Problems and Solutions on Electricity and Magnetism

8/22/2019 Problems and Solutions on Electricity and Magnetism

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CHAPTER 1

ELECTROSTATICS

1-1 COULOMBS LAWPROBLEM 1-1

A point charge C61012.3 + is cm3.12 distant from a second

charge of C61048.1 . Calculate the magnitude of the force on

each charge.

SOLUTION

The magnitude of the force between two point charges is given by

221

041

rqqF

=

22

669

)103.12(

)1048.1)(1012.3)(10998.8(

=

NF 743.2=PROBLEM 1-2

What Coulombs force exists between two protons inside a nucleus

which are separated by a distance of 15104

metres from eachother? B.U. B.Sc. 1993S

SOLUTION

According to Coulombs law

2

2

0

2

21

0 4

1

4

1

r

e

r

qqF

==

215

2199

)104(

)10602.1)(10988.8(

=

NF 418.14=

PROBLEM 1-3

The electron and proton in a hydrogen atom are pm9.52 apart.

What is the magnitude of the electric force between them?

SOLUTION


2

2

0

221

0 41

41

re

rqqF

==

212

2199

)109.52(

)10602.1)(10988.8(

=

NF 810243.8 =


2/15

CHAPTER 01 ELECTROSTATICS 2

PROBLEM 1-4

In an ionized helium atom, the electron and nucleus are separated

by a distance of pm5.25 . What electric force is being experienced

by the electron due to nucleus? F.P.S.C. 2005

SOLUTION

The magnitude of the force experienced by the electron due to helium

nucleus is given by

Ceeq19

1 10602.1===

CCeq 19192 10204.3)10602.1(22 ==+=

mmpmr1112 1055.2105.255.25 ===

Hence

211

19199

)1055.2(

)10204.3)(10602.1)(10988.8(

=F

NF 710095.7 =PROBLEM 1-5

Find the separation between two point charges of C9 and C5 , if

the Coulomb force between them is N62.1 . B.U. B.Sc.

2004A

SOLUTION


221

04

1

r

qqF

=

F

qqr 21

0

2

4

1

=

2/1

221

041

= r

qqr

2/1669

62.1

)105)(109)(10988.8(

=

r

mr 500.0=

PROBLEM 1-6

What must be the distance between point charges Cq 3.261 = and

point charge Cq 1.472 = in order that the attractive electrostaticforce between them has a magnitude of N66.5 ?

K.U. B.Sc. 2000A

SOLUTION



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2

21

04

1

r

qqF

=

F

qqr 21

0

2

4

1

=

2/1

2

21

04

1

=

r

qqr

2/1669

66.5

)101.47)(103.26)(10988.8(

=

r

mr 403.1=

PROBLEM 1-7

What must be the distance between point charges of C36 andC26 if the force of attraction is N96.3 ?

B.U. B.Sc. 2010A

SOLUTIONAccording to Coulombs law

2

21

04

1

r

qq

F =

F

qqr 21

0

2

4

1

=

2/1

2

21

04

1

=

r

qqr

2/1669

96.3

)1026)(1036)(10988.8(

=

r

mr 46.1=

PROBLEM 1-8

How far away an electron be kept from the nucleus of hydrogen

atom where the Coulombs force on the electron is equal to its

weight? B.U. B.Sc. 2011A

SOLUTION

Now TICELECTROSTAFW =

2

2

04

1

r

egme

=


4/15


gm

er

e

2

0

2

4

1

=

2/12

04

1

=

gm

er

e

mr 083.5)8.9)(10109.9(

)10602.1)(10988.8(2/1

31

2169

=

=

This separation clearly indicates that the force of gravity is not

important on the molecular scale.

PROBLEM 1-9

How far apart must two protons be if the magnitude of the

electrostatic force acting on either one due to the other is equal to

the magnitude of the gravitational force on a proton at Earths

surface?

SOLUTION

Under given condition we have

TICELECTROSTAFW =

2

2

04

1

r

egm

p =

gm

er

p

2

0

2

4

1

=

gm

er

p

2

04

1

=

mr 119.0

)8.9)(10673.1(

)10602.1)(10988.8(27

2169

=

=

PROBLEM 1-10

A proton is at the origin and an electron is located at the point

nmx 41.0= and nmy 36.0= . Find the electric force on the proton.

