Chapter 2 Electric Charges and Fields (Electrostatics Part 2) – Physics free study material by TEACHING CARE online tuition and coaching classes
(ii) Charged circular ring : Suppose we have a charged circular ring of radius R and charge Q. On it’s axis electric field and potential is to be determined, at a point ‘x’ away from the centre of the ring.
 Electric field : Consider an element carrying charge dQ . It’s electric field
dE = (
KdQ
) directed

as shown. It’s component along the axis is
dE cosq
and perpendicular to the axis is
dE sinq . By
symmetry
dE sinq
= 0 , hence E =
dE cosq =
kdQ . x
ò ò ò (R ^{2} + x ^{2} )
E = kQx
(R ^{2} + x ^{2} )3 2
directed away from the centre if Q is positive
(b) Potential :
V = 1 . Q
4pe _{0}
Note : At centre x = 0 so E_{centre}= 0 and
Vcentre
= kQ
R
□ At a point on the axis such that x >> R
E = kQ

x 2
and
V = kQ
x
□ At a point on the axis if
(3) Surface charge :
x = ± R ,
Emax =
 Infinite sheet of charge : Electric field and potential at a point P as shown
E = s
2e _{0}

s r
(E µ r ^{o} )
and
V = – 2e + C
 Electric field due to two parallel plane sheet of charge : Consider two large, uniformly charged Plates A and B, having surface charge
densities are s _{A}
and s _{B}
respectively. Suppose net electric
field at points P, Q and R is to be calculated.
At P,
At Q,
EP = (EA
E = (E
 EB ) =
– E ) =
1
2e _{0}


1
(s _{A}

(s
+ s _{B}

– s
)
); At R, E
= (E
+ E ) = –
1 (s
+ s )
Q A B
2e _{0}
R A B
2e _{0}
Note : If
s _{A} = +s
and
s _{B} = –s
then
Ep = 0, EQ
= s , E

e _{0}
= 0 . Thus in case of two infinite
plane sheets of charges having equal and opposite surface charge densities, the field is nonzero only in the space between the two sheets and is independent of the distance between them i.e., field is uniform in this region. It should be noted that this result will hold good for finite plane sheet also, if they are held at a distance much smaller then the dimensions of sheets i.e., parallel plate capacitor.
(iii) Conducting sheet of charge :
E = s
e _{0}
sr

V = – + C
0
 Charged conducting sphere : If charge on a conducting sphere of radius R is Q as shown in figure then electric field and potential in different situation are –
(a) Out side the sphere : P is a point outside the sphere at a distance r from the centre at which electric field and potential is to be determined.
Electric field at P
E = 1 . Q
= sR ^{2}
and V
= 1 . Q = sR ^{2}
ìQ = s ´ A
out
4pe _{0} r ^{2}
e _{0}r ^{2}
out
4pe _{0} r
e _{0}r î
=s ´ 4pR ^{2}

 At the surface of sphere : At surface r = R
So,
E = 1 . Q = s
and V
= 1 . Q = sR




^{s} 4pe R ^{2} e ^{s} 4pe R e
(c) Inside the sphere : Inside the conducting charge sphere electric field is zero and potential remains constant every where and equals to the potential at the surface.
Ein = 0
and
Vin = constant = Vs
Note : @Graphical variation of electric field and potential of a charged spherical conductor with distance
E – r graph
V – r graph
r r
(4) Volume charge (charged nonconducting sphere) :
Charge given to a non conducting spheres spreads uniformly throughout it’s volume.
(i) Outside the sphere at P
Eout
= 1
4pe _{0}
. Q and
r 2
Vout
= 1 . Q
4pe _{0} r
by using
r = Q
4 pR ^{3}
3
Eout
= rR 3 and
3e _{0}r ^{2}
Vout
= rR ^{3}
3e _{0}r
 At the surface of sphere : At surface r = R
E = 1 . Q = rR

and
V = 1 . Q =
rR ^{2}

^{s} 4pe R ^{2} 3e
0
^{s} 4pe R
3e _{0}
 Inside the sphere : At a distance r from the centre
E = 1 . Qr
= rr {E
µ r}

and
V = 1
Q[3R ^{2} – r ^{2} ] = r(3R ^{2} – r ^{2} )

^{in} 4pe R ^{3}
3e _{0}
^{in} 4pe
2R ^{3}
6e _{0}

Note : @ At centre
r = 0
So,
V = 3 ´ 1 . Q = 3 V
i.e.,

Vcentre
 Vsurface
 Vout
centre
2 4pe _{0} R 2
@ Graphical variation of electric field and potential with distance
Er graph
Vr graph
r r
r = R
(5) Electric field and potential in some other cases
 Uniformly charged semicircular ring : l = charge
length
At centre :
E = 2Kl = Q
+Q
+ + +
+ +
+ +

R 2p ^{2}e _{0} R ^{2}

V = KQ = Q
R 4pe _{0} R
(iii) Charged cylinder of infinite length
(a) Conducting (b) Nonconducting
P
r
R
+ + + +
+ + + +
^{+} + + + P

+ + + r
+ + + +
For both type of cylindrical charge distribution E
= l , and E
= l
but for conducting
out
2pe _{0}r
suface
2pe _{0} R
E = 0 and for nonconducting E
= lr . (we can also write formulae in form of r i.e., E
= rR ^{2}
etc.)


in in 2pe R 2 out 2e r
(ii) Hemispherical charged body :
s
+ + + +
At centre O, E = + + + +
+ +
4e _{0}
+ + + +
+ + +
V = sR
2e _{0}
+ + O · + + +
(iv) Uniformly charged disc
At a distance x from centre O on it’s axis
= s é
x ù s
E 2e
ê1 –
0 êë
x ^{2} + R
ú
_{2} úû R
O

