Elesolution manual multivariable calculus fourth editionctrostatics Field Theory

Outline Introduction Coulombs Law Electric field Intensity Electric flux Electric Potential Materials Energy Capacitors

Electrostatics

Steven Armoo

RUCST

April 13, 2015

Steven Armoo (RUCST) Electrostatics April 13, 2015 1 / 32


Outline

1 Introduction

2 Coulombs Law

3 Electric field Intensity

4 Electric flux

5 Electric Potential

6 Materials

7 Energy

8 Capacitors



Introduction

1 Static electric fields(electrostatics) - due to charges at rest

2 A charge can be either be concentrated at a point or distributed - inany case the charge is assumed to be constant in time

3 Coulombs law of electrostatic force between two point charges fixedin space

4 Electric field intensity as the force per unit charge5 We would also establish that

I the electric field intensity is irrotational or conservativeI the work done in moving a charge from one point to another in an

electrostatic field is independent of the path taken

6 Some aspects discussed here may be a repetition



Coulombs Law

1 Coulombs law pertaining to the electric force that one chargedparticle exerts on the other

I directly proportional to the product of the chargesI inversely proportional to the square of the distance between themI directed along the line between them andI attractive(repulsive) for unlike(like) charges

2 If q1 and q2 are two charged particles situated at points P(x , y , z)and Q(x ′, y ′, z ′), electric force acting on q1 due to q2 is

F = Kq1q2

R2(1)

3 In SI Unit, the constant of proportionality is

K =1

4πε0(2)



Coulombs Law

1 where ε0 = 8.85× 10−12 ≈ 10−9

36π farad/meter (F/m) is the permitivityof free space(vacuum)

2

~F12 =1

4πε0

q1q2

R212

~a12 (3)

3 ~a12 is the unit vector from 2 to 1

4 expand ~a12 ***5 Equation is not only valid for charged particles but also for charged

bodies can be considered point chargesI As long as their sizes are much less than the distance between them

6 ~F12 = −~F21



Coulombs Law

1 Coulombs law obeys the principle of superpositionI The total force ~Ft acting on a point charge q due to a system of n

point charges is the vector sum of the forces exerted individually byeach charge on q.

~Ft =n∑

i=1

qqi (~r − ~ri )

4πε0|~r − ~ri |3(4)

2 Example 1: Two point charges of 0.7mC and 4.9µC are situated infree space at (2,3,6) and (0,0,0). Calculate the force acting on the0.7mC charge. [0.18~ax + 0.27~ay + 0.54~azN]

3 Example 2: Three equal charges of 200nC are placed in free space at(0,0,0), (2,0,0) and (0,2,0). Determine the total force acting on acharge of 500nC at (2,2,0). [304.6(~ax + ~ay )µN]



Electric field Intensity

1 A charge will always exert force on another charge - action at adistance

2 If one charge is moved toward the other the force must changeinstantaneously

3 By the theory of relativity, increase in the force is not instantaneous

4 Energy and momentum associated with the system of charges istemporarily out of balance

5 Useful to define the force acting on a charge in the presence ofanother charge in terms of a field.

6 There exists an Electric Field or Electric Field Intensity everywhere inspace surrounding the charge

7 When another charge is brought into the field, it experiences a forceacting on it - action by contact




1 To detect the electric field intensity at a point P, place a unit positivecharge qt at that point and measure the force on it

Theorem

The Electric Field Intensity is then defined as the force per unit charge

2 qt creates its own electric fieldI distorts the initial electric fieldI its value must be decreased, qt → 0I generate exponential curve of |Ft | and qt

Theorem

The Electric Field Intensity, ~E , the force per unit charge exerted on a testcharge, qt as the magnitude of qt → 0, is

~E = limqt→0

~F

qt(5)




1 Electric field intensity, a vector field, has the units of N/C, which isdimensionaly equal to V/m

2 If ~E is the electric field intensity at a point, P in space, the forceacting on a charge q is

~Fq = q ~E (6)

~E =q

4πε0

~r1 − ~r2|~r1 − ~r2|3

=q

4πε0R2~aR (7)

3 The Electric field intensity due to n point charges is

~E =n∑

i=1

qi(~r − ~ri )

4πε0|~r − ~ri |3(8)




1 Example 3: Two point charges of 20nC and -20nC are situated at(1,0,0) and (0,1,0) in free space. Determine the electric field intensityat (0,0,1). [63.67(−~ax + ~ay )V /m]

2 Thus far we have assumed that the charge is concentrated at a point.

3 Cases of complexity involve the continuous distribution of charges onlinear elements, on surfaces and in volumes.

Line Charge Density

Line Charge Density, the charge per unit length is defined as

ρl = lim∆l→0

∆q

∆l(9)

where ∆q is the charge on a linear element ∆l




Surface Charge Density

Surface Charge Density, the charge per unit area is defined as

ρs = lim∆s→0

∆q

∆s(10)

where ∆q is the charge on a surface element ∆s

Volume Charge Density

Volume Charge Density, the charge per unit volume is defined as

ρv = lim∆q→0

∆q

∆v(11)

where ∆q is the charge contained in a volume element ∆v



Assignment

A semi-infinite line extending from -∞ to 0 along the z axis carries auniform charge distribution of 100nC/m. Find the electric field intensity atpoint P(0,0,2). If a charge of 1µC is placed at P, calculate the force actingon it.