SOLUTION

Now

mnmr 922 102977.02977.0)036.0()041.0( ==+=

The desired electric force on proton is given by

2

2

0

2

21

0 4

1

4

1

r

e

r

qqF

==


5/15


NF 1029

2199

10748.7)102977.0(

)10602.1)(10988.8(

=

=

PROBLEM 1-11

Two charged spheres are separated by m1 and attract each otherwith a force of N1 . Calculate the size of the charge. How many

electrons are in this charge?

SOLUTION

Assume that the charges are equal in magnitude i.e. qqq == 21 and

opposite in sign. The magnitude of the force of attraction between two

point charges is given by

2 2104

1r

qqF

=

2

2

0 )1(4

11

q

=

02

4 =q

04 =q

Cq 512 10055.1)10854.8(4 ==

PROBLEM 1-12Two small plastic spheres are given positive electrical charges.

When they are cm0.15 apart, the repulsive force between them

has magnitude N220.0 . What is charge on each sphere

(a) if the charges are equal and

(b) if one sphere has four times the charge on the other?

SOLUTION

The magnitude of force of repulsion between two charges is given by

2

21

04

1

r

qqF

=

(a) Now qqq == 21 therefore

2

2

04

1

r

qF

=

Frq 202

4 =

Frq 204 =

)220.0)(100.15)(10854.8(4 212 = q

Cq 71042.7 =

(b) For present case qq =1 and qq 42 = , therefore

2

2

0

4

4

1

r

qF

=


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Frq 202

=

Frq 20=

)220.0)(100.15)(10854.8( 212 = q

Cq 71071.3 =

Hence Cqq7

1 1071.3==

CCqq 672 1048.1)1071.3(44 ===

PROBLEM 1-13

Calculate the force of electrostatic repulsion between two alpha

particles when they are separated by a distance of m1310 .

Compare this force with the gravitational force between these

alpha particles. Given that

Charge on alpha particle = Ce 1910204.32 =+

Mass of alpha particle = kg2710645.6

SOLUTION

Now

Ceqq19

21 10204.32=+==

kgmm 2721 10645.6 == , mr13101 =

229

0

10988.84

1 = CmN

and 221110673.6 = kgmNG


221

04

1

r

qqFe

=

NFe2

213

2199

10227.9)101(

)10204.3)(10988.8(

=

=

According to Newtons law of gravitation

2

21

r

mmGFg =

NFg

37

213

22711

10946.2)101(

)10645.6)(10673.6(

=

=

3537

2

10310946.210227.9

=

g

e

FF

It is clear that gravitational force is negligible as compared with the

electrostatic force.

PROBLEM 1-14

(a) Two protons in a molecule are m101080.3 apart. Find theelectrical force exerted by one proton on the other.


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(b) State how the magnitude of this force compares with the

magnitude of the gravitational force exerted by one proton on the

other.

(c) What must be a particles charge-to-mass ratio if the magnitude

of the gravitational force between two of these particles is equal to

the magnitude of electrical force between them?

SOLUTION

(a) The electrostatic force between two point charges is given by

2

21

04

1

r

qqFe

=

Now Ceqq 1921 10602.1 =+== and mr101080.3 = .

Therefore

NFe9

210

2199

10597.1)1080.3(

)10602.1)(10988.8(

=

=

The nature of above force is repulsive because both charges (i.e.

protons) are positively charged.

(b) According to Newtons law of gravitation

2

21

r

mmGFg =

Now kgmmm p27

21 10673.1=== , therefore

NFg45

210

22711

10293.1)1080.3(

)10673.1)(10673.6(

=

=

36

45

9

10235.1

10293.1

10597.1=

=

g

e

F

F

i.e. the electrostatic force is 3610235.1 times larger than thegravitational force.

(c) If ge FF = then 221

2

21

04

1

r

mmG

r

qq=

210

21

4mmG

qq=

But qqq == 21 and mmm == 21 , therefore above equation reduces to

2

0

2

4mG

q=

Gm

q02

2

4 =

Gm

q04 =


8/15


)10673.6)(10854.8(4 1112 = m

q

kgCm

q/10617.8 11=

PROBLEM 1-15

A certain charge Q is to be divided into two parts )( qQ and q

. What is the relation of Q to q if the two parts, placed a givendistance apart, are to have maximum Coulomb repulsion?