V = s é
2e _{0} ë
x ^{2} + R ^{2}
x

– xù
û
Note : @ Total charge on disc Q = spR^{2}
@ If x ® 0,
E ~– s
2e _{0}
i.e. for points situated near the disc, it behaves as an infinite sheet of charge.
Concepts
No point charge produces electric field at it’s own position.
Since charge given to a conductor resides on it’s surface hence electric field inside it is zero.
++ + +
+ + +
+ + +
+
E = 0 +
+
+
+ + +
+ + + +
+ +
+ E = 0 +
+ +
+ +
+ + +
The electric field on the surface of a conductor is directly proportional to the surface charge density at that point i.e, E µ s
Two charged spheres having radii r_{1} and r_{2} charge densities s _{1} and s _{2} respectively, then the ratio of electric field on their
E1 s _{1} r 2 ì Q
surfaces will be
= = 2
ís = 2


E2 s _{2} 2
î 4pr
In air if intensity of electric field exceeds the value 3 ´ 10^{6} N/C
air ionizes.
A small ball is suspended in a uniform electric field with the help of an insulated thread. If a high energy x–ray bean falls on the ball, xrays knock out electrons from the ball so the ball is positively charged and therefore the ball is deflected in the direction of electric field.
r
E
F= QE
X–Ray
Electric field is always directed from higher potential to lower potential.
A positive charge if left free in electric field always moves from higher potential to lower potential while a negative charge moves from lower potential to higher potential.
The practical zero of electric potential is taken as the potential of earth and theoretical zero is taken at infinity. An electric potential exists at a point in a region where the electric field is zero and it’s vice versa.
A point charge +Q lying inside a closed conducting shell does not exert force another point charge q placed outside the shell as
shown in figure
+ +
+Q + +
+ – – – + +q
+ – +Q – +
+ – – +

+ +
+
Actually the point charge +Q is unable to exert force on the charge +q because it can not produce electric field at the position of
+q. All the field lines emerging from the point charge +Q terminate inside as these lines cannot penetrate the conducting medium (properties of lines of force).
The charge q however experiences a force not because of charge +Q but due to charge induced on the outer surface of the shell.
æ 1 ö
Example: 21 A half ring of radius R has a charge of l per unit length. The electric field at the centre is ç k = ÷
è 4pe_{0} ø
[CPMT 2000; CBSE PMT 2000; REE 1999]
Solution: (c)
 Zero (b)
dl = Rdq
Charge on dl = lRdq .
kl (c)
R
2kl
R
dl
dq
q
q
(d)
C
q
kpl R
Field at C due to dl = k lRdq . = dE
R2
We need to consider only the component
dl¢
dE cos q , as the component
dE sin q
dE
will cancel out because
of the field at C due to the symmetrical element dl¢,

p 2
The total field at C is = 2 dE cos q
= 2 kl òp 2 cosqdq = 2k l
ìï= Q üï
0 R 0
í


R î 2pe _{0} R
_{2} ýï
Example: 22 What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2
newton/coulomb[1 / 4pe _{0} = 9 ´ 10^{9} Nm^{2} ]
[MP PMT 1996]
(a)
2 ´ 10 ^{–}^{11} coulomb
9 ´ 10 ^{–}^{11} coulomb
(b)
3 ´ 10 ^{–}^{11} coulomb
(c)
5 ´ 10 ^{–}^{11} coulomb
(d)
Solution: (a) By using
E = 1
4pe_{0}
. Q ;
r 2
2 = 9 ´ 10^{9} ´
Q Þ Q = 2 ´ 10^{–}^{11}C

30 ´ 10^{2}
Example: 23 Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q
is E, then at the locality of – 3Q, it is
 – E
 E/3 (c)
3E
 – E/3
Solution: (b) Let the charge Q and – 3Q be placed respectively at A and B at a distance x
Now we will determine the magnitude and direction to the field produced by charge – 3Q at A, this is E
as mentioned in the Example. A B
\ E = 3Q
x2
(along AB directed towards negative charge)
Q – 3 Q
x
Now field at location of – 3Q i.e. field at B due to charge Q will be away from positive charge)
E’ = Q = E
x^{2} 3
(along AB directed
Example: 24 Two charged spheres of radius
R1 and
R_{2} respectively are charged and joined by a wire. The ratio of
electric field of the spheres is [CPMT 1999 Similar to CBSE 2003]
(a)
R1
R2
(b)
R2
R1
2



 ^{1} 2
2



 ^{2} 1
Solution: (b) After connection their potential becomes equal i.e., k . Q1
R1
= k . Q2 ; Þ
R2
Q1 = R1 Q2 R2
E Q æ R ö ^{2} R
Ratio of electric field ^{1} = ^{1} ´ ç^{ } ^{2} ÷
= 2 .
E2 Q2
è R1 ø R1
Example: 25 The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of
0.036 N/C just above its surface is [MNR 1994]
(a)
2.7 ´ 10^{5}
(b)
2.6 ´ 10^{5}
(c)
2.5 ´ 10^{5}
(d)
2.4 ´ 10^{5}
Solution: (c) By using
E = k
Q , where R = radius of sphere so 0.036 = 9 ´ 10^{9} ´
R2
ne
(0.1)2
Þ n = 2.5 ´ 10^{5}
Example: 26 Eight equal charges each +Q are kept at the corners of a cube. Net electric field at the centre will be
æ 1 ö
çk = ÷
è 4pe_{0} ø
(a)
kQ (b)
r 2
8kQ (c)
r 2
2kQ (d) Zero
r 2
Solution: (d) Due to the symmetry of charge. Net Electric field at centre is zero.

