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
According to Coulombs law
2
2
0
221
0 41
41
re
rqqF
==
212
2199
)109.52(
)10602.1)(10988.8(
=
NF 810243.8 =
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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
According to Coulombs law
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
According to Coulombs law
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CHAPTER 01 ELECTROSTATICS 3
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
=
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CHAPTER 01 ELECTROSTATICS 4
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
==
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CHAPTER 01 ELECTROSTATICS 5
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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CHAPTER 01 ELECTROSTATICS 6
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
According to Coulombs law
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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CHAPTER 01 ELECTROSTATICS 7
(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 =
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CHAPTER 01 ELECTROSTATICS 8
)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 =
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CHAPTER 01 ELECTROSTATICS 9
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
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CHAPTER 01 ELECTROSTATICS 10
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.
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CHAPTER 01 ELECTROSTATICS 11
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 .
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CHAPTER 01 ELECTROSTATICS 12
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 =
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CHAPTER 01 ELECTROSTATICS 13
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
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CHAPTER 01 ELECTROSTATICS 14
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
=
=
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CHAPTER 01 ELECTROSTATICS 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
=
=
=
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