which are parallel to thecharge. From Gauss theorem

one face0 0

(q / 8) q3 24

And from three faces meeting at point where charge is placed = 0 as E A

TRY IT YOURSELFQ.1 A charge is placed at the centre of a cube with side L. The electric flux linked with cubical surface is

(A) 20(Q / 6L ) (B) 2

0(Q / L ) (C) 0(Q / ) (D) zeroQ.2 A charge Q is situated at the centre of a cube. The electric flux through one of the faces of the cube is –

(A) 0(Q / ) (B) 0(Q / 2 ) (C) 0(Q / 4 ) (D) 0(Q / 6 )Q.3 A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the

vessel is –(A) zero (B) 0(q / ) (C) 0(q / 2 ) (D) 0(2q / )

Q.4 A hemispherical surface of radius R is placed with its cross-sectionperpendicular to a uniform electric field E as shown in fig. flux linked with itscurved surface is –

(A) zero (B) 22 R E

E

R(C) 2R E (D) 0(E / 2 )

Q.5 The application of Gauss's theorem gives rise to an easy evolution of electric field in the case of –(A) A charged body of any geometrical configuration (B) A charged body of regular geometrical configuration(C) Revolving charged bodies (D) Charges forming dipoles

Q.6 Three charges q1 = 1µc, q2 = 2µc and q3 = – 3µc and four surface S1, S2,S3 and S4 are shown. The flux emerging through surface S2 in N-M2/Cis(A) 36 × 103 (B) – 36 × 103 q1

s1

s3

s4

–q2

q3

s2(C) 36 × 109 (D) – 36 × 109

Q.7 A cubical box of side 1m is immersed a uniform electric field of strength 104N/C. The flux through the cube is(A) 104 (B) 6 × 104 (C) 2 × 104 (D) zero

Q.8 If an insulated non-conducting sphere of radius R has charge density . The electric field at a distance r from the centre of sphere(r < R) will be –

(A) 0

r3 (B)

0

R3 (C)

0

r (D)

0

R

25GyaanSankalp

Electric charges and fieldsQ.9 The inward and outward electric flux from a closed surface are respectively 8 × 10³ and 4 × 103 units. Then the net charge inside

the closed surface is –

(A) 3

0

4 10 coulomb (B) – 4 × 103 0 coulomb (C)

3

0

4 10 coluomb (D) 4 × 103 0 coulomb

ANSWERS(1) (C) (2) (D) (3) (C) (4) (C) (5) (B)(6) (B) (7) (D) (8) (A) (9) (B)

ELECTRIC DIPOLEIn some molecules, the centre of +ve and -ve charge do not coincide. This results in the formation of electric dipole. Atom is non-polar because in it the centre of +ve and -ve charges coincide. Polarity can be induced in an atom by the application of electricfield. Hence it can be called as induced dipole.An electric dipole is a system formed by two equal and opposite charges placed at a short distance apart. Product of one of thetwo charges and the distance between them is called "electric dipole moment"

p .

p q 2

(1) It is a vector quantity, directed from - ve to + ve charge.(2) Dimension [L T A], unit Cb x mt.(3) Practical unit is Debye

p of two equal and opposite point charges each having charge 10–10 frankline and separation of 1 Å.

i.e. 1D = 10–10 x 10–10 = 10–20 Fr x mt. = 103 10

20

9

x Cb x mt. 1D 3.3 x 10–30 Cb x mt.

Electric field at an axial point :Electric field at P due to negative charge

1 20

1 q ˆE ( r)4 dr

2

P+q-q

O

rd

p

Electric field at P due to positive charge 2 20

1 q ˆE ( r)4 dr

2

Total electric field at axial point P is a 1 2E E E

aE

= 0

q4 2 2

1 1 r(r d / 2) (r d / 2)

or aE

= 0

q4 2 2 2

2.r.d(r d / 4)

r = 2 2 20

2pr r4 (r d / 4)

When r >> d aE

= 0

14 3

2pr

r, The axial field is parallel to dipole moment

Electric field at an equatorial point :

Electric field at P due to negative charge E1 = 0

14 2 2

q(r d / 4)

Electric field at P due to positive charge 2 2 20

1 qE4 r d / 4

P

E sin2

E cos1

E cos2

E1

E sin1

BOpA

d

4/dr 224/dr 22

E2

r

q-qFields E1 and E2 are equal in magnitudeResolving E1 and E2 into two components one along OP and other perpendicular to OPWe find E1sin = E2 sinTotal field E = E1cos + E2cos = 2E1cos = 2E2cos

26 GyaanSankalp

Electric charges and fields

E = 2 0

q4 2 2

1(r d / 4) 2 2

d / 2

r d / 4 = 2 2 3/2

0

q.d4 (r d / 4)

= 2 2 3/20

p4 (r d / 4)

If r >> d E

= 0

14 3

pr

r i.e. field at equatorial point is antiparallel to dipole moment.

From this it is clear that :1. Intensity due to a dipole varies as (1/r³) and can never be zero unless r or p 0 .

2. E will be maximum when 2cos max 1 , i.e., 0 , for end on, axial or tan A position E is maximum and is,

max 30

1 2pE4 r

3. E will be minimum when 2cos min 0 , i.e., 90 , i.e., for broad on, equatorial or tan B position E is minimum and is,

min 30

1 pE4 r

4. The electric field at axial point is parallel to dipole moment vector.5. The electric field at equatorial point is antiparallel to dipole moment vector.6. The ratio of field at axial point to field at equatorial point is Ea : E = 2 : 1.

Electric field at an arbitrary point :We resolve dipole moment p in two components one along r and another perpendicular to r.

The radial component of electric field Er = 0

14 3

2pcosr

The magnitude of resultant field is E = 0

14

3psin

r

x

+qE

E

Er

Py

p

psin

B -q

A rpcos

The magnitude of resultant field is E = 2 2rE E =

0

14 3

pr

21 3cos

The direction of resultant field is tan = r

E 1E 2 tan

Case I at axial point =0° so Ea = 0

14 3

pr

21 3cos 0 = 30

1 2p4 r

Case II at equatorial point = /2 so E = 0

14 3

pr

21 3cos / 2 = 0

14 3

pr

Dipole in a field :(1) When an electric dipole is placed in an uniform electric field.

A torque acts on it which subjects the dipole to rotatory motionF net = [q E + (-q E )] = 0

Torque = q E x 2 sin = P x E

2l

F=qE

E

+q

2 sin l

–qThere is no net force acting on the dipole in a uniform electric field.(2) Work : Work done in rotating an electric dipole from to (uniform field)

dw = dw = z dw = z d.

W =

1

2z pE sin d= pE (cos – cos )

e.g. W180 = pE (1– (–1)] = 2pE W 90º = pE (1 – 0) = pE

27GyaanSankalp

Electric charges and fieldsIf a dipole is rotated from field direction (= 0) to , then W = pE (1 – cos )

= min = 0 = max. = pE = min = 0 W = min = 0 W = pE W = max. = 2pE

(3) Electrostatic potential energy :In case of dipole (in uniform field) potential energy of dipole is defined as work done in rotating a dipole from direction to thefield to the given direction i.e.

U = w0 – w90º = pE (1 – cos ) – pE = – pE cos .

U = – P.E

E is a conservative field so what work done in rotating a dipole from 1to 2is just equal to change in electrostatic P.E.

W 1 2 = 2U – 1U = pE (cos 1– cos 2 )

(4) If = 0º, i.e. P

|| E

, = 0 and U = – pE,

Dipole is in the minimum potential energy state and no torque acting on it and hence it is in the stable equilibrium state.

For = 180º, i.e. P

and E

are in opposite direction then = 0 but U = pE which is max. potential energy state. Although it is inequilibrium, but it is not a stable state and a slight displacement can disturb it.

(5) If dipole is placed in a non-uniform electric field, it preforms rotatory as well as translatory motion, because now a net force alsoacts on the dipole along with the torque.In Uniform electric field, Total force = 0, Torque may or may not be zero. ( )0if0( In Non-uniform electric field, Total force 0, Torque may or may not be zero.

–2+2

FF'

QFor situation shown in figure, Torque = 0 (Force along same axis)(6) Angular SHM :

When a dipole is suspended in uniform field, it will align itself parallel to the field. Now if it is given a small angular displacement about its angular position, the restoring couple will be

= – pE sin . if is small sin . = – pE – (Angular SHM).for balanced condition : deflecting = restoring

I = – pE = – pEI

= – 2 =

pEI

T = 2

= 2 I

pE [I moment of inertia]

Example 25 :An electric dipole consists of charges 2.0 × 10–8 C separated by distance of2mm. It is placed near a long line charge of density 4.0 × 10–4 cm–1 as shown infigure, such that the negative charge is at a distance of 2 cm from the linecharge. Calculate the force acting on the dipole.

Sol. We know that electric field intensity at a distance r from the line charge of

–q +q

2cm 2mm

Line chargedensity is given by

0E

2 r

So, the field intensity on negative charge is given by 4

91

4.0 10E (2 9 10 )

0.02

= 3.6 × 108 N/C

Force on negative charge,F1 = qE1 = (2 × 10–8) (3.6 × 108) F1 = 7.2 N

This is directed towards line charge.

28 GyaanSankalp

Electric charges and fields

Similarly, field intensity E2 on positive charge4

92

4.0 10E (2 9 10 )

0.022

= 3.27 × 108 N/C

Force on positive charge, F2 = qE2. F2 = (2 × 10–8) × (3.27 × 108) = 6.54 N (away)So, net force F on dipole, F = F1 – F2 = (7.2 – 6.54) N = 0.66 NThe force is towards the line charge.