Note : q
q a q
a a
+q +q
Equilateral triangle
q q
Square
Example: 27 q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre
O of the square has the direction along.
[CPMT 1989]
(a) AB (b) CB (c) AC (d) BD
Solution: (b) By making the direction of electric field due to all charges at centre. Net electric field has the direction along
CB
Example: 28 Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is
(a)
Q
4pe_{0}a^{2}
(b)
4pe_{0}a^{2}
(c)
4pe_{0}a^{2}
(d)
Q
2pe_{0}a^{2}
Solution: (c) As shown in figure Net electric field at A
E =
1 Q

EB = EC = 4pe . a2
So,
E = 3Q
4pe_{0}a^{2}
+ B a C +
Example: 29 Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q is one micro coulomb, then electric field intensity at centre will be [RPET 1999]
(a)
(b)
(c)
(d)
1.02 ´ 10^{7} N / C upwards
2.04 ´ 10^{7} N / C downwards
2.04 ´ 10^{7} N / C upwards
1.02 ´ 10^{7} N / C downwards
Q – 2Q
– Q + 2Q
Solution: (a)
 EC >  E A so resultant of EC & EA is
ECA = EC – EA directed toward Q Q
– 2Q
Also  E
A B
 >  E so resultant of E and E i.e.
B D B D
EBD = EB – ED directed toward – 2Q charge hence Net electric field at centre is
E = .… (i)
By proper calculations  EA = 9 ´ 10^{9} ´
æ
10 6
5
D
– Q
= 0.72 ´ 10^{7} N/C
ö 2
C
+2Q
ç ´ 10 ^{2} ÷
è ø
 EB
= 9 ´ 10^{9} ´
æ
2 ´ 10 ^{6}
5 ö 2
= 1.44 ´ 10^{7} N/C ;
 EC
= 9 ´ 10^{9} ´
æ
2 ´ 10 ^{6}
5 ö 2
= 1.44 ´ 10^{7} N/C
ç ´ 10 ^{2} ÷
è ø
ç ´ 10 ^{2} ÷
è ø
 E = 9 ´ 10^{9} ´
10 6
= 0.72 ´ 10^{7} N/C ; So, E
= E –E = 0.72 ´ 10^{7} N/C

^{D} æ ö ^{2}
ç ´ 10 ^{2} ÷
è ø
CA C A
and  EBD = EB  ED = 0.72 ´ 10^{7} N/C. Hence from equation – (i) E = 1.02 ´ 10^{7} N/C upwards
Example: 30 Infinite charges are lying at x = 1, 2, 4, 8…meter on Xaxis and the value of each charge is Q. The value of intensity of electric field and potential at point x = 0 due to these charges will be respectively
(a) 12 ´ 10^{9} Q N/C, 1.8 ´ 10^{4} V (b) Zero, 1.2 ´ 10^{4}V
(c)
6 ´ 10^{9} Q N/C, 9 ´ 10^{3} V (d)
4 ´ 10^{9} Q
N/C , 6 ´ 10^{3} V
Solution: (a) By the superposition, Net electric field at origin
E = kQé 1 + 1 + 1 + 1 + …¥ù
êë12 22 4 2 82 úû
E = kQé1 + 1 + 1 + 1 + …¥ù
ëê 4 16 64 úû
1 + 1 + 1 + 1
4 16 64
+ …¥
is an infinite geometrical progression it’s sum can be obtained by using the

formula S = a
1 – r
; Where a = First term, r = Common ratio.
Here a = 1 and r = 1
4
so, 1 + 1 + 1 + 1
4 16 64
+ ….. ¥ =
1
1 – 1 / 4
= 4 .
3
Hence E = 9 ´ 10^{9} ´ Q ´ 4 = 12 ´ 10^{9} Q N/C
3



Electric potential at origin V = 1 é1 ´ 106 + 1 ´ 106 + 1 ´ 106 + 1 ´ 106 +…………………. ¥ù
4pe _{0} ê 1
2 4 8 ú
é ù
= 9 ´ 10^{9} ´ 10 ^{6} é1 + 1 + 1 + 1 + …………¥ù = 9 ´ 10^{3} ê 1 ú = 1.8 ´ 10^{4} volt

ëê 2 4 8
ú ê

û ê1 –
ë
ú
ú
2 úû
Note : @ In the arrangement shown in figure +Q and – Q are alternatively and
equally spaced from each other, the net potential at the origin O is V = 1 . Q log e 2
4pe _{0} x
+Q – Q +Q – Q
O x 2x 3x 4x
Example: 31 Potential at a point xdistance from the centre inside the conducting sphere of radius R and charged with charge Q is [MP PMT 2001]
(a)
Q (b)
R
Q (c)
x
Q (d) xQ
x 2
Solution: (a) Potential inside the conductor is constant.
Example: 32 The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9 ´ 10^{5} V is
 80 V (b)
8 ´ 10^{6} V
 9 V (d)
9 ´ 10^{5} V
Solution: (b)
V = 9 ´ 10^{9} ´ ne = 9 ´ 10^{9} ´ 50 ´ 1.6 ´ 10 19
= 8 ´ 10^{6} V
r 9 ´ 10 ^{15}
Example: 33 Eight charges having the valves as shown are arranged symmetrically on a circle of radius 0.4m in air.
Potential at centre O will be
(a)
63 ´ 10^{4} volt
(b)
63 ´ 10^{10} volt
(c)
63 ´ 10^{6} volt
 Zero
Solution: (a) Due to the principle of superposition potential at O
V = 1 ´ 28 ´ 10 ^{6}
= 9 ´ 10^{9} ´ 28 ´ 10 6
= 63 ´ 10^{4} volt
4pe _{0} 0.4
0.4
Example: 34 As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is [MP PET 2000]
(a)
1 . 2q
4pe _{0}
(b)
1 . q
4pe _{0}
(c)
1 .
4pe _{0}
(q)
 Zero
Solution: (d) Potential at A = Potential due to (+q) charge + Potential due to (– q) charge
= 1 . q + 1
4pe _{0}
(q) = 0
Example: 35 A conducting sphere of radius R is given a charge Q. consider three points B at the surface, A at centre and C at a distance R/2 from the centre. The electric potential at these points are such that
 V_{A} = V_{B} = V_{C} (b) V_{A} = V_{B} ¹ V_{C} (c) V_{A} ¹ V_{B} ¹ V_{C} (d) V_{A} ¹ V_{B} = V_{C}
Solution: (a) Potential inside a conductor is always constant and equal to the potential at the surface.
Example: 36 Equal charges of
10 ´ 10 ^{9}
3
coulomb are lying on the corners of a square of side 8 cm. The electric
potential at the point of intersection of the diagonals will be
(a) 900 V (b)
Solution: (d) Potential at the centre O
900 V
(c)
150 V
(d)
1500 V
V = 4 ´
1 . Q
given Q =
Q Q
10 ´ 10 ^{–}^{9} C Þ a = 8 c+m = 8– ´ 10 ^{–}^{2} m
4pe _{0} 3
10 ´ 10 ^{9}
+ –
+ –
+ C + Q
V = 5 ´ 9 ´ 10^{9} ´ 3 = 1500