Electric Flux

1 Place a point charge in an electric field and allow it to move

2 The force acting on it will move it along a certain path3 This path is called line of force or a flux line

I we can create many lines by repeating the process

4 In order not to crowd a region with infinite lines of forceI arbitrarily state that the number of lines due to a charge is equal to the

magnitude of the charge in coulombs

5 The field lines are said to represent the flux

6 For an isolated positive point charge, the electric flux points radiallyoutwards

7 The electric field intensity at any point is tangential to the lines ofelectric flux



Electric Flux

1 Early investigators established the following properties for electricflux:

I It must be independent of the mediumI Its magnitude solely depends on the charge from which it originatesI If a point charge is enclosed in an imaginary sphere of radius, R, the

electric flux must pass perpendicularly and uniformly through thesurface of the sphere.

I The electric flux density, flux per unit area, is then inverselyproportional to R2

2 The elctric flux density, ~D can be defined interms of electric fieldintensity ~E as

~D = ε0~E (12)

Substituting for ~E due to a point charge, the elctric flux density, withunits [C/m2] at a radius, r is

~D =q

4πε0r2~ar (13)



Definition of Electric Flux

1 We can now define electric flux, Ψ interms of elctric flux density, ~D as

Ψ =

∫s

~D. ~ds (14)

where ~ds is the surface differential element on surface, s



Gauss’s Law

Gauss’s law states that the net outward flux passing through a closedsurface is equal to the toal charge enclosed by that surface. That is∮

s

~D. ~ds = Q (15)

Gauss’s law can also be expressed in terms of electric field intensity in freespace as ∮

s

~E . ~ds =Q

ε0(16)

If the charges are distributed in a volume bounded by a surface, thenequation (16) can be written as∮

s

~D. ~ds =

∫vρv dv (17)

Equation (17) is the integral form of Gauss’s lawSteven Armoo (RUCST) Electrostatics April 13, 2015 16 / 32


Gauss Law

1 Gauss law can be used to determine the total charge enclosed if theelectric field intensity or the electric flux density is known at all pointson the surface.

2 By Applying the divergence theorem, equation (17) can be written as∫v5. ~D dv =

∫vρv dv (18)

3 At any point in space5 . ~D = ρv (19)

Equation (19) is called the point or differential of Gauss’s law



Electric Potential

1 If we place a positive test charge, q in an electric field, ~E , there willbe a force on the charge given by ~F = q ~E

2 The charge moves a differential distance, ~dl , The amount of workdone by the ~E field, is given by

dWe = ~F .~dl = q ~E .~dl

The subscript e signifies that the work is done by the ~E field.3 If the test charge is moved against the direction of the ~E by an

external force, ~Fext , the differential work done by the external force is

dW = −~Fext .~dl

4 Assume that the external force just balances the electric force toavoid any kinetic energy considerations,

dW = −q ~E .~dl



Electric Potential

1 The total work done by the external force in moving the test chargefrom point b to a is:

Wab = −q∫ a

b

~E . ~dl (20)

2 If we move the test charge around a closed path, the work done mustbe zero ∮

c

~E . ~dl = 0 (21)

3 Equation (21), states that, the ~E field under static conditions isirrotational or conservative. However, a field is conservative if its curlis zero, thus

5x ~E = 0 (22)



Electric Potential

1 If the curl of a vector field is zero, the vector field can be representedin terms of the gradient of a scalar field as

~E = −5 V (23)

2 Equation (20) can be expressed as

Wab = −q∫ a

b

~E . ~dl = q

∫ a

b5V .~dl

Substituting 5V .~dl = dV , we have

Wab = −q∫ a

b

~E . ~dl = q

∫ Va

Vb

dV = q[Va − Vb] = qVab (24)

3 Va and Vb are the values of the scalar field, V at points a and b.They are the electric potentials at points a and b.



Electric Potential

1 Vab = Va − Vb, defines the potential of point a wrt point b ⇒potential difference between two points

2 If the work done is positive then the potential at point a is higherthan that at point b.

3 When the external force is pushing the positive test charge q againstthe ~E , the potential energy of the charge is increasing.

4 The work done in moving the test charge against the ~E field is equalto the increase in the potential energy of the charge.