Example 26 :Two points masses, m each carrying charge –q nad +q are attached to the ends of a massless rigid non-conducting rod of length. The arrangement is placed in a uniform electric field E such that the rod makes a small angle = 5° with the field direction. Show

that the minimum time needed by the rod to align itself along the field (after it is set free) is mt

2 2qE

Sol. The situation is shown in figure.The torque on rod AB is given by qE( sin ) qE The moment of inertia of the rod AB about O is given by

2 2 2mI m m2 2 2

We know that, I or I

qE

–qE –qA

B+q

O

2

2qE 2qE

m(m / 2)

where 2 2qEm

As acceleration is directly proportional to , hence the motion of rod is S.H.M. The time period T is given by

2 mT 22qE

The rod will become parallel to E in a time

T 2 m mt4 4 2qE 2 2qE

TRY IT YOURSELFQ.1 When a test charge is brought from infinity along the perpendicular bisector of the electric dipole the work done is –

(A) Positive (B) Negative (C) Zero (D) None of the aboveQ.2 An electric dipole has charges +q and –q at a separation r. At distance d >>r along the axis of the dipole, the field is proportional

to –

(A) 2d/q (B) 2d/qr (C) 3d/q (D) 3d/qrQ.3 In case of a dipole field –

(A) Intensity can be zero (B) Potential can be zero (C) Both can be zero (D) None can be zeroQ.4 When an electric dipole is placed in a uniform electric field a couple acts on it. The moment of couple will be maximum when the

dipole is placed –(A) along the direction of the field (B) perpendicular to the direction of the field(C) against the direction of the field (D) inclined at an angle of 45° to the direction of the field

Q.5 The electric intensity due to a dipole of length 10cm. and having a charge of 500µC, at a point on the axis 20cm. from one of thecharges in air is –(A) 9.28 × 107 N/C (B) 20.5 × 107 N/C (C) 6.25 × 107 N/C (D) 13.1 × 1011 N/C

Q.6 An electric dipole placed in a non-uniform electric field experiences –(A) A force but not a torque (B) A torque but not a force (C) A force and a torque (D) Neither a force nor a torque

Q.7 An electric dipole consists of two opposite charges each of magnitude 1.0 µC separated by a distance of 2.0cm. The dipole isplaced in an external field of 1.0 × 105 N/C. The maximum torque on the dipole is –(A) 0.2 × 10–3 N-m (B) 2.0 × 10–3 N-m (C) 4.0 × 10–3 N-m (D) 1.0 × 10–3 N-m

29GyaanSankalp

Electric charges and fieldsQ.8 The force of attraction between two coaxial electric dipoles whose centres are r m apart varies with distance as –

(A) r–1 (B) r–2 (C) r–3 (D) r–4

Q.9 An electric dipole is kept on the axis of a uniformly charged ring at distance R / 2 from the centre of the ring. The direction ofthe dipole moment is along the axis. The dipole moment is P, charge of the ring is Q and radius of the ring is R. The force on thedipole is nearly –

(A) 24kPQ

3 3R (B) 34kPQ

3 3R (C) 32kPQ

3 3R (D) zero

Q.10 Point P lies on the axis of a dipole. If the dipole is rotated by 90° anticlockwise, the electric field vector E at P will rotate by–

(A) 90° clockwise (B) 180° clockwise (C) 90° anticlockwise (D) noneANSWERS

(1) (C) (2) (D) (3) (B) (4) (B) (5) (C)(6) (C) (7) (B) (8) (D) (9) (D) (10) (A)

USEFUL TIPS1. The dipole p

is parallel to the z-axis. Then, the zyx E,E,E components are

x 50

p 3xzE .4 r

, y 5

0

p 3yzE .4 r

,

2 2

z 50

p (3z r )E4 r

2. The distance dependence of the electric field due to (i) monopole (ii) dipole, and (iii) quadrupole is as follows :

(i) 21

E long ranger

(ii) 31

E short ranger

(iii) 41

E short ranger

3. The coulomb force between two point charges depends only on the charges, their separation and the medium. It is independentof other charges present.

4. The number of lines of force coming out of a unit positive charge is 11

0

1 1.11 10

5. If a cube is placed in uniform electric field the net flux through it will be zero. This also follows from Gauss’ theorem.6. The electric field (E) due to a line of charge is proportional to 1/r.7. The electric field (E) due to a point charge is proportional to 1/r2.8. The electric field (E) due to uniformly charged flat sheet is constant at all points. This means it does not depend on distance.9. The electric field is uniform in a region, if (a) the number of lines of force crossing unit area normally, is same at all points and (b)

the lines of force are parallel. The first condition (a) makes the magnitude of the field to be the same, while the second condition(b) makes the direction of the field to be the same at all points.

10. To find the direction of electric field at a point, imagine a unit positive charge at the point. Find the magnitude of force on it. Thisgives the magnitude of field. Find the direction of motion of that charge. This gives the direction of electric field.

MISCELLANEOUS SOLVED EXAMPLESExample 1 :

A point charge +Q is placed at the centroid of an equilateral triangle. When a second charge +Q is placed at a vertex of the triangle,the magnitude of the electrostatic force on the central charge is 4N. What is the magnitude of the net force on the central chargewhen a third charge when a third charge +Q is placed at another vertex of the triangle.

Sol.2

2

QF k 4Nr

Q

Q

r

In II case : Charge at centre experiences two forces eachof magnitude 4N at 120°.

2 2F 4 4 2 4 4cos120 , F = 4N

Q

F=4Nr

F=4N

Q

r120°

Q

30 GyaanSankalp

Electric charges and fieldsExample 2:

Two electrons are a certain distance apart from one another. What is the order of magnitude of the ratio of the electric forcebetween them to the gravitational force between them.

Sol.9 19 2 29

1 2e 2 2 2

kq q 9 10 (1.6 10 ) 9 1.6 1.6 10Fr r r

11 31 2 731 2

g 2 2 2Gm m 6.7 10 (9.1 10 ) 6.7 9.1 9.1 10F

r r r

42e

g

F10

F

Example 3 :A particle of mass m and charge q is lying at the mid point of two stationary particles distant 2 and each carrying a charge q. Ifthe free charged particle is displaced from its equilibrium position through distance x(x <<), then the particle will(A) move in the direction of displacement (B) stop at its equilibrium position(C) oscillate about its equilibrium position (D) execute S.H.M. about its equilibrium position

Sol. (D). When the charge q is displaced through small displacement x, then the resultant force acting on it isqq qx

P OQ

F = 2

2kq

( x) – 2

2kq

( x) 2

34kq

x = ma or a = 2

34kqm

x = 2x or a x (Towards O)

The motion of the particle will be S.H.M.Example 4 :

Millikan’s drop experiment attempts to measure the charge on a single electron, e, by measuring the charge of tiny oil dropssuspended in an electrostatic field. It is assumed that the charge on the oil drop is due to just a small number of excess electrons.The charges 3.90 × 10–19 C, 6.50 × 10–19 C and 9.10 × 10–19 C are measured on three drops of oil. Find the charge of an electron.

Sol. 3.9 ×10–19 = n1e, 6.5 ×10–19 = n2e, 9.1 ×10–19 = n3en1, n2 and n3 are integers for e = 1.3 × 10–19 .

Example 5 :Four charges each of magnitude q, are lying at the four corners of a square of side a. How much charge should be placed at thecentre of the square so that the whole system remains in equilibrium?

Sol. Charge Q always remains in equilibriumCondition of equilibrium of charge C

FO = FBD + FA or 2KQqa / 2

= 2 2

2Kqa

+ 2

2Kq2a

q

qA

•O–Q

Bq

DC

qF

FF

F

D

A

BDB

or Q = 1

(2 2 1) q4

Example 6 :The radius of a soap bubble is r and its surface tension is T. If the surface charge density on the bubble is and the excess ofpressure inside it is p, then the value of will be

Sol. In the state of equilibrium

pex + pel = pST or p + 2

02

= 4Tr

or = 1/2

04T2 pr

Example 7 :

An electric field line emerges from a positive point charge +q1 at anangle to the straight line connecting it to a negative point charge –q2(figure). At what angle will the field line enter the charge –q2 ?

31GyaanSankalp

Electric charges and fieldsSol. In the immediately proximity of each of point charges, the contribution from the other charge to the total field strength is negligibly

small, and hence the electric field lines emerge from (enter) the charge in a spatially homogeneous bundle, their total number beingproportional to the magnitude of the charge. Only a fraction of these lines gets into a cone with an angle 2 at the vertex near thecharge +q1. The ratio of the number of these lines to the total number of the lines emerging from the charge +q1 is equal to the ratioof the areas of the corresponding spherical segments.

22 RR (1 cos ) 1 (1 cos )

24 R

Since electric field lines connect the two charges of equal magnitude, the number of lines emerging from the charge +q1 within theangle 2 is equal to the number of lines entering the charge –q2 at an angle 2. Consequently,

1 2| q | (1 cos ) | q | (1 cos )

whence1

2

| q |sin sin

2 2 | q |

If 1 2| q | / | q | sin( / 2) 1 , an electric field line will not enter the charge – q2.Example 8 :

The strength of the electric field produced by charges uniformly distributedover the surface of a hemisphere at its centre O is E0. A part of the surface isisolated from this hemisphere by two planes passing through the same diameterand forming an angle with each other.Determine the electric field strength E produced at the same point O by thecharges located on the isolated surface (on the “mericarp”).

Sol. It can easily be seen from symmetry considerations that the vector of the electric

field strength produced by the “lobule” with an angle lies in the planes of longitudinal and transverse symmetry of the lobule.Let the magnitude of this vector be E. Let us use the superposition principle and complement the lobule to a hemisphere chargedwith the same charge density. For this purpose, we append to the initial lobule another lobule with an angle – . Let themagnitude of the electric field strength vector produced by this additional lobule at the centre of the sphere be E’. It can easily be

seen that vectors E

and E

are mutually perpendicular, and their vector sum is equal to the electric field vector of the hemisphere

at its centre. By hypothesis, this sum is equal to E0. Since the angle between vectors E

and 0E

is / 2 / 2 ,

we obtain 0E E sin2

Example 9 :Two very large thin conducting plates having same cross-sectional area areplaced as shown in figure they are carrying charges Q and 3Q respectively.The variation of electric field as a function of x (for x = 0 to x = 3d) will be bestrepresented by –

Y

X

Q 3Q

(d,0) (2d,0) (3d,0)

(A)

E

xd 2d 3d

(B)

E

xd 2d 3d

(C)

E

x

d

2d 3d(D)

E

x

d

2d 3d

Sol. (C).Using the formula for electric field produced by large sheet 0

QE2A

We get A0

4Q ˆE ( i)2A

; B

0

2Q ˆE ( i)2A

; C

0

4Q ˆE ( i)2A

32 GyaanSankalp

Electric charges and fieldsExample 10 :

The minimum strength of a uniform electric field which can tear a conducting uncharged thin-walled sphere into two parts isknown to be E0. Find minimum electric field strength capable of tearing the conducting shell of twice as large radius.

Sol. The density of charges induced on the sphere is proportional to the electric field strength : E . The force acting on thehemispheres is proportional to the field strength :

2 2F SE R E Where S is the area of the hemisphere, and R is its radius. As the radius of the sphere changes by a factor of n, and the fieldstrength by a factor of k, the force will change by a factor of k²n². Since the thickness of the sphere walls remains unchanged, theforce tearing the sphere per unit radius must remain unchanged,

i.e. 2 2k n

1n

and 1 1

kn 2

.

Consequently, the minimum electric field strength capable of tearing the conducting shell of twice as large radius is 01

EE

2 .

Example 11 :The thickness of a flat sheet of a metal foil is d, and its area is S. A charge q is located at a distance from the centre of the sheetsuch that d S .Determine the force F with which the sheet is attracted to the charge q, assuming that the straight line connecting the charge tothe centre of the sheet is perpendicular to the surface of the sheet.

Sol. Since the sheet is metallic, the charges must be redistributed over its surface so that the field in the bulk of the sheet is zero. In thefirst approximation, we can assume that this distribution is uniform and has density – and over the upper and the lower surfacerespectively of the sheet. According to the superposition principle, we obtain the condition for the absence of the field in the bulk

of the sheet : 200

q 04

Let us now take into consideration the nonuniformity of the field produced by the point charge since it is the single cause of the

force F of interaction. The upper surface of the sheet must be attracted with a force 20

Sq4

, while the lower surface must be

repelled with a force 20

Sq4 ( d)

.