8 ´ 10 ^{2}
2 volt
+ –
+ R –
+ – r Q Q
Tricky example: 4 
A point charge q is placed at a distance of r from the centre of an uncharged conducting sphere of radius R (< r). The potential at any point on the sphere is
(a) Zero (b) 1 . q (c) 1 . qR (d) 1 . qr 2 4pe _{0} r 4pe _{0} r 2 4pe _{0} R Solution: (c) Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the centre of the sphere. \ V = 1 . q + V ‘ ; where V¢ = potential at centre due to induced charge = 0 (because net 4pe _{0} r induced charge will be zero) \ V = 1 . q . 4pe _{0} r 
Potential Due to Concentric Spheres.
To find potential at a point due to concentric sphere following guideline are to be considered
Guideline 1: Identity the point (P) at which potential is to be determined.
Guideline 2: Start from inner most sphere, you should know where point (P) lies w.r.t. concerning sphere/shell (i.e. outside, at surface or inside)
Guideline 3: Then find the potential at the point (P) due to inner most sphere and then due to next and so
on.
Guideline 4: Using the principle of superposition find net potential at required shell/sphere.
Case (ii) : The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Q_{a}, Q_{b} and Q_{c} respectively what will be the potential of each shell
After following the guidelines discussed above Qc Potential at A; V = 1 é Qa + Qb + Qc ù Q Qb ^{A} 4pe ê a b c ú a 0 ë û c 1 é Q Q Q ù b a A B Potential at B; VB = ê a + ^{b} + ^{c} ú 4pe _{0} ë b b c û 

Potential at C;  V = 1 é Q_{a} + Q_{b} + Q_{c} ù C 4pe _{0} ê c c c ú
ë û 

Case (iii) : The figure shows two concentric spheres having radii r_{1} and r_{2} respectively (r_{2} > r_{1}). If charge on inner sphere is +Q and outer sphere is earthed then determine.
(a) The charge on the outer sphere (b) Potential of the inner sphere Q¢ +Q 1 Q 1 Q’ r2 (i) Potential at the surface of outer sphere V2 = 4pe . r + 4pe . r = 0 r 0 2 0 2 1 Þ Q’ = –Q (ii) Potential of the inner sphere V = 1 . Q + 1 (–Q) = Q é 1 – 1 ù 1 4pe _{0} r1 4pe _{0} r2 4pe _{0} ê r1 r2 ú ë û 

Case (iv) : In the case III if outer sphere is given a charge +Q and inner sphere is earthed then
(a) What will be the charge on the inner sphere (b) What will be the potential of the outer sphere (i) In this case potential at the surface of inner sphere is zero, so if Q’ is the charge induced on inner sphere then V = 1 é Q’ + Q ù = 0 i.e., Q’ = – r1 Q ^{1} 4pe _{0} ê r_{1} r_{2} ú r ë û 2 r2 +Q (Charge on inner sphere is less than that of the outer sphere.) (ii) Potential at the surface of outer sphere r1 V2 = 1 . Q’ + 1 . Q 4pe 0 r2 4pe 0 r2 V = 1 é Q r_{1} + Qù = Q é – r_{1} ù 2 4pe r ê r ú 4pe r ê1 r ú 0 2 ë 2 û 0 2 ë 2 û 
Example: 37 A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 volts. The potential at the centre of the sphere is
 Zero (b) 10 V
(c) Same as at a point 5 cm away from the surface (d) Same as at a point 25 cm away from the surface
Solution: (b) Inside the conductors potential remains same and it is equal to the potential of surface, so here potential at the centre of sphere will be 10 V
Example: 38 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is
(a) 54 e.s.u. (b)
1 e.s.u. (c) 30 e.s.u. (d) 36 e.s.u.
4
Solution: (d) Let charge on inner sphere be +Q. charge induced on the inner surface of outer sphere will be –Q. So potential at the surface of inner sphere (in CGS)
3 = Q – Q
4 6
Þ Q = 36 e.s.u.
Example: 39 A charge Q is distributed over two concentric hollow spheres of radii r and (R > r) such that the surface densities are equal. The potential at the common centre is [IITJEE 1981]
(a)
Q(R^{2} + r ^{2})
4pe_{0}(R + r)
(b)
Q R + r
 Zero (d)
Q(R + r) 4pe_{0}(R^{2} + r ^{2})
Solution: (d) If q_{1} and q_{2} are the charges on spheres of radius r and R respectively, in accordance with conservation of charge
Q = q1
+ q2
….(i) q2
and according to the given problem s_{1} = s _{2}
q1 q2
q1 r ^{2}
i.e.,
4pr ^{2} = 4pR^{2} Þ
q = R^{2}
…. (ii)

So equation (i) and (ii) gives
q1 =
Qr^{2}
(R^{2} + r ^{2})
and q2
= QR^{2}
(R^{2} + r ^{2})
Potential at common centre
V = 1 é q_{1} + q_{2} ù = 1 é Qr + QR ù = 1 . Q(R + r)