5 The potential difference is the change in potential energy per unitcharge in the limit q → 0

Vab = limq→0

Wab

q= −

∫ a

b

~E .~dl (25)

6 Units of joules per coulomb, [J/C ] or volts, [V ]



Example

Determine the potential difference between two points due to a pointcharge, q at the origin



Conductors

1 A conductor is a material, such as a metal, that posseses a relativelylarge number of free electrons.

2 An electron is said to be free if itI is loosely associated with its nucleusI is free to wander through the conductorI responds to almost an infinitesmal electric fieldI contimues to move as long as it experiences a force

3 As the number of valence electrons increases the, the conductivity ofthe material decreases

4 Under steady-state(equilibrium) conditions, the net volume chargedensity within the conductor is zero.

ρv = 0 (26)

5 The net electric field inside a conductor under steady-state is zero

~E = 0 (27)



Dielectrics

1 An ideal dielectric(insulator) is a material with no free electrons in itslattice structure

2 An ideal dielectric is a material in which positive and negative chargesare so sternly bound that they are inseparable. It has zeroconductivity.

3 Under the influence of an electric field, the molecules of a dielectricmaterial experience distortion ⇒ polarized.

4 We can generalize the definition of electric flux density for anymedium as

~D = ε0~E + ~P (28)

where ~P is the polarization vector to include the effect of polarizationin a dielectric material.



Dielectrics

~P = ε0χ~E (29)

Equation (28) can now be expressed as

~D = ε0(1 + χ)~E (30)

The quantity (1 + χ) is called relative permittivity or dielectric constant ⇒εr . Electric flux density finally becomes

~D = ε0εr ~E = ε~E (31)

The maximum ~E field that a dielectric can withstand prior to breakdown iscalled the dielectric strength of the material



Semiconductors

1 In materials like Si and Ge, a small fraction of the total number ofvalence electrons are free to move about randomly within the spacelattice.

2 Free electrons impart some conductivity to the material.

3 This type of material, semiconductor is a poor conductor

4 Under steady-state conditions, the net electric field inside an isolatedsemi conductor will be zero



Energy in an Electric field

1 Let us consider a region devoid of electric field⇒ charges, if any, mustbe located at infinity

2 Let us assume n point charges at infinite distance from the regionunder consideration

3 The energy required to bring a charge, q1 from infinity and place it apoint a is W1 = 0. Why ?

4 Let us bring another charge, q2 from infinity to a point b, the energyrequired to do so is

W2 = q2Vb,a =q1q2

4πεR(32)

where Vb,a is the potential established at point b due to the charge atpoint a and R is the distance between the two charges.

TotalEnergy ,W = W1 + W2 =q1q2

4πεR(33)




1 Let us now extend to a system of three charges, q1, q2 and q3,brought from infinity to the points a, b and c.

2 The enrgy required to do this would be

W = W1 + W2 + W3 = 0 + q2Vb,a + q3(Vc,a + Vc,b)

=1

4πε

[q2q1

R2,1+

q3q1

R3,1+

q3q2

R3,2

]3 Assume the charges were brought in the reverse order, then total

energy expended would have been

W = W3 + W2 + W1 = 0 + q2Vb,c + q1(Va,c + Va,b)

=1

4πε

[q2q3

R2,3+

q1q3

R1,3+

q1q2

R1,2

]same as previous equation




1 Adding the two previous equations, we obtain

W =1

2[q1(Va,c + Va,b) + q2(Vb,a + Vb,c) + q3(Vc,a + Vc,b)] (34)

Because Va,c + Va,b is the total potential at point a due to charges atpoints b and c , we can write

V1 = Va,c + Va,b =1

4πε

[q3

R1,3+

q2

R1,2

](35)

Similarly, the potentials at b and c are V2 = Vb,a + Vb,c andV3 = Vc,a + Vc,b

2 The total energy can now be written as

W =1

2[q1V1 + q2V2 + q3V3] =

1

2

3∑i=1

qiVi (36)



Capacitors

1 Two insulated conductors of any arbitrary shape adjacent to eachother form a capacitor

2 By applying an external charge, charges are transfered from oneconductor to the other. All times, the conductors will have equal butopposite charges.

3 This separation of charges establishes and ~E field in the dielectricmedium ⇒ potential difference between the two conductors. Thepotential difference between the conductors is proportional to thecharge transfered.

4 Capacitance is defined as the ratio of the charge on one conductorto its potential wrt the other.

C =Qa

Vab(37)



Equations

1 The potential of plate a wrt plate b is

Vab =Qd

εA(38)

2 Thus the capacitance of the parallel plate capacitor is

C =Q

Vab=εA

d(39)

3 The energy stored in the system is

W =1

2CQ2 =

1

2CV 2

ab (40)



Question

A spherical capacitor is formed by two concentric metallic spheres of radiia and b. The charge of the inner sphere is +Q and that on the outersphere is −Q. Determine the capacitance of the system. What is thecapacitance of an isolated sphere? Assuming the earth to be an isolatedsphere of radius 6.5X106 meters, calculate its capacitance. Deduce anapproximate expression for the capacitance when the separation betweenthe spheres is very small as compared to their radii.


Elesolution manual multivariable calculus fourth editionctrostatics Field Theory

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