Consequently, the force of attraction of the sheet to the charge is

2

2 2 2 50 0

Sq 1 q SdF 14 (1 d / ) 8

Example 12 :In the figure shown, initially the spring of the negligible mass is in underformed state and the block has zero velocity E is a uniformelectric field. Then –

(i) The maximum speed of the block will be QEmK

(ii) The maximum speed of the block will be 2QEmK

(iii) The maximum compression of the spring will be QEK

(iv) The maximum compression of the spring will be 2QE

K(A) only (i) and (ii) are correct (B) only (i) and (iv) are correct(C) only (ii) and (iii) are correct(D) only (ii) and (iv) are correct

m

QE

K

smoothSol. (B). Speed will be maximum when acceleration becomes zero

i.e. , when KX = EQ EQ

XK

By work-energy theorem : 2 2all

1 1W KE EQX KX mv

2 2

33GyaanSankalp

Electric charges and fields

Substituting EQ

xK

: maxQE

VmK

compression will be maximum when velocity becomes zero.

2all

1W KE EQX KX 0

2 ; max

2EQX

K

Example 13 :Two balls of charge q1 and q2 initially have a velocity of the same magnitude and direction. After a uniform electric field has beenapplied during a certain time, the direction of the velocity of the first ball changes by 60°, and the velocity magnitude is reducedby half. The direction of the velocity of the second ball changes thereby by 90°. In what proportion will the velocity of the secondball change ? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to k1 for the first ball. Theelectrostatic interaction between the balls should be neglected.

Sol. Let v1 and v2 be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the anglebetween the velocity v is 60°. Therefore, the change in the momentum of the first ball is 1 1 1p q E t m vsin 60

Here we use the condition that v1 = v/2, which implies that the change in the momentum 1p of the first ball occurs in a directionperpendicular to the direction of its velocity v1.Since E || 1p and the direction of variation of the second ball momentum is parallel to the direction of 1p , we obtain for thevelocity of the second ball. (it can easily be seen that the charges on the balls have the same sign.)

2v

v v tan 303

The corresponding change in the momentum of the second balls is

22 2

m vp q E t

cos30

Hence we obtain, 1 1

2 2

q m sin 60q m / cos30

; 2 1

12 1

q q4 4 km 3 m 3

Example 14 :An infinite long plate has surface charge density . As shown in the figure apoint charge q is moved from A to B. Net work done by electric field is –

B(x ,0)1 A(x ,0)2(A) 1 20

q (x x )2

(B) 2 1

0

q (x x )2

(C) 2 10

q (x x )

(D) 0

q (2 r r)

Sol. (A). netW qE.d

where 0

ˆE i2

, d = (x1 – x2)

Example 15 :Estimate the upper limit of the error made in calculating the force of interaction between charged spherical conductors with the aidof the Coulomb law. The radii of the spheres are r0, the distance between their centres is r.

Sol. It is evident from figure for the case of charges of differentsigns that the actual of interaction of the charges is greaterthan it would have been, if the charges were concentratedin the centres of the spheres, and less than if the chargeswere concentrated at the nearest points of the spheres :

2 2

2 20 0 0

q qF4 r 4 (r 2r )

The absolute error is less than the difference between the boundling values, the relative error being less than the ratio of thisdifference to the minimum force. Hence,

34 GyaanSankalp

Electric charges and fields2 2

0 0 02 12 2

1 0 0

r (r 2r ) 4r (r r )F FF (r 2r ) (r 2r )

Example 16 :A dipole is placed in the field of a point charge, the distance between the dipole and the field source being much greater than thedipole separation. Find the force acting on the dipole and the torque, if the dipole is arranged :(a) perpendicular to the line of force, (b) in the direction of the lines of force.

Sol. (a) It is evident from figure (a) that in this case a force couple acts on the dipole.The resultant of the force is zero. To obtain the torque, multiply the force bythe dipole separation.

(b) It is evident from figure (b) that in this case to forces act on the dipole in thedirection of the radius vector. Hence it is clear that the torque is zero, and theresultant force acts in the direction of the radius to the source.The resultant can be found by two methods. One may use the Coulomb law :

2 20 0

Qq QqF F F4 (r / 2) 4 (r / 2)

= 2 2 20

2Qrq4 (r / 4)

Nothing that the problem stipulates that << r, we obtain

e4 3

0 0

2Qp2QrqF4 r 4 r

The same result may be obtained from formula, if the derivative is substituted for the ratio of increments. We have

ee e 2 3

0 0

2QpdE d QF p pdr dr 4 r 4 r

.

Alternatively, use electric field expression due to dipole.Example 17 :

A cube of edge a metres carries a point charge q at each corner. Calculate the resultant force on any one of the charges.Sol. Let us take one corner of cube as origin O(0, 0, 0) and the opposite corner as P(a, a, a). We will calculate the electric field at P due

to the other seven charges at corners.Expressing the field of a point charge in vector form

rr4

qE3

0

(i) Field at P due to A, B, C

CPBPAPa4

qE3

01

iakaja

a4q

30

C

E

DF

AP

B

ZY

X

O(ii) Field at P due to D, E, FNow that DP = EP = FP = 2a

FPEPDP

2a4

qE 30

2

kaiajaiakaja)a22(4

q3

0

kjia24

q2

0

(iii)Field at P due to O : 3aOP

OP

3a4

qE 30

3

; )kajaia()a33(4

qE2

03

; kji

)a33(4qE

20

3

Resultant Field at P

321 EEEE ;

33

12

11a4

)kji(qE 20

outward along OP

35GyaanSankalp

Electric charges and fields

Force on charge at P is F = q E 20

2

a43qF

331

211

Outwards along diagonal OPExample 18 :

An electron is projected along the axis midway between the plates of a cathode ray tube with an initial velocity of2 × 107 ms–1. The uniform electric field between the plates has an intensity of 20,000 NC–1 and is upward.(a) How war below the axis has the electron moved when it reaches the end of the plates ?(b) At what angle axis is it moving as it leaves the plates ?

Sol. The electric field is upward, so the electric force on the electron is downward. Let the x-axis lie along the initial velocity. The x-component of the electron’s velocity remains constant, and the time for the electron to traverse the plates is

1907 1

0x

(x x ) 0.04mt 2.0 10 sv 2 10 ms

In the downward (y) direction the acceleration of the electron is produced by the electric force.

1 19y

y 31e e

F E | q | (20,000NC ) (1.60 10 C)am m 9.11 10 kg

= 15 23.51 10 ms

The downward displacement of the electron as it passes between theplates is then given by

20 0y y

1y y v t a t

2

15 2 9 20

1y y (3.51 10 ms ) (2.0 10 s)

2 = 7.02 × 10–3 m = 0.702 m

y

x

vtan

v ; 7 1

xv 2 10 ms , 6 1y oy yv v a t 7.02 10 ms

y

x

vtan 0.351

v 19.3

Example 19 :Two small balls with equal but opposite charges are secured in a horizontal plane at a distance a from each other. A third chargedball is suspended on a string. The point of suspension is so move that the third ball, when in a state of equilibrium, is preciselyabove the first ball at a distance a from it, and then it is so moved that the third ball is above the second one. Find the anglesthrough the string is deflected the balls if in second case it is double the first case.

Sol. The conditions of equilibrium of the suspended ball give the following equations for the two cases being considered :

s1 1 2

KQQ 2T sin 022a

s s1 1 2 2

KQQ KQQ2T cos mg 022a a

; s2 2 2

KQQ 2T sin 022a

s s2 2 2 2

KQQ KQQ 2T cos mg 02a 2a

where T1 and T2 are the tensions of the thread, 1 and 2 the angles ofdeflection of the thread, +Q and – Q the charges of the fixed balls, +QSthe charge of the suspended ball, and mg is the weight of the suspendedball.

Upon excluding the unknowns from the above simultaneous equations, we get

1 2 1 1cot cot cot cot 2 2 ( 2 1)

36 GyaanSankalp

Electric charges and fields

whence, 1cot 2(2 2 1) 35 16 2

Thus, 1 7 56' and 2 15 52 ' (From Trigonometric Table)

whens

2KQQ 2mg 1

4a

and 1 82 04' and 2 164 08'

whens

2KQQ 2mg 1

4a

Example 20 :The negative charge –q2 is fixed while positive charge q1 as wellas the conducting sphere ‘S’ is free to move. If the system isreleased from rest(A) both S and q1 move towards left(B) q1 moves towards right while S moves towards left(C) q1 remains at rest, S moves towards left(D) both q1 and S remain at rest

/////////////q1

q2Conducting shell 'S'

Sol. (C).Net force on q1 is zero, while that on the conducting sphere is towards the left due to the attraction of –q2.After sometime : q1 will not be at the centre and due to this the charge on the inner surface will not remain uniformly charged.q1 will experience net force towards rightmore negative charge will be present on that portion of the shell which is near to q1.

Example 21 :Find the electric field inside a sphere which carries a charge density proportional to the distance from the origin = r(is a constant).

Sol. We can consider all the charge inside the sphere to be concentratedon the center of sphere.Consider an elementary shell of radius x and thickness dx.

2 2

2 20

K dq K 4 x dx ( x) rE4r r

x

dx

Example 22 :Figure shows two large cylindrical shells having uniform linear chargedensities + and –. Radius of inner cylindrical is ‘a’ and that of outercylinder is ‘b’. A charged particle of mass m, charge q revolves in acircle of radius r. Find its speed v. (Neglect gravity and assume the radiiof both the cylinders to be very small in comparison to their length)

r

va

b

Sol. Electric field between the two cylinders = 2k

r

Force on charge 2k q

qr

This force is centripetal force 22k q mv

r r

0 0

1 q qv 24 m 2 m

37GyaanSankalp

Electric charges and fieldsExample 23 :

Three identical positive charges Q are arranged at the vertices of an equilateral triangle. The side of the triangle is a. Find theintensity of the field at the vertex of a regular tetrahedron of which the triangle is the base.

Sol. Each charge creates at point D a field intensity of 1 2KQEa

. The total intensity

will be the sum of three vectors (fig.). The sum of the horizontal components ofthese vectors will be zero, since they are equal in magnitude and form angles of120° with each other. The vectors from angles of 90° – with the vertical,where is the angle between the edge of the tetrahedron and the altitude h oftriangle ABC.

The vertical components are identical each being equal to 2KQ sina

.

It follows from triangle ADE that 2sin3

.

Therefore, the sought intensity of the field is 2KQE 6a

Example 24 :A molecule is at a distance r from the axis of a charged infinitely long metallic cylinder. Find the force acting on the molecule if the

intensity of the cylinder field is expressed by the formula 2q

Er

(q is the charge per unit of cylinder length) and the molecule has

the form of a “dumb-bell” with a length and with charges +Q and – Q at its ends.Sol. The molecule will be attracted to the charged cylinder. The force of attraction is

1 1 2qQF 2qQr r r (r )

In this expression we want neglect the quantity 8( 10 cm) as compared with r (r cannot be smaller than the cylinder radius).