SOLUTION

Let the given charges qQq =1 and qq =2 be placed at a given

distance rapart, then Coulomb repulsive force is given by

20

2

2

0

2

21

0

)(44

)(

4

1

r

qQq

r

qqQ

r

qqFq

=

==

The value of Q for which )(qF will be maximum can be achieved by

equating )/( dqdF to zero i.e.

04

22

0

=

=r

qQ

dq

dF

02 = qQ

qQ 2=

PROBLEM 1-16

Each of two small spheres is charged positively, the total charge

being C6.52 . Each sphere is repelled from the other with a force

of N19.1 when the spheres are m94.1 apart. Calculate the

charge on each sphere.

SOLUTIONLet 1q and 2q be the charge on two small spheres respectively, then

6

21 106.52=+ qq

16

2 )106.52( qq =

(1)

The force of repulsion the two charges 1q and 2q is given by

2

0

21

4 r

qqF

=

2

1

6

1

9

)94.1(

])106.52[()10988.8(19.1

qq =

])106.52[()10988.8(

)94.1)(19.1(1

6

19

2

qq =

2

11

610)106.52(10983.4 qq =


9/15


010983.4)106.52(10

1

62

1 =+ qq (2)

Eq.(2) is quadratic in 1q and the solution can be obtained using the

quadratic formula

a

acbbq2

42

1 =

with )106.52(,1 6== ba and 1010983.4 =c . Hence

)1(2

)10983.4)(1(4)106.52()106.52( 10266

1

=q

Cq6

1 10206.40= and C610394.12

Now )106.52(6

12

= qqIf Cq

6

1 10206.40= , then

)10206.40()106.52(66

2

=q Cq 62 10394.12

= and)10394.12()106.52(

66

2

=q Cq

6

2 10206.40= for Cq 61 10394.12

= .Hence the charge on two spheres are Cq

6

1 10206.40= and

Cq 62 10394.12 = respectively.PROBLEM 1-17

Two free point charges q+ and q4+ are a distance apart. Athird charge is so placed that the entire system is in equilibrium.

Find the location, magnitude and sign of the third charge.

B.U. B.Sc. 1991S

SOLUTION

As the given charges qq =1 and qq 42 = are both positive, thereforethe third charge 3q has to be placed in between 1q and 2q

to achieve the equilibrium i.e. the force on each charge is zero. The

situation is illustrated in the figure.

Now

2313 FF =

2

23

32

0

2

13

31

04

1

4

1

r

qq

r

qq

=

x

1q

2q

3q

)( xl

l


10/15


2

23

2

2

13

1

r

q

r

q =

22)(

4

x

q

x

q

=

xx

=

21

xx 2=

=x3 or3

=x

For calculation of 3q , we take the force on 1q due to 2q and 3q as

zero because of equilibrium i.e. 01213 =+= FFF

04

4

1

2

2

2

3

0

=

+

q

x

qq

04

22

3=+

q

x

q

2

2

3

4

qxq =

qq

q9

4)3/(42

2

3==

Hence the system of three charges will be in equilibrium if qq9

43 =

and located at3

=x from 1q .

PROBLEM 1-18Two point charges are placed on the x-axis as follows: Charge

nCq 00.41 += is located at mx 200.0= and charge nCq 00.52 += is at mx 300.0= . What are the magnitude and direction of the

total force exerted by these two charges on a negative charge

nCq 00.63 = that is placed at the origin?

SOLUTION

Observe that like charges repel and unlike charges attract each other.Taking into account this fact, the charges 1q , 2q and their

corresponding forces on 3q are illustrated in figure given below.


11/15


The magnitudes of 1F and 2F are calculated as under

2

1

31

0

14

1

r

qqF

=

NF 62

999

1 10393.5)200.0(

)1000.6)(1000.4)(10988.8(

=

=

2

2

32

0

14

1

r

qqF

=

NF6

2

999

1 10996.2)300.0(

)1000.6)(1000.5)(10988.8(

=

=

As 1F and 2F are acting in opposite directions with 21 FF > ,

therefore the magnitude of net force will be

21 FFF =

NF 666 10397.2)10966.2()10393.5( ==

The direction of the net force is along 1F i.e. from charge 3q to 1q .


