DC Pandey Solutions: Electrostatics- 1 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


In tro duc tory Ex er cise 21.1
1. No, be cause charged body can at tract an
un charged by in duc ing charge on it.
2. Yes.
3. On clear ing, a pho no graph re cord be comes
charged by fric tion.
4. No. of electrons in 3 g mole of hy dro gen atom
     = ´ ´ 3 10
23
6.022
\ q ne = = ´ ´ ´ ´
-
3 6.022 1.6 10 10
23 19
     = ´ 2.9 10
5
 C
In tro duc tory Ex er cise 21.2
1. F
q q
r
e
r
e
= × = ×
1
4
1
4
0
1 2
2
0
2
2
pe pe
  F
Gm m
r
g
=
1 2
2
 
F
F
e
Gm m
e
g
=
×
2
0 1 2
4pe
     =
´ ´ ´
´ ´ ´ ´ ´
-
- - -
9 10 10
10 10 10
9 19 2
11 31 2
( ) 1.6
6.67 9.11 1.67
7
   = ´ 2.27 10
39
2.     F
q q
r
= ×
1
4
0
1 2
2
pe
     e
p
0
1 2
2
4
=
q q
Fr
   [ ]
[ ]
[ ][ ]
e
0
2
2
=
q
F r
e
       =
-
[ ] IT
[MLT ][L]
2
2
2
       =
- -
[ ] M L T I
1 3 4 2
SI units of e
0
=
- -
C N m
2 1 2
.
3. Let us find net force on charge at A.
   F
q
a
AB
= ×
1
4
0
2
2
pe
  F
q
a
AC
= ×
1
4
0
2
2
pe
Net force on charge at A
F F F
A AB AC
= ° + ° cos cos 30 30
         =
×
3
4
2
0
2
q
a pe
A
q
B
q
C
q
Electrostatics
21
60°
F
AC
F
AB
A
F sin 30°
AB 
F sin 30°
AC 
F cos 30°
AB 
F cos 30°
AC 
Page 2


In tro duc tory Ex er cise 21.1
1. No, be cause charged body can at tract an
un charged by in duc ing charge on it.
2. Yes.
3. On clear ing, a pho no graph re cord be comes
charged by fric tion.
4. No. of electrons in 3 g mole of hy dro gen atom
     = ´ ´ 3 10
23
6.022
\ q ne = = ´ ´ ´ ´
-
3 6.022 1.6 10 10
23 19
     = ´ 2.9 10
5
 C
In tro duc tory Ex er cise 21.2
1. F
q q
r
e
r
e
= × = ×
1
4
1
4
0
1 2
2
0
2
2
pe pe
  F
Gm m
r
g
=
1 2
2
 
F
F
e
Gm m
e
g
=
×
2
0 1 2
4pe
     =
´ ´ ´
´ ´ ´ ´ ´
-
- - -
9 10 10
10 10 10
9 19 2
11 31 2
( ) 1.6
6.67 9.11 1.67
7
   = ´ 2.27 10
39
2.     F
q q
r
= ×
1
4
0
1 2
2
pe
     e
p
0
1 2
2
4
=
q q
Fr
   [ ]
[ ]
[ ][ ]
e
0
2
2
=
q
F r
e
       =
-
[ ] IT
[MLT ][L]
2
2
2
       =
- -
[ ] M L T I
1 3 4 2
SI units of e
0
=
- -
C N m
2 1 2
.
3. Let us find net force on charge at A.
   F
q
a
AB
= ×
1
4
0
2
2
pe
  F
q
a
AC
= ×
1
4
0
2
2
pe
Net force on charge at A
F F F
A AB AC
= ° + ° cos cos 30 30
         =
×
3
4
2
0
2
q
a pe
A
q
B
q
C
q
Electrostatics
21
60°
F
AC
F
AB
A
F sin 30°
AB 
F sin 30°
AC 
F cos 30°
AB 
F cos 30°
AC 
4. F F
® ®
= OA OC
and F F
® ®
= - OB OD
Hence, net force on charge at centre is zero.
5. No. In case of in duc tion while charge co mes
closer and like charge moves fur ther from
the source.
The cause of attraction is more attractive
force due to small distance. But if
electrostatic force becomes independent of
distance, attractive force will become equal
to repulsive force, hence net force becomes
zero.
6. When the charged glass rod is brought near
the metal sphere, neg a tive in duces on the
por tion of sphere near the charge, hence it
get at tracted. But when the sphere touches
the rod it be comes pos i tively charged due to
con duc tion and gets re pelled by the rod.
7. Yes as q e
min
= 
F
e
r
min
= ×
1
4
0
2
2
pe
8. No. Elec tro stati c force is in de pend ent of
pres ence or ab sence of other charges.
9. F F
21 12
4 3
® ®
= - = - + ( )
^ ^
i j N.
Introductory Exercise 21.3
1. False.  E
q
r
= ×
1
4
0
2
pe
2. V V
A B
> as elec tric lines of force move from
higher po ten tial to lower po ten tial.
3. False. Pos i tively charged par ti cle moves in
the di rec tion of elec tric field while neg a tively 
charged par ti cle moves op po site to the
di rec tion of elec tric field.
4. False. Di rec tion of mo tion can be dif fer ent
from di rec tion of force.
5. E =
s
e
0
 Þ s e = = ´ ´
-
E
0
12
10 3.0 8.85
         = ´
-
2.655 10
11
 C/m
2
6. q
1
 and q
3
 are pos i tively charged as lines of
force are di rected away from q
1
 and q
3
. q
2
 is
neg a tively charged be cause elec tric field
lines are to wards q
2
.
7. If a charge q is placed at A also net field at
cen tre will be zero.
Hence net field at O is same as produced by 
A done but in opposite direction,02 i.e.,
E
q
a
= ×
1
4
2
pe
8. Net field at the cen tre (O) of wire is zero. If a
small length of the wire is cut-off, net field
will be equal to the field
due to cut-off por tion, i.e.,
  dE
dq
R
= ×
1
4
0
2
pe
     = ×
1
4
2
0
2
pe
p
q
R
dl
R
     =
qdl
R 8
2
0
3
p e
31  
O
A
q
B
q
q
D
q
C
–q
+
+
+
+
–
–
–
–
+++++
A
E
B
D
C
q
q
q
O
q
O
R
Page 3