We finally obtain 22qQF

r

Example 25 :Two molecules of equal mass are at a certain distance from the axis of a charged cylinder. One molecule has a constant electricmoment p = Q . An elastic force acts between the charges of the other molecule, i.e., the distance is determined from theexpression QE = k , where E is the mean intensity of the field acting on the molecule and k is a proportionality factor. First theelectric moments of the molecules are the same and their velocities are zero.Which molecule will reach the surface of the cylinder quicker under action of the force of attraction.

Sol. At the initial moment the forces acting on both molecules are identical. When the molecules approach the cylinder, the force F1acting on the molecule with a constant electric moment grows in proportional to 1/r2.

1 22qQF

r

The force F2 that acts on the elastic molecule grows faster, in proportion to 1/r³, owing to the continuous increase in the electric

moment of this molecule 2 2 32(F 4q Q / kr )

The masses of the molecules are the same, and for this reason the acceleration of the second molecule when it approaches thecylinder grows faster than that of the first one and it will reach the surface of the cylinder quicker.

Example 26 :A charged particle of mass m and having a charge Q is placed in an electric field E which varies with time as E = E0 sin t. Whatis the amplitude of the S.H.M. executed by the particle.

Sol. For a particle undergoing SHM with an amplitude A and angular frequency , the maximum acceleration = 2AHere the maximum force on the particle = QE0

maximum acceleration = 20QEA

m 0

2

QEA

m

38 GyaanSankalp

Electric charges and fields

QUESTION BANK

EXERCISE - 1CONCEPTUAL QUESTIONSQ.1 Under what conditions can a brass rod be charged ?Q.2 An electric charge on a conducting sphere has to be

divided into three equal parts. How can this be done ?Q.3 Why is a metal chain that reaches the ground fixed to a

lorry for transporting petrol ?Q.4 Can a positive charged be obtained on the electroscope

using a negatively charged ebonite rod ?Q.5 What will happen to the surface charge density on a metal

sheet rolled into a cylinder ?Q.6 An elder ball is tied to a silk thread. What will happen

when an electrostatically charged rod is brought close toit ?

Q.7 Do the electric field vector and the vector of the forceexerted on a charge by an electric field always have thesame directions ?

Q.8 Why is metal cap sometimes put on a vacuum tube ?Q.9 Can electric charges be separated on

(a) conductor, (b) a dielectric ?

Q.10 A thin insulator rod is placed between two unlike pointcharges +q1 and –q2 (fig.). How will the force acting on thecharges change ?

Q.11 A sphere carries a uniformly distributed electric charge.Prove that the field inside the sphere is zero.

Q.12 If only one charged body is available, can it be used toobtain a charge exceeding many times in absolute magni-tude that which it itself has ?

Q.13 Can two likely-charged balls be attracted to each other ?Q.14 A small metallic sphere is brought in contact in turn with

points A, B, C of a charged body. Find the approximatecharge on the sphere after each contact as would be indi-cated by bringing the sphere in contact with an electro-scope. Do the leaves of the electroscope diverge at thesame angle in all three cases?

EXERCISE - 2ONLY ONE OPTION IS CORRECTQ.1 A semi-infinite insulating rod has linear charge density .

The electric field at the point P shown in figure is

B

+ + + + ++ + + + ++ + + + +A

P

r

(A) 2

20

2(4 r)

at 45° with AB (B) 2

20

24 r

at 45° with ABAB

(C) 0

24 r

at 45° with AB

(D) 0

24 r

at perpendicular to AB

Q.2 A certain charge Q is divided into two parts q and (Q-q).For the maximum coulomb force between them, the ratio(q/Q) is :(A) 1/16 (B) 1/8(C) 1/4 (D) 1/2

Q.3 The force between two point charges + Q and – Q placed‘r’ distance apart is f1 and force between two sphericalconductors, each of radius R placed with their centres ‘r’distance apart charged with charge +Q and –Q is f2. If theseparation ‘r’ is not much larger than R, then :(A) f1 > f2 (B) f1 = f2(C) f1 < f2 (D) f2 = (r/R) f1

Q.4 A point charge +Q is placed at the centroid of an equilat-eral triangle. When a second charge +Q is placed at a ver-

tex of the triangle, the magnitude of the electrostatic forceon the central charge is 8N. The magnitude of the net forceon the central charge when a third charge +Q is placed atanother vertex of the triangle is –(A) zero (B) 4N(C) 4 2N (D) 8N

Q.5 Two identical metallic blocks resting on a frictionless hori-zontal surface are connected by a light metallic spring hav-ing the spring constant 100 N/m and an unstretched lengthof 0.2m, as shown in figure 1. A total charge Q is slowlyplaced on the system, causing the spring to stretch to anequilibrium length of 0.3m, as shown in figure 2. The valueof charge Q, assuming that all the charge resides on theblocks and that the blocks are like point charges, is–

m mK

Figure 1

Figure 2

m mK

(A) 10 µC (B) 15 µC(C) 20 µC (D) 30 µC

Q.6 A particle of charge – q and mass m moves in a circle ofradius r around an infinitely long line charge on linearcharge density +. Then time period will be –

[where 0

1k4

]

39GyaanSankalp

Electric charges and fields

(A) mT 2 r

2k q

(B) 2

2 34 mT r2k q

(C) 1 2k qT

2 r m

r –q

(D) 1 mT

2 r 2k q

Q.7 A large uniformly charged(negative) plate is placedin xz plane and a positivepoint charge is fixed onthe y-axis. A dipole is po-sitioned in between withits axis along y-axis, asshown. The dipole initiallymoves in –(A) negative y-direction

y

x

+Q

–

–

+

plate

(B) negative x-direction (C) positive x-direction(D) positive y-direction

Q.8 A ring of radius R is made out of a thin metallic wire of areaof cross section A. The ring has a uniform charge Qdistributed on it. A charge q0 is placed at the centre of thering. If Y is the young’s modulus for the material of the ringand R is the change in the radius of the ring, then :

(A) 0

0

q QR

4 RAY

(B) 0

0

q QR

4 RAY

(C) 0

2 20

q QR

8 RAY

(D) 0

20

q QR

8 RAY

Q.9 A charge is situated at a certain distance from an electricdipole in the end-on position experiences a force F. If thedistance of the charge is doubled, the force acting on thecharge will be :(A) F/4 (B) F/8(C) 2F (D) F/2

Q.10 The dipole moment of thegiven charge distribution is :

(A) 4Rq

i

(B) 4Rq

i

R X

q

Y

–q + + ++++–

–––– – –

(C) 2Rq

i

(D) 2Rq

i

Q.11 Six charges are placed at thevertices of a regular hexagonas shown in the figure.The electric field on the line

O

+Q

+Q

+Q

–Q

–Q

–Q

a

passing through point O andperpendicular to the plane ofthe figure at a distance ofx ( >>a) from O is

(A) 30

Qax (B) 3

0

2Qax

(C) 30

3Qax (D) zero

Q.12 When three electric dipoles are near each other, they eachexperience the electric field of the other two, and the threedipole system has a certain potential energy. Figure belowshows three arrangements (A), (B) and (C) in which threeelectric dipoles are side by side. All three dipoles have thesame electric dipole moment magnitude and the spacingsbetween adjacent dipoles are identical. If U1, U2 and U3are potential energies of the arrangements (A) , (B) and (C)respectively, then –

(a) (b) (c)(A) U1 > U3 > U2 (B) U1 > U2 > U3(C) U1 > U2 = U3 (D) U1 = U2 = U3

Q.13 In a region of space the electric field is given byˆ ˆ ˆE 8i 4j 3k

. The electric flux through a surface of area

of 100 units in x-y plane is :(A) 800 units (B) 300 units(C) 400 units (D) 1500 units

Q.14 A horizontal electric field (E = (mg)/q) exists in space asshown in figure and a mass m is attached at the end of alight rod. If mass m is released from the position shown infigure find the angular velocity of the rod when it passesthrough the bottom most position :

(A) g

(B) 2g

(C) 3gl

(D) 5gl

=45ºq

mgE

mm

+q

Q.15 A dipole of dipole moment p is kept at the centre of radiusR and charge Q. The dipole moment has direction alongthe axis of the ring. The resultant force on the ring due tothe dipole is –

(A) zero (B) 3kPQR

(C) 32kPQ

R

(D) 3kPQR

only if the charge is uniformly distributed on the

ringQ.16 Two point charges +q and –q are held fixed at (-d, 0) and (d,

0) respectively of a (X, Y) coordinate system. Then –(A) The electric field E

at all points on the X-axis has the

40 GyaanSankalp

Electric charges and fieldssame direction.

(B) E

at all points on the Y-axis is along i(C) Work has to be done in bringing a test charge frominfinity to the origin.(D) The dipole moment is 2qd directed along i

Q.17 A particle of charge q and mass m moves rectilinearly un-der the action of electric field E = A – Bx, where A and B arepositive constants and x is distance from the point whereparticle was initially at rest then the distance traveled bythe particle before coming to rest and acceleration of par-ticle at that moment are respectively :

(A) 2A

,0B

(B) qA

0,m

(C) 2A qA

,B m

(D) 2A qA

,B m

Q.18 A charge q is placed at O inthe cavity in a sphericaluncharged conductor. Point Sis outside the conductor.If the charge is displaced from O towards S, still remainingwithin the cavity,(A) electric field at S will increase(B) electric field at S will decrease(C) electric field at S will first increase and then decrease(D) electric field at S will not change

Q.19 Which of the following statements is true –(A) The electric field due a point charge can be same at twopoints.(B) The electric field increases continuously as one goesaway from centre of a solid uniformly charged sphere(C) The electric field of force of the electric field producedby the static charges from closed loops(D) The magnetic lines of force of magnetic field producedby current carrying wire from closed loops

Q.20 An electron of mass me initially at rest moves through acertain distance in a uniform electric field in time t1. A pro-ton of mass mp also initially at rest takes time t2 to movethrough an equal distance in this uniform electric field.Neglecting the effect of gravity the ratio of t2/t1 is nearlyequal to :(A) 1 (B) (mp/me)

1/2

(C) (me/mp)1/2 (D) 1836

Q.21 The variation of electric field between two point chargesalong the line joining the charges is given in figure. Thenwhich is/are correct ?

(A) Q1 is +ve and Q2 is –ve (B) Q1 is +ve and Q2 is +ve(C) 1 2| Q | | Q | (D) 1 2| Q | | Q |

Q.22 A largest sheet carriesuniform surface chargedensity A rod of length 2has a linear charge densityon one half and – on theother half.The rod is hinged at mid point +

)O

+

–

O and makes angle with the normal to the sheet. Thetorque experienced by the rod is

(A) 2

0cos

2

(B) 2

0cos

(C) 2

0

sin2

(D) 2

0

sin

Q.23 Find the force experienced by thesemicircular wire charged with acharge q, placed as shown in figure.Radius of the wire is R and the line

+

++

+

+

+

+

+

+

+

+

+

+ + +

+++++

++

+

+

+R

of charge with linear charge density is passing through its centre andperpendicular to the plane of wire.