4pe0 êë r R úû 4pe0 ê(R^{2} + r ^{2}) (R^{2} + r ^{2})ú 4pe0 (R^{2} + r ^{2})
Example: 40 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is
(a) V (b) 2V (c) 4V (d) –2V
Solution: (a) If a and b are radii of spheres and spherical shell respectively, potential at their surfaces will be
Vsphere = 1 . Q and Vshell = 1 . Q
4pe_{0} a 4pe_{0} b
and so according to the given problem.
V = Vsphere – Vshell
= Q é1 4pe_{0} êë a
1

– b úû
…. (i)
Now when the shell is given a charge –3Q the potential at its surface and also inside will change by
V0 = 1 é 3Q ù
4pe_{0} êë b úû
So that now
Vsphere = 1 é Q – 3Q ù
and
Vshell = 1 é Q – 3Q ù
hence
Vsphere – Vshell = Q é1 – 1 ù = V
4pe_{0} êë a
b úû
4pe_{0} êë b
b úû
4pe_{0} êë a b úû
Example: 41 Three concentric metallic spheres A, B and C have radii a, b and c
(a < b < c)
and surface charge
densities on them are s , – s
and s respectively. The valves of V_{A}
and
VB will be


s s æ a^{2} ö
(a)
(a – b – c), ç – b + c ÷
(b)
e_{0}
(a – b – c), a2
c
e_{0}
e0 ç b ÷
s


e0 æ a2 ö
(c)
(a – b – c),
ç – b + c ÷
s
s æ a^{2}
ç ÷
è ø

b^{2} ö s

(d)
ç –
è c
 c ÷
ø
and
(a – b + c)

0
Solution: (a) Suppose charges on A, B and C are qa , qb and qc
Respectively, so s _{A} = s = qa
4pa^{2}
Þ qa = s ´ 4pa^{2} , s _{B} = –s = qb Þ qb = –s ´ 4pb^{2}
4pb^{2}
and s _{C} = s =
qc
4pc^{2}
Þ qc = s ´ 4pc^{2}
Potential at the surface of A
V = (V )
+ (V )
+ (V )
= 1 é q_{a} + q_{b} + q_{c} ù
= 1 és ´ 4pa^{2} + (s )´ 4pb^{2} + s ´ 4pc^{2} ù
 A surface B in C in
VA = s [a – b – c]]
4pe_{0}
êë a b
c úû
4pe
ê
_{0} êë a b
ú
c úû
e_{0}
Potential at the surface of B
V = (V )
+ (V )
+ (V )
= 1 é q_{a} + q_{b} + q_{c} ù = 1 és ´ 4pa^{2} – s ´ 4pb^{2} + s ´ 4pc^{2} ù
 A out
 surface
 in
4pe_{0}
êë b b
c úû
4pe
ê
_{0} êë b b
ú
c úû


= s é a2 – b + cù
e0 ê b ú
Electric Lines of Force.
(1) Definition : The electric field in a region is represented by continuous lines (also called lines of force). Field line is an imaginary line along which a positive test charge will move if left free.
Electric lines of force due to an isolated positive charge, isolated negative charge and due to a pair of charge are shown below
 Properties of electric lines of force
(i) Electric field lines come out of positive charge and go into the negative charge.
 Tangent to the field line at any point gives the direction of the field at that
(iii) Field lines never cross each other.
 Field lines are always normal to conducting
(v) Field lines do not exist inside a conductor.
 The electric field lines never form closed (While magnetic lines of forces form closed loop)
(vii) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In the following figure electric lines of force are originating from A and terminating at B hence Q_{A} is positive while Q_{B} is negative, also number of electric lines at force linked with Q_{A} are more than those linked with Q_{B} hence QA >QB 
 Number of lines of force per unit area normal to the area at a point represents magnitude of intensity (concept of electric flux e., f = EA )
(ix) If the lines of forces are equidistant and parallel straight lines the field is uniform and if either lines of force are not equidistant or straight line or both the field will be non uniform, also the density of field lines is proportional to the strength of the electric field. For example see the following figures.
 Electrostatic shielding : Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external electric Sensitive instruments and
appliances are affected seriously with strong external electrostatic fields. Their working suffers and they may start misbehaving under the effect of unwanted fields.