12/15


1-2 QUANTIZATION OF CHARGE

PROBLEM 1-19

Find the total charge in coulombs of kg0.75 of electrons.

SOLUTION

Total mass = M = 75.0 kg

Mass of an electron = kgme 3110109.9 =The total number of electrons N in total mass is calculated as under

31

3110234.8

10109.9

0.75=

==

em

MN

Hence the total charge is given by eNq =

Cq 131931 10319.1)10602.1)(10234.8( ==

PROBLEM 1-20

A charge equal to the charge of Avogadros number of protons is

called a faraday. Calculate the number of coulombs in a faraday.

SOLUTION

Now

AeNq =


13/15


Cq 42319 10647.9)10022.6)(10602.1( ==

PROBLEM 1-21

How many electrons would have to be removed from a coin to leaveit with a charge of C7100.1 + .

SOLUTION

Now qne =

1119

7

10242.610602.1

101=

==

e

qn

The desired number of electrons is 1110242.6 .

PROBLEM 1-22

What is the total charge, in coulombs, of all the electrons in

mol8.1 of hydrogen atoms?

SOLUTION

The number of electrons (or atoms) in 1.8 mol of Hydrogen is

calculated as under.

2423

1008396.11

)10022.6)(8.1( ===M

mNn A

The desired charge q is

Cneq51924 10737.1)10602.1)(1008396.1( ===

PROBLEM 1-23

The electrostatic force between two identical ions that are

separated by a distance of m10105 is N9107.3 .

(a) Find the charge on each ion.(b) How many electrons are missing from each ion?

SOLUTION

(a) According to Coulombs law

221

04

1

r

qqF

=

Now mrqqq10

21 105,

=== and NF 9107.3 = , therefore

210

29

9

)105()10988.8(107.3

= q

Cq 199

2109

10208.310988.8

)105)(107.3(

=

=

(b) qne = or 210602.1

10208.319

19

=

==

e

qn


14/15


Hence two electrons are missing from each ion.

PROBLEM 1-24

You have a pure (24-karat) gold ring with mass grams7.17 . Gold

has an atomic mass of molg/197 and an atomic number of79 .

(a) How many protons are in the ring and what is their total

positive charge?

(b) If the ring carries no net charge, how many electrons are in it?

SOLUTION

(a) The number of gold atoms in 17.7 g is

2223

10411.5

197

)10022.6)(7.17( ===M

mNN A

The number of protons in the ring is

2422

10275.4)79)(10411.5( === NZn

The total positive charge on the gold ring is

Cenq51924 10848.6)10602.1)(10275.4( ===

(b) The number of electrons will be equal to the number of protons i.e.2410275.4

PROBLEM 1-25

Excess electrons are placed on a small lead sphere with massg00.8 so that its net charge is C91020.3 .

(a) Find the number of excess electrons on the sphere.

(b) How many excess electrons are there per lead atom?

The atomic number of lead is 82 and its atomic mass ismolg/207 .

SOLUTION(a) The number of excess electron needed to produce the given net

charge q is obtained by

electronse

qn 10

19

9

10998.110602.1

1020.3=

==

(b) First the number of atoms '' 1N present in the given mass

gM 00.81 = is calculated using the relation

MMNN A 11 =

atomsN 223

322

1 10327.210207

)1000.8)(10022.6( =

=

The desired number of excess electrons per lead atom is

atomelectronsN

n/10586.8

10327.2

10998.1 1322

10

1

=

=


15/15


PROBLEM 1-26

Two tiny, spherical water drops, with identical charges of

C161000.1 , have a centre-to-centre separation of cm00.1 .(a) What is the magnitude of the electrostatic force acting between

them?

(b) How many excess electrons are on each drop, giving it its

charge imbalance?

SOLUTION

(a) The electrostatic force between two point charges is given by

2

2

02

21

0 4

1

4

1

r

q

r

qqF

==

22

2169

)1000.1(

)1000.1)(10988.8(

=F

NF 1910988.8 =

(b) electronse

qn 624

10602.1

1000.119

16

=

=

=

Problems and Solutions on Electricity and Magnetism

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