(A) 20

q2 R

(B) 20

qR

(C) 20

q4 R

(D) 0

q4 R

Q.24 Five styrofoam balls are suspended from insulating threads.Several experiments are performed on the balls and thefollowing observations are made –(i) Ball A repels C and attracts B(ii) Ball D attracts B and has no effect on E(iii) A negatively charged rod attracts both A and E.An electrically neutral styrofoam ball gets attracted ifplaced nearby a charged body due to induced charge.What are the charges, if any, on each ball ?

////////////////

A

////////////////

B

////////////////

C////////////////

D

////////////////

E A B C D E(A) + – + 0 +(B) + – + + 0(C) + – + 0 0(D) – + – 0 0

41GyaanSankalp

Electric charges and fieldsQ.25 The electric force on 2C

charge placed at the centre Oof two equilateral triangleseach of side 10cm, as shownin figure is F.If charges A, B, C are each 2C and charges D, E and F areeach – 2C, then F is :(A) 2 1.6N (B) 64.8N(C) 0 (D) 43.2N

Q.26 The electric field in a region of space is given byˆ ˆE 5i 2 j N / C

. The electric flux due to this field through

an area 2m2 lying in the YZ plane, in S.I. unit, is –(A) 10 (B) 20(C) 10 2 (D) 2 29

Q.27 Two point dipoles ˆpk and p

k2

are located at (0, 0, 0) and

(1m, 0, 2m) respectively. The resultant electric field due tothe two dipoles at the point (1m, 0, 0) is –

(A) 0

9p k32 (B)

0

7p k32

(C) 0

7p k32 (D) None of these

Q.28 Four charges are rigidlyfixed along the Y-axis asshown. A positive chargeapproaches the systemalong the X-axis with ini-tial speed just enough tocross the origin. Then itstotal energy at the originis –

–q

2

–q

2q

2q

v–1

–2

y

Q x

(A) zero (B) positive(C) negative (D) data insufficient

Q.29 A circle of radius r has a linear charge density = 0 cos2along its circumference. Total charge on the circle is(A) 0(2r) (B) 0(r)

(C) 0r

2 (D) 0

r4

Q.30 Find the electric flux cross-ing the wire frame ABCDof length , width b andwhose centre is at a dis-tance OP = d from an infi-nite line of charge with lin-ear charge density .Consider that the plane

O P

AB

CD

ld

b

of frame is perpendicular to the line OP.

(A) 1

0

btan2d

(B)

1

0

btan2 2d

(C) 1

0

btan4d

(D)

1

0

btan2 4d

Q.31 Two point point charges exert on each other a force F whenthey are placed r distance apart in air. When they are placedR distance apart in a medium of dielectric constant K, theyexert the same force. The distance R equals –

(A) rK (B)

rK

(C) rk (D) r KQ.32 A ring of radius R is placed in the plane with its centre at

origin and its axis along the x-axis and having uniformlydistributed positive charge. A ring of radius r (<< R) andcoaxial with the ring is moving along the axis with constantvelocity then the variation of electric flux () passingthrough the smaller ring with position will be best repre-sented by –

rxv

y

R

(A) (B)

(C) t (D) t

Q.33 A point charge q is placed at origin. Let AE

, BE

and CE

be

the electric field at three points A (1,2,3), B (1, 1, –1) andC (2, 2, 2) due to charge q. Then

(A) A BE E

(B) B C| E | 4 | E |

(C) A BE || E

(D) B C| E | 2 | E |

Q.34 A small sphere of mass m and carrying a charge q isattached to one end of an insulating thread of length ‘a’,the other end of which is fixed at (0,0) as shown in the

figure. There exists a uniform electric field 0ˆE E j

in the

42 GyaanSankalp

Electric charges and fieldsregion.

vy

xO

The minimum velocity which shouldbe given to the sphere at (a,0) in thedirection shown so that it is able tocomplete the circle around the originis (There is no gravity) :

(A) 05qE am

(B) 03qE am

(C) 0qE am

(D) 0qE a2

mQ.35 A charge Q is placed at the

centre of an imaginary hemi-spherical surface. Using sym-metry arguments and theGauss’s law.The flux of the electric field

Q

due to this charge through the surface of the hemisphere(figure) is –

(A) 0

Q2 (B)

0

Q

(C) 0

2Q (D)

0

2Q3

Q.36 A particle of mass m and charge Q is placed in an electricfield E which varies with time t as E = E0sin wt. It willundergo simple harmonic motion of amplitude –

(A) 202

QEm

(B) 02

QEm

(C) 02

QEm

(D) 0QEm

Q.37 An equivalent triangle wireframe of side L having 3 pointcharges at its vertices is keptin x-y plane as shown.Component of electric

q y

x

q –2qfield due to the configurationin z direction at (0, 0, L) is [origin is centroid of triangle]

(A) 29 3 kq

8L(B) zero

(C) 29 kq8L

(D) None

Q.38 A particle of mass m and charge q is placed at rest in auniform electric field E and then released. The KE attainedby the particle after moving a distance y is –(A) qEy2 (B) qEy(C) qE2y (D) q2Ey

Q.39 If electric flux entering and leaving on enclosed surface is1 and 2 respectively, the electric charge inside theenclosed surface will be –

(A) 1 2 0( )/ (B) 2 1 0( )/

(C) 1 2 0( )/ (D) 2 1 0( ) / 2 Q.40 Two particles A and B having equal charges are placed at a

distance d apart. A third charged particle placed on theperpendicular bisector at a distance x will experience themaximum Coulomb’s force when(A) x d / 2 (B) x = d/2

(C) x d / 2 2 (D) x d / 3 2Q.41 A charged particle of mass m and charge q initially at rest is

released in an electric field of magnitude E. Its kinetic energyafter time t will be –

(A) 2 22E t

mq (B) 2 2 2E q t2m

(C) 2

2Eq m

2t(D)

Eqmt

Q.42 A square surface ofside L metres is in theplane of the paper. Auniform electric field

E (volt/m), also in the

E

plane of the paper, is limited only to lower half of the squaresurface, (see figure). The electric flux in SI units associ-ated with the surface is –(A) EL²/2 (B) zero(C) EL² (D) EL²/(20)

Q.43 A mass particle (mass = m and charge = q) is placed be-tween two point charge q. If these charge displaced 2L.distance the frequency of oscillation of mass particle, if itis displaced for a small distance

(A) q

2 30

1m L (B)

q2 3

0

12m L

(C) q

2 3

0

14m L (D)

q2

30

116 m L

Q.44 Two equal negative charges -q are placed at point (0, ± a)on y-axis, one positive charge q is placed at x = 2a, thischarge will-(A) Execute S.H.M. about the origin(B) Oscillate but not execute S.H.M.(C) Move towards origin and will become stationary(D) S.H.M. along x axis

Q.45 An electric charge q is placedat the centre of the open endof a hollow cylindrical vesselof length and radius r.Then the electric flux coming

q

r

out of the surface of the cylinder is-

(A) 0

q

43GyaanSankalp

Electric charges and fields

(B) 2

0

q ( r 2 r )

(C) 0

q2

(D) ZeroQ.46 The electrical breakdown of air occurs at E = 3000 volt/mm.

Then the maximum charge that can be given to a sphere ofdiameter 5 metre will be nearly- (in coulomb)(A) 2.1 × 10–3 (B) 0.83 × 10–3

(C) 4.6 × 10–2 (D) 2.1 × 10–6

Q.47 The electric field in the regionshown here is given by

ˆE x i volt/m. Then thetotal electric flux through thecube of side ‘a’ is- X

Y

aO

(A) Zero (B) a2

(C) a5/2 (D) a3/2

Q.48 Three identical spheres each having a charge q and radiusR, are kept in such a way that each touches the other two.Find the magnitude of the electric force on any sphere dueto other two.

(A) 2

0

1 3 q4 4 R

(B) 2

0

1 3 q4 2 R

(C) 2

0

1 3 q2 4 R

(D) None

Q.49 If two infinite oppositelycharged plates with surfacecharge density || on eachplate are kept perpendicularto each other as shown inabove figure, then a charge q at P will experience a force

(A) 0

q2

towards the

(B) 0

q2

away from horizontal plate

(C) 0

q2

towards the origin O

(D) 0

q2

making an angle 45° with respect to vertical

Q.50 The electric dipole of moment ˆp pi is kept at a point

(x, y) in an electric field 2 2ˆ ˆE 4xy i 4x y j

. The force onthe dipole is :

(A) zero (B) 2 2

pyx

x y

(C) 2 24py y 4x (D) can not be calculatedQ.51 Figure shows a metal body of

mass M charged positively. Pis a point in front of the body. The electric field at point Pdue to the body M is EP .Now a negative charge q is placed at point P and itexperiences a force F . Then :(A) EP = F/q (B) EP < F/q(C) EP > F/q (D) None of these

Q.52 There is a non-uniform electricfield along x-axis as shown infigure. The field increases ata uniform rate along +ve x-axis. A dipole is kept insidethe field as shown.Which one of the following statements is correct for dipole ?(A) dipole moves along positive x-axis and rotatesclockwise(B) dipole moves along negative x-axis and rotatesclockwise(C) dipole moves along positive x-axis and rotatesanticlockwise(D) dipole moves along negative x-axis and rotates anti-clockwise

EXERCISE - 3ONE OR MORE THAN ONE CHOICE MAYBE CORRECTQ.1 A thin walled spherical conducting shell S of radius R is

given charge Q. The same amount of charge is also placedat its centre C. Which of the following are correct –

(A) On the outer surface of S charge density = 2

Q2 R

(B) The electric field is zero at all points inside S

(C) At a point just outside S, the electric field is double, thefield at a point just inside S.(D) At any point inside S, the electric field is inverselyproportional to the square of distance from C.

Q.2 Electric field, due to an infinite line of charge, as shown infigure at a point P at a distance r from the line is E. If onehalf of the line of charge is removed from either side ofpoint A, then :

44 GyaanSankalp

Electric charges and fields

(A) Electric field at P will have magnitude E/2(B) Electric field at P in x direction will be E/2.(C) Electric field at P in y direction will be E/2.(D) none of these

Q.3 An electric dipole moment ˆ ˆp (2.0i 3.0j) C . m is

placed in a uniform electric field 5 1ˆ ˆE (3.0i 2.0k) 10 NC

(A)The torque that E exerts on p is

ˆ ˆ ˆ(0.6i 0.4j 0.9k)Nm (B) The potential energy of the dipole is –0.6 J(C The potential energy of the dipole is 0.6 J(D) If the dipole is rotated in the electric field, the maximumpotential energy of the dipole is 1.3 J

Q.4 Mark the correct options –(A) Gauss’s law is valid only for uniform charge distribu-tions(B) Gauss’s law is valid only for charges placed in vacuum(C) The electric field calculated by Gauss’s law is the fielddue to all the charges.(D) The flux of electric field through a closed surface dueto all the charges is equal to the flux fue to the chargesenclosed by the surface.