The electrostatic shielding can be achieved by protecting and enclosing the sensitive instruments inside a hollow conductor because inside hollow conductors, electric fields is zero.
(i) It is for this reason that it is safer to sit in a car or a bus during lightening rather than to stand under a tree or on the open ground.
 A high voltage generator is usually enclosed in such a cage which is This would prevent the electrostatic field of the generator from spreading out of the cage.
(iii) An earthed conductor also acts as a screen against the electric field. When conductor is not earthed field of the charged body C due to electrostatic induction
continues beyond AB. If AB is earthed, induced positive charge neutralizes and the field in the region beyond AB disappears.
Equipotential Surface or Lines.
If every point of a surface is at same potential, then it is said to be an equipotential surface
or
for a given charge distribution, locus of all points having same potential is called “equipotential surface” regarding equipotential surface following points should keep in mind :
 The density of the equipotential lines gives an idea about the magnitude of electric Higher the density larger the field strength.
(2) The direction of electric field is perpendicular to the equipotential surfaces or lines.
 The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric
lines.
 For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field
(5) A metallic surface of any shape is an equipotential surface e.g. When a charge is given to a metallic
surface, it distributes itself in a manner such that its every point comes at same potential even if the object is of irregular shape and has sharp points on it.
If it is not so, that is say if the sharp points are at higher potential then due to potential difference between these points connected through metallic portion, charge will flow from points of higher potential to points of lower potential until the potential of all points become same.
 Equipotential surfaces can never cross each other
(7) Equipotential surface for pair of charges
Example: 42 Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in
(a)
(b)
(c)
(d)
Solution (c) Option (a) shows lines of force starting from one positive charge and terminating at another. Option (b) has one line of force making closed loop. Option (d) shows all lines making closed loops. All these are not correct. Hence option (c) is correct
Example: 43 A metallic sphere is placed in a uniform electric field. The lines of force follow the path (s) shown in the figure as
(a) 1 (b) 2 (c) 3 (d) 4
Solution: (d) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due to induced charges on its surface the field is distorted close to its surface and a line of force must deviate near the surface outside the sphere.
Example: 44 The figure shows some of the electric field lines corresponding to an electric field. The figure suggests
[MP PMT 1999]
Solution: (c)
(a)
EA > EB > EC
(b)
EA = EB = EC
(c)
EA = EC > EB
(d)
EA = EC < EB
Example: 45 The lines of force of the electric field due to two charges q and Q are sketched in the figure. State if
 Q is positive and Q > q
 Q is negative and
 q is positive and
Q > q
Q < q
 q is negative and Q < q
Solution: (c) q is +ve because lines of force emerge from it and lines terminate at Q.
Q < q
because more lines emerge from q and less
Example: 46 The figure shows the lines of constant potential in a region in which an electric field is present. The magnitude of electric field is maximum at
 A (b) B (c) C (d) Equal at A, B and C
Solution: (b) Since lines of force are denser at B hence electric field is maximum at B
Example: 47 Some equipotential surface are shown in the figure. The magnitude and direction of the electric field is
(a) 100 V/m making angle 120^{o} with the xaxis (b) 100 V/m making angle 60^{o} with the xaxis
(c) 200 V/m making angle 120^{o} with the xaxis (d) None of the above
Solution: (c) By using dV = E dr cosq
suppose we consider line 1 and line 2 then
(30 – 20) = E cos 60^{o} (20 – 10) × 10^{–2} So E = 200 volt / m making in angle 120^{o} with xaxis
y
Relation Between Electric Field and Potential.
In an electric field rate of change of potential with distance is known as potential gradient. It is a vector quantity and it’s direction is opposite to that of electric field. Potential
gradient relates with electric field according to the following
relation E = – dV ;
dr
This relation gives another unit of electric field is
volt .
meter
In the above relation negative sign indicates that in the direction of electric field potential decreases.
In space around a charge distribution we can also write
r ˆ ˆ ˆ
E = Ex i + Ey j + Ez k
where
E = – dV ,
^{x} dx
Ey = – dV
dy
and
E = – dV
^{z} dz
With the help of formula
E = – dV , potential difference between any two points in an electric field can be
dr
determined by knowing the boundary conditions dV = òr2 E . dr = òr2 E . dr cosq .
r1 r1
For example: Suppose A, B and C are three points in an uniform electric field as shown in figure.
 Potential difference between point A and B is
B
VB – VA = òA E × dr


Since displacement is in the direction of electric field, hence q = 0^{o}
So,
VB – VA = òA
E × dr cos 0 = – òA
E × dr = –Ed
In general we can say that in an uniform electric field
E = – V
d
or  E = V
d
Another example
E = V
d
(ii)