In tro duc tory Ex er cise 21.1
1. No, be cause charged body can at tract an
un charged by in duc ing charge on it.
2. Yes.
3. On clear ing, a pho no graph re cord be comes
charged by fric tion.
4. No. of electrons in 3 g mole of hy dro gen atom
     = ´ ´ 3 10
23
6.022
\ q ne = = ´ ´ ´ ´
-
3 6.022 1.6 10 10
23 19
     = ´ 2.9 10
5
 C
In tro duc tory Ex er cise 21.2
1. F
q q
r
e
r
e
= × = ×
1
4
1
4
0
1 2
2
0
2
2
pe pe
  F
Gm m
r
g
=
1 2
2
 
F
F
e
Gm m
e
g
=
×
2
0 1 2
4pe
     =
´ ´ ´
´ ´ ´ ´ ´
-
- - -
9 10 10
10 10 10
9 19 2
11 31 2
( ) 1.6
6.67 9.11 1.67
7
   = ´ 2.27 10
39
2.     F
q q
r
= ×
1
4
0
1 2
2
pe
     e
p
0
1 2
2
4
=
q q
Fr
   [ ]
[ ]
[ ][ ]
e
0
2
2
=
q
F r
e
       =
-
[ ] IT
[MLT ][L]
2
2
2
       =
- -
[ ] M L T I
1 3 4 2
SI units of e
0
=
- -
C N m
2 1 2
.
3. Let us find net force on charge at A.
   F
q
a
AB
= ×
1
4
0
2
2
pe
  F
q
a
AC
= ×
1
4
0
2
2
pe
Net force on charge at A
F F F
A AB AC
= ° + ° cos cos 30 30
         =
×
3
4
2
0
2
q
a pe
A
q
B
q
C
q
Electrostatics
21
60°
F
AC
F
AB
A
F sin 30°
AB 
F sin 30°
AC 
F cos 30°
AB 
F cos 30°
AC 
4. F F
® ®
= OA OC
and F F
® ®
= - OB OD
Hence, net force on charge at centre is zero.
5. No. In case of in duc tion while charge co mes
closer and like charge moves fur ther from
the source.
The cause of attraction is more attractive
force due to small distance. But if
electrostatic force becomes independent of
distance, attractive force will become equal
to repulsive force, hence net force becomes
zero.
6. When the charged glass rod is brought near
the metal sphere, neg a tive in duces on the
por tion of sphere near the charge, hence it
get at tracted. But when the sphere touches
the rod it be comes pos i tively charged due to
con duc tion and gets re pelled by the rod.
7. Yes as q e
min
= 
F
e
r
min
= ×
1
4
0
2
2
pe
8. No. Elec tro stati c force is in de pend ent of
pres ence or ab sence of other charges.
9. F F
21 12
4 3
® ®
= - = - + ( )
^ ^
i j N.
Introductory Exercise 21.3
1. False.  E
q
r
= ×
1
4
0
2
pe
2. V V
A B
> as elec tric lines of force move from
higher po ten tial to lower po ten tial.
3. False. Pos i tively charged par ti cle moves in
the di rec tion of elec tric field while neg a tively 
charged par ti cle moves op po site to the
di rec tion of elec tric field.
4. False. Di rec tion of mo tion can be dif fer ent
from di rec tion of force.
5. E =
s
e
0
 Þ s e = = ´ ´
-
E
0
12
10 3.0 8.85