Q.5 An electric field converges at the origin whose magnitudeis given by the expression E = 100rNt/Coul., where r is thedistance measured from the origin.(A) total charge contained in any spherical volume with itscentre at origin is negative.(B) total charge contained at any spherical volume, irre-spective of the location of its centre, is negative.(C) total charge contained in a spherical volume of radius 3cmwith its centre at the origin has magnitude 3×10–13 C.(D) total charge contained in a spherical volume of radius3 cm with its centre at the origin has magnitude 3 × 10–9 C.

Q.6 Select the correct alternative –(A) The charge gained by the uncharged body from acharged body due to conduction is equal to half of thetotal charge initially present.(B) The magnitude of charge increases with the increase invelocity of charge.(C) Charge cannot exist without matter although mattercan exist without charge(D) Between two non-magnetic substances repulsion isthe true test of electrification (electrification means bodyhas net charge)

Q.7 The electric field intensity at a point in space is equal inmagnitude to –(A) Magnitude of the potential gradient there

(B) The electric charge there(C) The magnitude of the electric force, a unit charge wouldexperience there(D) The force, an electron would experience there

Q.8 Figure shows a charge qplaced at the centre of a hemi-sphere. A second charge Q isplaced at one of the positionsA, B, C and D.In which position(s) of this C Aq

B D

second charge, the flux of the electric field through thehemisphere remains unchanged ?(A) A (B) B(C) C (D) D

Q.9 An electric dipole is placed at the centre of a sphere, markthe correct options –(A) The flux of the electric field through the surface is zero(B) The electric field is zero at every point of the sphere(C) The electric field is not zero anywhere on the sphere(D) The electric field is zero an a circle on the sphere.

Q.10 A large nonconducting sheet M is given a uniform chargedensity. Two uncharged small metal rods A and B areplaced near the sheet as shown in figure.

BA

(A) M attracts A (B) M attracts B(C) A attracts B (D) B attracts A

Q.11 A uniform electric field of strength Ej exists in a region.An electron (charge –e, mass m) enters a point A with

velocity Vj . It moves through the electric field and exits atpoint B. Then –

y

x

V

V

B(2a, d)

A(a, 0)(0, 0)

(A) 2

22amv ˆE j

ed

(B) Rate of work done by the electric field at B is 2 3

34ma v

d(C) Rate of work by the electric field at A is zero

(D) Velocity at B is 2av ˆ ˆi vjd

Q.12 An oil drop has a charge –9.6 × 10–19 C and has a mass1.6 × 10–15 gm. When allowed to fall, due to air resistance

45GyaanSankalp

Electric charges and fieldsforce it attains a constant velocity. Then if a uniform elec-tric field is to be applied vertically to make the oil ascendup with the same constant speed, which of the followingare correct. (g = 10 m/s) (Assume that the magnitude ofresistance force is same in both the cases)(A) the electric field is directed upward(B) the electric field is directed downward

(C) the intensity of electric field is 2 11 10 NC3

(D) the intensity of electric field is 5 11 10 NC6

Q.13 An electric dipole is kept in the electric field produced bya point charge –

(A) dipole will experience a force(B) dipole can experience a torque(C) dipole can be in stable equilibrium(D) it is possible to find a path (not closed) in the field onwhich work required to move the dipole is zero.

Q.14 Point charges are located onthe corner of a square asshown.Find the components of electric

–1µC

–1µC

+1µC

+1µC

y

xfield at any point on the z-axiswhich is axis of symmetry ofthe square –(A) Ez = 0 (B) Ex = 0(C) Ey = 0 (D) None of these

EXERCISE - 4ASSERTION AND REASON QUESTIONS

Note : Each question contains STATEMENT-1 (Assertion)and STATEMENT-2 (Reason). Each question has 5 choices(A), (B), (C), (D) and (E) out of which ONLY ONE is cor-rect.(A) Statement-1 is True, Statement-2 is True; Statement-2is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2is NOT a correct explanation for Statement-1.(C) Statement -1 is True, Statement-2 is False.(D) Statement -1 is False, Statement-2 is True.(E) Statement -1 is False, Statement-2 is False.

Q.1 Statement 1 : A positive point charge is brought in anelectric field, the electric field at nearby point will increase.Statement 2 : Electric field produce by charge may favourthe existing electric field.

Q.2 Statement 1 : Flux through a closed surface is zero.Statement 2 : Total charge inside the surface must be zero.

Q.3 Statement 1 : Consider twoidentical charges placed dis-tance 2d apart, along x-axis.The equilibrium of a positive test charge placed at the point O midway between them isstable for displacements along the x-axis.Statement 2: Force on test charge is zero.

Q.4 Statement 1 : A parallel beam of electrons is shot into auniform strong electric field initially parallel to and thenagainst the field with a small initial speed. Then the beamtends to spread out at the beginning and narrows down later.Statement 2: The total energy of the beam is conserved.

Q.5 Statement 1 : An ellipsoidal cavity is carved within a per-fect conductor. A positive charge q is placed at the centreof the cavity. The points A and B are on the cavity surface.Potential at A = potential at B.Statement 2 : Surface of charge conductor is always equi-potential.

Q.6 Statement 1 : A deuteron and an -particle are placed in anelectric field. If F1 and F2 be the forces acting on them anda1 and a2 be their accelerations respectively then, a1 = a2.Statement 2 : Forces will be same in electric field.

Q.7 Statement 1 : Charges Q1 and Q2 are placed inside andoutside respectively of an uncharged conducting shell,their separation is r. Then the force on Q1 is zero.Statement 2 : Lines of force cannot enter conducting shell.

Q.8 Statement 1 : When a charged comb is brought near asmall piece of paper, it attracts the piece.Statement 2 : Because the paper becomes charged.

Q.9 Statement 1 : A point charge is placed in a cavity in a metalblock. If another charge is brought outside the metal, thecharge in the cavity does not feel any electric forceStatement 2 : There is no electric field line in the cavity ofa metal block.

Q.10 Statement 1 : A small metal ball is suspended in a uniformelectric field with an insulated thread. If high energy X-raybeam falls on the ball, the ball will be deflected in the elec-tric field.Statement 2 : X-rays beam falls on the ball, the ball will bedeflected in the magnetic field.

Q.11 Statement 1 : The tyres of aircrafts are slightly conducting.Statement 2 : If a conductor is connected to ground, theextra charge induced on conductor will flow to ground.

Q.12 Statement 1 : Four pointcharges q1, q2, q3 and q4 areas shown in figure. The fluxover the shown Gaussian sur-face depends only on chargesq1 and q2.Statement 2 : Electric field at all points on Gaussian surface depends only on chargesq1 and q2.

Q.13 Statement 1 : An insulator does not conduct electricityusually.Statement 2 : The number of electrons in an insulator isvery small in comparison to that in a conductor.

Q.14 Statement 1 : The positive charge particle is placed infront of a spherical uncharged conductor. The number oflines of forces terminating on the sphere will be more thanthose emerging from it.Statement 2 : The surface charge density at a point on thesphere nearest to the point charge will be negative and maxi-mum in magnitude compared to other points on the sphere.

46 GyaanSankalp

Electric charges and fieldsEXERCISE - 5

MATCH THE COLUMN TYPE QUESTIONSEach question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column Ihave to be matched with statements (p, q, r, s) in column II.

Q.1 Match the column :Column I Column IIArrangement Flux through the entire surface of the body.

(A) (Cube) (p) 0

q

(B) (hemisphere) (q) 0

q8

(C) (r)0

q4

(D) (sphere) (s) 0

3q

Q.2 An electric dipole is placed in an electric field. The column I gives the description of electric field and the angle between thedipole moment p and the electric field intensity E

and the column II gives the effect of the electric field on the dipole. Match

the description in Column I with the statements in column II.Column I Column II

(A) Uniform electric field, = 0 (p) force = 0(B) Electric field due to a point charge, = 0 (q) Torque = 0

(C) Electric field between the two oppositely (r) p.E 0

charged large plates, = 90°(D) Dipole moment parallel to uniformly (s) Force 0

charged long wire.Q.3 Column II describe graph for charge distribution given in column I match the description.

Column I Column II

(A) Uniformly charged ring (p) r

E (Electricfield intensities)

47GyaanSankalp

Electric charges and fields

(B) Infinitely large charge conducting sheet (q) r

V (Electricpotential)

(C) Infinite non conducting thin sheet. (r)

E

r

(D) Hollow non conducting sphere. (s)

V

r

Q.4 Two points, like charges QA and QB are positioned at points A and B. Theelectric field strength to the right of charge QB on the line that passesthrough the two charges varies according to a law that is representedshematically in the figure accompanying the problem without employing adefinite scale. Assume electric field to be positive if its direction coincideswith the positive direction on the x-axis. Distance between the charges is .

Column I Column II (A) Charge QA (p) –ve(B) Charge QB (q) +ve

(C) |QA/QB| (r)2

1

1

xx

(D) x2 (s) 1/ 3A B(Q / Q ) 1

Q.5 Match the columnColumn I Column II

(A) Like charges repels and unlike attracts. (p) Dr. William Gilbert(B) Numerical value of force between two charges (q) Thomos Brown(C) Methods of charging (r) Coulomb by (torsion balance)(D) Amount of induced charge (s) Faraday ice pail exp.

Q.6 Match the columnColumn I Column II

(A) e/m of electron (p) J.J. Thomson(B) Charge and mass (indirectly) of electron (q) R.A. millikan (by oil drop exp.)

and quanta of charge.(C) Concept of line of force (r) M. Faraday(D) Highest common factor method (s) Max Plank

48 GyaanSankalp

Electric charges and fields

PASSAGE BASED QUESTIONSPASSAGE 1 (Q.1-Q.5)

A very large, charged plate floats in deep space. Due tothe charge on the plate, a constant electric field E existseverywhere above the plate. An object with mass m andcharge q is shot upward from the plate with a velocity vand an angle . It follows the path shown reaching a heighth and a range R. Assume the effects of gravity to be neg-ligible.

vE

h

R

Q.1 Which of the following must be true concerning the object(a) q must be positive (b) q must be negative(c) m must be large (d) m must be small

Q.2 Which of the following gives the vertical velocity of theobject in terms of h just before colliding with the plate atthe end of its flight –

(a) 2gh (b) 2Eqh

(c) 2mhEq (d)

2qhEm

Q.3 Which of the following is true concerning all objects thatfollow the path shown when propelled with a velocity v atan angle –(a) They must have the same mass(b) they must have the same charge(c) they must have the same mass and the same charge(d) their mass to charge ratios must be the same

Q.4 Suppose E is 10 N/C, m is 1 kg, q is –1C, v is 100 m/s and is 30°. What is h –(a) 25m (b) 45m(c) 80 m (d) 125

Q.5 Which of the following is true concerning the flight of theprojectile shown –(a) Increasing the mass m decreases the maximum height h(b) Increasing the charge q increases the maximum height h(c) Increasing the mass m decreases the downward accel-eration(d) Increasing the charge q decreases the downward ac-celeration

PASSAGE 2 (Q.6-Q.8)Related to the following diagram of two charges, +Q and– 4Q.