Potential difference between points A and C is :
VC – VA
= òA E dr cosq
= –E(AC) cosq
= –E(AB) = – Ed
Above relation proves that potential difference between A and B is equal to the potential difference between A and C i.e. points B and C are at same potential.
+ –
Example: 48 The electric field, at a distance of 20 cm from the centre of a dielectric sphere of radius 10 cm is 100 V/m.
The ‘E’ at 3 cm distance from the centre of sphere is [RPMT 2001]
(a) 100 V/m (b) 125 V/m (c) 120 V/m (d) Zero
Solution: (c) For dielectric sphere i.e. for nonconducting sphere Eout = k.q
r 2
and
Ein = kqr
R 3
Eout
= 100 KQ
(20 ´ 10^{2} )^{2}
Þ KQ = 100 ´ (0.2)^{2} so
Ein
= 100 ´ (0.2)^{2} ´ (3 ´ 10^{2} )^{2} = 120 V/m
(10 ´ 10^{2} )^{3}
Example: 49 In x–y coordinate system if potential at a point P(x, y) is given by V = axy ; where a is a constant, if r is
the distance of point P from origin then electric field at P is proportional to [RPMT 2000]
(a) r (b) r^{–1} (c) r^{—2} (d) r^{2}
Solution: (a) By using
E = – dV
dr
Ex = – dV = –ay ,
dx
Ey = – dV = –ax dy
Electric field at point P E =
= a = ar
i.e., E µ r
Example: 50 The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x^{2} volt. The electric
field at the point (1m, 0, 2m) in volt/metre is [MP PMT 2001; IITJEE 1992; RPET 1999]
 8 along negative Xaxis (b) 8 along positive Xaxis
 16 along negative Xaxis (d) 16 along positive Zaxis
Solution: (a) By using
xaxis.
E = – dV
dx
Þ E = – d (4 x ^{2} ) = 8 x . Hence at point (1m, 0, 2m). E = – 8 volt/m i.e. 8 along – ve
dx
Example: 51 The electric potential V is given as a function of distance x (metre) by V = (5x^{2} + 10x – 9) volt. Value of electric field at x = 1m is [MP PET 1999]
(a) – 20 V/m (b) 6 V/m (c) 11 V/m (d) – 23 V/m
Solution: (a) By using
E = – dV ;
dx
E = – d (5x ^{2} + 10x – 9) = (10x + 10) ,
dx
at x = 1m E = 20 V/m
Example: 52 A uniform electric field having a magnitude E_{0} and direction along the positive Xaxis exists. If the electric potential V, is zero at X = 0, then, its value at X = +x will be [MP PMT 1987]
(a) V(x)= +xE_{0} (b) V(x)= – xE_{0} (c) V(x)= x^{2}E_{0} (d) V(x)= – x^{2}E_{0}
Solution: (b) By using
E = – DV = – (V_{2} – V_{1}) ; E = – {V(x) – 0}
Þ V(x) = – xE
Dr (r2 – r1 ) ^{0}
x – 0 ^{0}
Example: 53 If the potential function is given by V = 4x + 3y, then the magnitude of electric field intensity at the point (2, 1) will be [MP PMT 1999]
(a) 11 (b) 5 (c) 7 (d) 1
Solution: (b) By using i.e., E = ; E
= – dV = – d (4 x + 3y) = 4
^{x} dx dx
and
Ey = – dV = – d (4 x + 3y) = 3
dy dy
\ E =
= 5 N/C
Work Done in Displacing a Charge.
(1) Definition : If a charge Q displaced from one point to another point in electric field then work done
in this process is
W = Q ´ DV
where DV = Potential difference between the two position of charge Q.
( DV = E .D r
charge).
= EDr cosq
where q is the angle between direction of electric field and direction of motion of
 Work done in terms of rectangular component of E and r : If charge Q is given a
displacement
r = (r ˆi + r ˆj + r kˆ) in an electric field E = (E ˆi + E ˆj + E
kˆ).
The work done is
1 2 3 1 2 3
W = Q( E . r ) = Q(E1r1 + E2 r2 + E3 r3 ) .
Conservation of Electric Field.
As electric field is conservation, work done and hence potential difference between two point is path independent and depends only on the position of points between. Which the charge is moved.
Example: 54 A charge (– q) and another charge (+Q) are kept at two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (– q) at A is moved to another point C such that ABC forms an equilateral triangle of side l. The network done in moving the charge (– q) is
(a)
1 Qq
4pe _{0} l
(b)
1 Qq
4pe _{0} l 2
(c)
1
4pe _{0}
Qql
 Zero
Solution: (d) Since VA
= VC
= kQ l
A – q
so W = q (VC – VA ) = 0
l l
B C
+ Q l
Example: 55 The work done in bringing a 20 coulomb charge from point A to point B for distance 0.2 m is 2 Joule. The potential difference between the two points will be (in volt) [RPET 1999 Similar to MP PET 1999]
(a) 0.2 (b) 8 (c) 0.1 (d) 0.4
Solution: (c)
W = Q.DV
Þ 2 = 20 ´ DV Þ DV = 0.1 volt
Example: 56 A charge +q is revolving around a stationary +Q in a circle of radius r. If the force between charges is F
then the work done of this motion will be
[CPMT 1975, 90, 91, 97; NCERT 1980, 83; EAMCET 1994; MP PET 1993, 95;
MNR 1998; AIIMS 1997; DCE 1995; RPET 1998]
 F × r (b)
F ´ 2pr
(c)
F
2pr
 0
Solution: (d) Since +q charge is moving on an equipotential surface so work done is zero.
+ q
Example: 57 Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in removing a charge – Q from its centre to infinity is [AIIMS 1995]
 0 (b)
4pe _{0} a
(c)
pe _{0} a
(d)
Q2
2pe _{0} a
Solution: (c) We know that work done in moving a charge is W = QDV
Q a Q
Here
W = Q(V
– V ) Q
A B
V = 0 \ W = Q × V
Also V
0 ¥ ¥ 0
= 4 ´ 1 . Q = =
a O a
^{0} 4pe _{0} a / 4pe _{0}a pe _{0}a
–Q
D a C Q Q
So, W =
pe _{0} a
Example: 58 Two point charge 100 mC and 5 mC are placed at point A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 mC from B to C, where BC = 30 cm, angle
ABC = p and 1 = 9 ´ 10^{9} Nm^{2} /C ^{2}
[MP PMT 1997]
2 4pe _{0}
 9 J (b)
81 J
20
(c)
9 J
25
(d)
 9 J
4
Solution: (d) Potential at B due to +100 mC charge is A
V = 9 ´ 10^{9} ´ 100 ´ 106
^{B} 40 ´ 10^{2}
= 9 ´ 10^{6} volt
4
+ 100 mC
Potential at C due to +100 mC charge is
VC = 9 ´ 10^{9}
´ 100 ´ 10^{6} 50 ´ 10^{2}
= 9 ´ 10^{6}
5
volt
50 cm
Hence work done in moving charge +5mC from B to C
W = 5 ´ 10^{–}^{6} (VC – VB )
p / 2
B
30 cm
+ 50 mC C
W = 5 ´ 10^{6} æ 9 ´ 10^{6} – 9 ´ 10^{+6} ö = – 9 J
ç ÷