Q.6 The net electric field is zero near which point?(A) A (B) B(C) C (D) D

Q.7 At which point does the net electric field vector point tothe left?(A) A (B) B(C) C (D) D

Q.8 At which point would a small positive charge q feel thegreatest force?(A) A (B) B(C) C (D) D

PASSAGE 3 (Q.9-Q.11)A thin insulating wire is stretched along the diameter of aninsulated circular loop of radius R. A small bead of mass mand charge –q is threaded on to the wire. Two small identicalcharges are tied to the hoop at points opposite to eachother, so that the diameter passing through them isperpendicular to the thread (see figure). The bead isreleased at a point which is a distance x0 from the centre ofthe loop. Assume that x0 << R.

Q.9 The resultant force acting on the charged bead is :

(A) 0

2 20

xk 2 Q qF .rx R

(B)

02 20

xk 2Q qF .rx R

(C) 0

2 20

xk 2Q qF .rx R

(D)

02 20

xk 2 Q qF .rx R

Q.10 The exact equation of motion of the bead along the threadis :

(A)

2

2 3/22 2

d x k 2Qq xm

dt x R

(B)

2

2 3/22 2

d x k Q q xm

dt x R

(C)

2

2 3/22 2

d x k 2Qq xmdt x R

(D)

2

2 3/22 2

d x k Qq xmdt x R

Q.11 The time when will the velocity of the bead vanish for thefirst time is :

(A) 2 3m R

k 2 Qq

(B) 2 3m Rk Qq

EXERCISE - 6

49GyaanSankalp

Electric charges and fields

(C) 2 3m R

k 4 Qq

(D) 2 32 m R

k Qq

PASSAGE 3 (Q.12-Q.14)In the figure shown mA = mB = 1 kg. Block A is neutral whileqB = –1C. Sizes of A and B are negligible. B is released fromrest at a distance 1.8 m from A. Initially spring is neithercompressed nor elongated.

A B

E = 10 N/CK = 18 N/m

smooth x = 0 x = 1.8m x-axis

Q.12 If collision between A and B is perfectly inelastic, what isvelocity of combined mass just after collision.(A) 6 m/s (B) 3 m/s(C) 9 m/s (D) 12 m/s

Q.13 Equilibrium position of the combined mass is at x = ..... m(A) –2/9 (B) –1/3(C) –5/9 (D) –7/9

Q.14 The amplitude of oscillation of the combined mass will be:

(A) 2

m3

(B) 124

m3

(C) 72

m9

(D) 106

m9

EXERCISE - 7SUBJECTIVE QUESTIONSQ.1 Two identical conducting of spheres (of negligible radius),

having charges of opposite sign, attract each other with aforce of 0.108 N when separated by 0.5 meter. The spheresare connected by a conducting wire, which is then removed(when charge stops flowing) and therefore repel each otherwith a force of 0.036 N keeping the distance same. Whatwere the initial charges on the spheres ?

Q.2 Two particles A and B having charges of +2.00 × 10–6 Cand of – 4.00 × 10–6 C respectively are held fixed at aseparation of 20.0 cm. Locate the point(s) on the line ABwhere the electric field is zero.

Q.3 Find the flux of the elec-tric field through a spheri-cal surface of radius Rdue to a charge of18.85 × 10–8 C at the cen-tre and another equalcharge at a point 2R awayfrom the centre.

2R

R+ +

Q.4 A clock face has negative charges –q, –2q, –3q, ........,–12qfixed at the position of the corresponding numerals on thedial. The clock hands do not disturb the net field due topoint charges. At what time does the hour hand point inthe same direction is electric field at the centre of the dial.

Q.5 The electric field in a region is given by 0E x ˆE i

. Find

the charge contained inside a cubical volume bounded bythe surfaces x = 0, y = 0, y = a, z = 0 and z = a.Take E0 = 5 × 103 N/C, = 2 cm. and a = 1 cm.

Q.6 A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge density,

0rR

, where 0 is a constant and r is the distance

from the centre of the sphere. Show that(a) the total charge on the sphere is Q = 0R3 and(b) the electric field inside the sphere has a magnitude

given by, 2

4KQrE

R

Q.7 An electron beam after being accelerated from rest througha potential difference of 500V in vacuum is allowed to im-pinge normally on a fixed surface. If the incident current is100 µA, determine the force exerted on the surface assum-ing that it brings the electrons to rest.(e = 1.6 × 10–19C, m = 9.0 × 10–31 kg)

Q.8 A charge Q is uniformly distributed over a rod of length .Consider a hypothetical cube of edge with the centre ofthe cube at one end of the rod. Find the minimum possibleflux of the electric field through the entire surface of thecube.

Q.9 A dipole is placed at ori-gin of coordinate systemas shown in figure, findthe electric field at pointP (0, y).

Q.10 An infinitely long string uni-formly charged with a lineardensity 1 and a segment oflength uniformly chargedwith linear density 2 lie in aplane at right angles to eachother and separated by a dis-tance r0.Determine the force with which these two interact.

Q.11 A very long charged thread (lying in the xy plane) which ishaving a linear charge density is having one of its end ata point P as shown in figure. What is electric field intensityat point Q.

50 GyaanSankalp

Electric charges and fields

P

r

45°45°

Q

45°x

y Q.12 Two like charged, infinitely long rod with the same linearcharge density of 3 × 10–8 C/cm are 2 cm apart. Find theelectric force per unit length on each rod due to the otherand the work done against that force per unit length to bedone in bringing them closer by 1 cm.

EXERCISE - 8PREVIOUS YEAR IIT-JEE QUESTIONSQ.1 A metallic solid sphere is placed in a uniform electric field.

The lines of force follow the path (s) shown in figure as –(1996)

(A) 1(B) 2(C) 3(D) 4

A

C

2

D

1

3

4

Q.2 A positively charged thin metal ring of radius R is fixed inthe xy-plane with its centre at the O. A negatively chargedparticle P is released from rest at the point (0, 0, z0), wherez0 > 0. Then the motion of P is – (1998)(A) Periodic for all values of z0 satisfying 00 z

(B) Simple harmonic for all values of satisfying 00 z R (C) Approximately simple harmonic provided z0 << R(D) Such that P crosses O and continues to move alongthe negative z-axis towards z

Q.3 A non-conducting solid sphere of radius R is uniformlycharged. The magnitude of the electric field due to thesphere at a distance r from its centre – (1998)(A) Increases as r increases for r < R(B) Decreases as r increases for 0 < r < (C) Decreases as r increases for R < r < (D) Is discontinuous at r = R

Q.4 The dimension of (1/2) 20 0E ( : permittivity of free space,

E : electric field) is – (2000)(A) MLT–1 (B) ML2T–2

(C) ML–1T–2 (D) ML2T–1

Q.5 Three positive charges of equal value q are placed at thevertices of an equilateral triangle. The resulting lines offorce should be sketched as in – (2001)

(A) (B)

(C) (D)

Q.6 A small ball of mass 2 × 10–3 kg having a charge of 1µC issuspended by a string of length 0.8m. Another identicalball having the same charge is kept at the point of suspen-sion. Determine the minimum horizontal velocity whichshould be imparted to teh lower ball so that it can makecomplete revolution. (2002)

Q.7 A metallic shell has a point charge q kept inside its cavity.Which one of the following diagrams correctly representsthe electric lines of forces – (2003)

(A) (B)

(C) (D)

Q.8 A charge +Q is fixed at the origin of the co-ordinate systemwhile a small electric dipole of dipole-moment p pointingaway from the charge along the x-axis is set free from apoint far away from the origin. (2003)(a) Calculate the K.E. of the dipole when it reaches to apoint (d, 0).(b) Calculate the force on the charge +Q at this moment.

Q.9 Six charges, three positive and three negative of equalmagnitude are to be placed at the vertices of a regularhexagon such that the electric field at O is double the elec-

51GyaanSankalp

Electric charges and fieldstric field when only one positive charge of same magni-tude is placed at R. Which of the following arrangementsof charges is possible for P, Q, R, S, T and U respectively –

(2004)(A) +, –, +, –, –, +(B) +, –, +, –, +, –(C) +, +, –, +, –, –(D) –, +, +, –, +, –

P Q

R

ST

U O

Q.10 Consider the charge configuration and spherical Gaussiansurface as shown in the figure. When calculating the fluxof the electric field over the spherical surface the electricfield will be due to – (2004)(A) q2(B) only the positive charges(C) all the charges(D) +q1 and –q1

+q1

-q1

q2

Q.11 Three infinitely long charge sheets are placed as shown infigure. The electric field at point P is (2005)

-2-

Pk z=a

z=-az=-2a

(A) 0

2 k (B)

0

2 k

(C) 0

4 k (D)

0

4 k

Q.12 Consider a neutral conducting sphere. A positive pointcharge is placed outside the sphere. The net charge on thesphere is then – (2007)(A) negative and distributed uniformly over the surface ofthe sphere(B) negative and appears only at the point on the sphereclosest to the point charge(C) negative and distributed non-uniformly over the entiresurface of the sphere(D) zero

Q.13 A spherical portion has been removed from a solid spherehaving a charge distributed uniformly in its volume asshown in the figure. The electric field inside the emptiedspace is – [2007](A) zero everywhere(B) non-zero and uniform(C) non-uniform(D) zero only at its center

Q.14 Positive and negative point charge of equal magnitude arekept at (0, 0, a/2) and (0, 0, –a/2) respectively. The workdone by the electric field when another positive pointcharge is moved from (–a, 0, 0) to (0, a, 0) is – [2007](A) positive (B) negative (C) zero

(D) depends on the path connecting the initial and finalpositions

Q.15 Consider a system of threecharges q/3,q/3 and 2q/3placed at points A, B andC, respectively, as shownin the figure.Take O to be the centre ofthe circle of radius R andangle CAB = 60°

[2008]

x

y

O

B

C

A

60°

(A) The electric field at point O is 20

q8 R directed along

the negative x-axis(B) The potential energy of the system is zero(C) The magnitude of the force between the charges at C

and B is 2

20

q54 R

(D) The potential at point O is 0

q12 R

Paragraph for Question Nos. 16 to 18 [2008]The nuclear charge (Ze) is non-uniformly distributed withina nucleus of radius R. The charge density (r) [charge perunit volume] is dependent only on the radial distance rfrom the centre of the nucleus as shown in figure Theelectric field is only along the radial direction.

d

O Rar

Q.16 The electric field at r = R is(A) independent of a(B) directly proportional to a(C) directly proportional to a2

(D) inversely proportional to aQ.17 For a = 0, the value of d (maximum value of ñ as shown in

the figure) is –

(A) 33Ze

4 R(B) 3

3ZeR

(C) 34Ze

3 R(D) 3

Ze3 R

Q.18 The electric field within the nucleus is generally observedto be linearly dependent on r. This implies.(A) a = 0 (B) a = R/2(C) a = R (D) a = 2R/3

52 GyaanSankalp

Electric charges and fields

HINTS & SOLUTIONSEXERCISE - 1

(1) This can be done if the end of the rod held in the hand iscoated with an insulating material.

(2) The sphere should be brought in contact with two similarneutral spheres.

(3) The static charge that may be formed as the car moves isconducted to the earth through the chain.

(4) Yes, it can.(5) The surface charge density will increase.(6) At first the ball will touch the electrically charged rod and

then will be repelled from it.(7) No, they do not.(8) This is done for electrostatic shielding.(9) (a) Yes, they can (b) No, the cannot.(10) Due to polarization of the insulator rod AB, the point charge

+q1 will be acted upon, in addition to the point charge–q2, by the polarization charges formed at the ends of therod.

The attractive force exerted by the negative charge inducedat the end A will be stronger than the repulsive force exertedby the positive charge induced at the end B. Thus, thetotal force acting on the charge +q1 will increase.

(11) Suppose that there is a field inside the sphere. It is obviousfrom considerations of symmetry that in this caseequipotential surfaces must be spherical surfacesconcentric with the charged sphere. Accordingly, the linesof force coincide with the charged sphere. Accordingly,the lines of force coincide with the radii, i.e. they musteither begin or end at the centre of teh sphere. This wouldhave been possible, if there were a positive or a negativecharge at the centre of the sphere. But since there is nocharge inside the sphere, the lines of force cannot begin orend there. Consequently there is no field inside the sphere.

(12) It can, if we use the phenomenon of electrostatic induction.Bring a conductor on an insulated support up to thecharged body and connect the conductor to the earth fora short time. The conductor will retain a charge opposite insign to the given one, while the like charge will pass intothe earth.The charge can be removed from the conductor byintroducing the latter into a metallic space. Teh operationmay be repeated many times with a charge of anymagnitude. Electrostatic machines operate on a similarprinciple.

(13) They can, if the charge of one ball is much greater thanthat of the other. The forces of attraction caused by theinduced charges may exceed the forces of repulsion.

Q 1 2 3 4 5 6 7 8 9 10 11A C D C D C A D D B B AQ 12 13 14 15 16 17 18 19 20 21 22A B B B B B C D D B C CQ 23 24 25 26 27 28 29 30 31 32 33A B C D A B B B A D C BQ 34 35 36 37 38 39 40 41 42 43 44A B A B B B B C B B A BQ 45 46 47 48 49 50 51 52A C A C A D C B D

EX ERCIS E - 2

(2) (D). 20

1 q (Q q)F4 r

for F to be maximum, dF

0dq

20

1 1F [Q q q( 1)] 04 r

Q – 2q = 0 q 1Q 2

(4) (D). 2 2R 8 8 2.8.8cos120 8N 8N

8N

R

60°

60°

120°

53GyaanSankalp

Electric charges and fields

(5) (C). 20

1 (Q / 2) (Q / 2). (100 N / m) (0.1m)4 (0.3m )

22

94 100 (0.1) (0.3)Q

9 10

Q = 20 × 10–6 C = 20 µC(6) (A). We have centripetal force equation

22k mvqr r

so2kqv

m

Now,2 r mT 2 rv 2kq

where0

1k4

(7) (D). The electric field due to the plate is uniform, exertingno net force on the dipole. The electric field due to thepoint charge Q is non-uniform, exerting net force along thepositive y-axis on the dipole, causing it to move in thatdirection.

(9) (B). 30

1 2pE4 r

; 3

1E

r 3

1F

r

Hence, the force will become F/8.(12) (B). U1 will be positive and greatest since all forces among

dipoles are repulsive, U2 is negative as potential energy offirst and second dipole pair cancels out potential energyof second and third pair, leaving only potential energy ofinteraction of first and third, that is negative. In (C), effectof attraction is greatest.

(15) (B). Electric field ateach point on the sur-face of ring due to di-pole is

3kp

ER

in direction

E

R

p

opposite to the dipole moment.

Hence net force on ring is3

kpQF QE

R

(16) (B)The diagrammaticrepresentation of thegiven problem isshown in fig.The electrical field E

at all points on the X-axis will not have thesame direction.The electrical field E

E

-q+q(-d, 0) O (d, 0)

X

iEE

E

at all points on the Y-axis will be parallel to the X-axis

(i.e. i direction).The electric potential at the origin due to both the chargeis zero, hence, no work is done in bringing a test chargefrom infinity to the origin.Dipole moment is directed from the –q charge to the +qcharge (i.e. –x direction).

(17) (C). F=qE = q (A – Bx) ma = q (A – Bx)

qam

(A – Bx) ...(1)

vdv q(A Bx)

dx m ; dxBxA

mqvdv

0 x

0 0

qvdv A Bx dxm

; 02

BxAx2

x = 0, x = BA2

...(2)

From eq. (1) and (2)

BA2BA

mqBxA

mq

mqAA2A

mq

.

(19) (D). Electric field vector due to a point charge is neversame at two points.(A) is falseElectric field increases as one goes away from centre ofsolid uniformly charged sphere.This is true till one reaches the surface.As one moves away from surface (away from centre aswell) the magnitude of electric field increases.(B) is false

(20) (B). Force on a charge particle in a uniform electric fieldF = q EThe acceleration imparted to the particle is

a = qE/mThe distance traveled by the particle in time t is

22 tmqE

21at

21d

For the given problem2 2p e

p e

t tm m

;2p p2

ee

t mmt

p p

ee

t mmt

(24) (C). From (i) A and C both are charged, either positively ornegatively.From (ii) Both D and E has no charge and from (iii), A ispositively charged.Therefore from (i), B is negatively charged.

(26) (A). Eˆ ˆ ˆE.A (5i 2 j).2i 10

54 GyaanSankalp

Electric charges and fields(27) (B). The given point

is at axis of P2

dipole

and at equatorial lineof P dipole so that

field at given point.

x

z

(1,0,0)(1,0,2)

P

P/2

(28) (B). There exists a point P on the x-axis (other than theorigin), where net electric field is zero. Once the charge Qreaches point P, attractive forces of the two -ve charge willdominate and automatically cause the charge Q to crossthe origin.Now if Q is projected with just enough velocity to reach P,its K.E. at P is zero, but while being attracted towards ori-gin it acquires KE and hence its net energy at the origin ispositive. (P.E. at origin = zero)

(31) (D). 2

1 20

1 qF4 r

,

2

2 20

1 qF4 K R

As F1 = F2

Hence, 2 2

2 2q 1 q

Kr R

R r / K

(32) (C). E.ds

since r << R so we can consider electric field is constantthrough out the surface of smaller ring, hence

2 2 3 / 2xE

(R x )

(35) (A). Let us imagine another identical hemispherical surfaceover given one.Both being symmetric with respect to Q, hence flux will besame through both the hemisphere (1 = 2).

1 20

Q

1 20

Q2

Q 1 2 3 4 5 6 7 8 9 10 11A ACD ABC ABD CD ABC CD BC AD AC CD ABCDQ 12 13 14A BC BCD ABC

EXERCISE - 3

(1) (ACD). – Q charge must beinduced on the inner surfaceof the sheet. Hence +2Qcharge appears on its outersurface. Hence charge den-sity on the outer surface

= 2 22Q Q

4 R 2 R

RC

+Q

+2Q

–Q

It is evident that at any point outside S is double the fieldat a point inside S the field is inversely proportional to thesquare of its distance from C as the charge Q is placed atthe centre.

(14) (ABC). Diagonally opposite charges will produce field inz-axis, but fields due to +ve & –ve charges will cancel.

Q 1 2 3 4 5 6 7 8 9 10 11A D A B A A C A C A C BQ 12 13 14A C C D

EXERCISE - 4

(1) (D). A positive charge broughtin an electric field may in-crease or decrease the elec-tric field.At point A electric field increase at B decrease.

(2) (A). Apply Gauss law.(3) (B). If +ve charge is displaced along x-axis, then net force

will always act in a direction opposite to that of displace-ment and the test charge will always come back to its origi-nal position.

(4) (A). In conservative electric field total energy remain con-stant.

(5) (A). Electric field inside the conductor is equal to zero,therefore, potential difference between two position onthe conductor is equal to zero. It means, the potential issame at every point of the conductor.

(6) (C). qd = e, md = 2mp = 2mq = 2e, m = 4mp = 4mF1 = F = eE, F2 = F = 2eE F1

Further, 11

F eEa2m 2m

and 22 1

F 2eE eEa a2m 4m 2m

55GyaanSankalp

Electric charges and fields(7) (A). Lines of force from outside charge Q2 cannot pen-

etrate into conducting shell, hence, force on Q1 = 0.(8) (C). There is redistribution of charge but the net charge

remains zero.(10) (C). When high energy X-ray beam falls on the ball, the

metal will emit photoelectrons, thus leaving the positivecharge on the ball. As a result of this, ball is deflected inthe direction of electric field.

(11) (B). During take off and landing, the friction between tyresand the run way may cause electrification of tyres. Due to

conducting nature of tyre, the charge so collected to aground and electrical sparking is avoided

(12) (C). Electric field at any point depends on presence of allcharges.

(13) (C). The number of electrons in insulator is of the sameorder as that in a conductor. But the number of free elec-trons is very small in an insulator. This is basic differencebetween a conductor and an insulator.

(14) (D). No. of lines entering the surface = No. of lines leavingthe surface.

(1) (A) q, (B) p, (C) p, (D) s(2) (A) p, q , (B) q, s (C) p, r (D) p, r(3) (A) p, q, (B) r, (C) r, (D) s

EXERCISE - 5(4) (A) q , (B) p, (C) r, (D) s(6) (A) p, (B) q (C) r, (D) s(5) (A) p, (B) r (C) q, (D) s

Q 1 2 3 4 5 6 7 8 9 10 11A B D D D C A D C A A AQ 12 13 14A B C D

EXERCISE - 6

(1) ± 1.0 × 10–6 C , 3 × 10–6 C

(2)20

48.3 cm.2 1

from A llong BA

(3)2

4 N m10

C

(4) 9.39

(5) 2.2 × 10–12 C (6) 2k

0, ,0r

EXERCISE - 7

(7) 7.5 × 10–9 N (8) 0

Q2

(9) 3kp ˆ ˆ( i 2j)2y

(10) 1 2

0 0F ln 1

2 r

(11)2k 2k ˆ| E | , E ( j)r r

(12)2

22kB 8.1 N / m, 2k , ln 2 0.1129 J / m

r

(1) (D) (2) (AC) (3) (AC)(4) (C) (5) (C) (6) 5.86 m/s(7) (C)

(8) 20

P QK.E.4 d

(b) 3

0

QP2 d along positive x-axis

(9) (D) (10) (C) (11) (B)(12) (ABCD) (13) (B) (14) (C)(15) (C). Net electric field to both charges q/3, will get cancelled.

Electric field due to (–2q/3) will be directed in –ve axis.

2 20

k (2q / 3) qE ER 6 R

P.E. of system = 2

q 2q q 2qK KK (q / 3) 3 3 3 3

2R 2R sin 60 2R cos60

EXERCISE - 8

(16) (A). 2KQ

ER

(Q = Total charge within the nucleus = Ze)

So, 2KZe

ER

So electric field is independent of a.

(17) (B). R

2

0

dq (R x) 4 x dx ZeR

33Ze

dR

(18) (C). If within a sphere is constant E r