è ø
Example: 59 There is an electric field E in xdirection. If the work done in moving a charge 0.2 C through a
distance of 2 metres along a line making an angle 60^{o} with the xaxis is 4J, what is the value of E [CBSE 1995]
 4 N/C (b) 8 N/C (c)
N/C
 20 N/C
Solution: (d) By using
W = q ´ DV
and DV = EDr cosq
So,
W = qE Dr cosq
W = 4 j = 0.2 ´ E ´ 2 ´ cos 60
Þ E = 20 N/C x
Example: 60 An electric charge of 20 mC is situated at the origin of X–Y coordinate system. The potential difference between the points. (5a, 0) and (– 3a, 4a) will be
 a (b) 2a (c) Zero (d) a
Solution: (c)
VA = kQ
5a
and VB = kQ
5a
B (–3a, 4a)
\ VA
 VB = 0
A (5a, 0)
Example: 61 Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q_{1} and Q_{2} are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is
æ Q1 ö
q(Q
– Q )(
– 1)
q(Q
 Q )
qç ÷(
Q
– 1)
(a) Zero (b)
1 2
4pe _{0} R
(c)
1 2
4pe _{0} R
(d)
è 2 ø
4pe _{0} R
Solution: (b) Potential at the centre of first ring VA =
Q1 +
4pe_{0}R
Q_{1} Q_{2}
Potential at the centre of second ring V
= Q2 + Q1
B
Potential difference between the two centres V
4pe _{0} R
 V
4pe _{0} 1 2
=
A B
R
\ Work done W =
Equilibrium of Charge.
(1) Definition : A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium.
 Type of equilibrium : Equilibrium can be divided in following type:
(i) Stable equilibrium : After displacing a charged particle from it’s equilibrium position, if it returns back then it is said to be in stable equilibrium. If U is the potential energy then in case of stable equilibrium
d ^{2}U dx ^{2}
is positive i.e., U is minimum.
 Unstable equilibrium : After displacing a charged particle from it’s equilibrium position, if it never
returns back then it is said to be in unstable equilibrium and in unstable equilibrium maximum.
d ^{2}U dx ^{2}
is negative i.e., U is
(iii) Neutral equilibrium : After displacing a charged particle from it’s equilibrium position if it neither comes back, nor moves away but remains in the position in which it was kept it is said to be in neutral
equilibrium and in neutral equilibrium
d ^{2}U dx ^{2}
is zero i.e., U is constant
 Guidelines to check the equilibrium
(i) Identify the charge for which equilibrium is to be analysed.
 Check, how many forces acting on that particular
(iii) There should be atleast two forces acts oppositely on that charge.
 If magnitude of these forces are equal then charge is said to be in equilibrium then identify the nature of
 If all the charges of system are in equilibrium then system is said to be in equilibrium
(4) Different cases of equilibrium of charge
Case – 1 : Suppose three similar charge
Q1 , q
and
Case – 2 : Two similar charge
Q1 and Q2
are
Q_{2} are placed along a straight line as shown below
placed along a straight line at a distance x from each other and a third dissimilar charge q is placed in between them as shown below
Charge q will be in equilibrium if  F1  =  F2 
Charge q will be in equilibrium if  F1  =  F2 


2
Q æ x ö 2
i.e.,
_{ }Q1 = æ x1 ö .
i.e.,
1 = ç 1 ÷
; This is the condition of
Q2 è x 2 ø
Q2 è x 2 ø
equilibrium of charge q. After following the guidelines we can say that charge q is in stable equilibrium and this system is not in equilibrium
Note : q Same short trick can be used here to find the position of charge q as we discussed in Case–1 i.e.,
Note : q
x = x
x = x
and x = x
^{1} 1 + Q /Q
2 1
^{1} 1 + Q /Q
2 1
^{2} 1 + Q /Q
1 2
and
x 2 = 1 +
x
Q1 /Q2
□ It is very important to know that magnitude of charge q can be determined if one of the extreme
charge (either Q1 or Q2 ) is in equilibrium i.e. if Q2

e.g. if two charges +4mC and +16 mC are
separated by a distance of 30 cm from each other
is in equilibrium then  q = Q æ x 2 ö and if Q is
then for equilibrium a third charge should be placed
_{1} ç x ÷ _{1}


between them at a distance
æ x ö ^{2}
in equilibrium then
 q = Q_{2} ç ^{1} ÷
(It should be
x = 30
= 10 cm or x
= 20 cm
è x ø
^{1} 1 +
16 / 4 2
remember that sign of q is opposite to that of Q_{1} (or Q_{2} ) )
(5) Equilibrium of suspended charge in an electric field
(i)
Freely suspended charged particle : To suspend a charged a particle freely in air under the influence of electric field it’s downward weight should be balanced by upward electric force for example if a positive charge is suspended freely in an electric field as shown then
In equilibrium QE = mg
Þ E = mg
Q
Note : @In the above case if direction of electric field is suddenly reversed in any figure then acceleration of charge particle at that instant will be a = 2g.
 Charged particle suspended by a massless insulated string (like simple pendulum) : Consider a charged particle (like Bob) of mass m, having charge Q is suspended in an electric field as shown under the influence of electric It turned through an angle (say q) and comes in equilibrium.
So, in the position of equilibrium (O¢ position)
T sinq
T cosq
= QE
= mg
….(i)
….(ii)
By squaring and adding equation (i) and (ii) T =
Dividing equation (i) by (ii)
tanq
= QE mg
Þ θ = tan ^{1} QE
mg
 Equilibrium of suspended point charge system : Suppose two small balls having charge +Q on each are suspended by two strings of equal length l. Then for equilibrium position as shown in
T sinq = Fe
….(i)
T cosq
= mg
….(ii)
T ^{2} = (Fe )2 + (mg)2
and
tanq = Fe
mg
; here
Fe =
1 Q^{2}
4pe _{0} x ^{2}
and
x = l sinq 2
 Equilibrium of suspended point charge system in a liquid : In the previous discussion if point charge system is taken into a liquid of density r such that q remain same then
In equilibrium
Fe’ = T ‘ sinq
Fe’
and (mg – Vrg) = T ‘ cosq
Q 2
\ tanq
= =
(mg – Vrg) 4pe _{0}
K (mg – Vrg)x ^{2}
Fe Q ^{2}
When this system was in air tanq
= =
mg 4pe _{0} mgx ^{2}
\ So equating these two gives us
1 = 1
m k(m – Vr)
Þ K =
m m – Vr
= 1
æ1 – V r ö

ç ÷
è ø
If s is the density of material of ball then K = 1
æ1 – θ ö

ç ÷
è ø
QE Q
mg
Example: 62 A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium. If q is equal to
[CPMT 1999; MP PET 1999, MP PMT 1999; CBSE 1995; Bihar MEE 1995; IIT 1987]
 – Q
2
 Q
4
æ x / 2 ö ^{2}
 + Q
4
 + Q
2
Solution: (b) By using Tricky formula
q = Q ç ÷
è x ø
Þ q = Q
4
since q should be negative so q = – Q .
4