         = ´
-
2.655 10
11
 C/m
2
6. q
1
 and q
3
 are pos i tively charged as lines of
force are di rected away from q
1
 and q
3
. q
2
 is
neg a tively charged be cause elec tric field
lines are to wards q
2
.
7. If a charge q is placed at A also net field at
cen tre will be zero.
Hence net field at O is same as produced by 
A done but in opposite direction,02 i.e.,
E
q
a
= ×
1
4
2
pe
8. Net field at the cen tre (O) of wire is zero. If a
small length of the wire is cut-off, net field
will be equal to the field
due to cut-off por tion, i.e.,
  dE
dq
R
= ×
1
4
0
2
pe
     = ×
1
4
2
0
2
pe
p
q
R
dl
R
     =
qdl
R 8
2
0
3
p e
31  
O
A
q
B
q
q
D
q
C
–q
+
+
+
+
–
–
–
–
+++++
A
E
B
D
C
q
q
q
O
q
O
R
9.             E r
®
®
= ×
1
4
0
3
pe
q
r
        =-
´ ´ ´
+
+
-
9 10 2 10
3 4
3 4
9 6
2 232
( )
( )
/
^ ^
i j =- + 1443 4 ( )
^ ^
i j N/C
Introductory Exercise 21.4
1. Gain in KE = loss of PE
1
2
1
4
1 1
2
0
1 2
1 2
mv q q
r r
= × -
æ
è
ç
ç
ö
ø
÷
÷
pe
1
2
10
4 2
´
-
v
= - ´ ´ ´ ´ ´ -
æ
è
ç
ö
ø
÷
- -
1 10 2 10 9 10
1
1
1
6 6 9
0.5
v
2
360 =
      v=6 10 ms
-1
2. W q V V
A B
= - ( )
= ´ ×
- ´
- ×
- ´ æ
è
ç
ç
ö
ø
÷
÷
-
- -
2 10
1
4
1 10
1
1
4
1 10
2
6
0
6
0
6
pe pe
=- ´
-
9 10
3
J 
=-9mJ
3. When ever work is done by elec tric force,
po ten tial en ergy is de creased.
W U = - D
U U W
2 1
8
1 0 = - = - ´
-
8 . 6 J
4. No. As U
q q
r
=
1 2
0
4pe
If there are three particles
U
q q
r
q q
r
q q
r
= × + +
é
ë
ê
ê
ù
û
ú
ú
1
4
0
1 2
12
2 3
23
3 1
31
pe
Here U may be zero.
In case of more than two particles PE of
systems may same as if they were separated
by infinite distance but not in case of two
particles.
Introductory Exercise 21.5
1. V
W
q
ba
a b
= = ´
®
12 10
2
 = 1200 V
2. l a = x
(a) SI Units of l = C/m
             a
l
=
x
Hence SI unit of a =
C/ m
m
 = C/m
2
.
(b) Consider an elementary portion of rod at
a distance x from origin having length
dx. Electric potential at P due to this
element.
dV
dx
x d
= ×
+
1
4
0
pe
l
Net electric potential at P
V
dx
x d
L
= ×
+
ò
1
4
0
0
pe
l
Þ
a
pe 4
0
0
×
+
ò
x dx
x d
L
      = × -
+
é
ë
ê
ê
ù
û
ú
ú
ò ò
a
pe 4
0
0 0
dx d
dx
x d
L L
      = × - +
a
pe 4
0
0 0
[[ ] [ln( )] ] x d x d
L L
      = × -
+
é
ë
ê
ù
û
ú
a
pe 4
0
L d
L d
d
ln
3. Con sider an el e men tary por tion of length d x
at a dis tance x fro my centre O of the rod.
Electric potential at P due to this element,
dV
dx
d x
= ×
+
1
4
0
2 2
pe
l
     V
dx
d x
l
l
= ×
+
-
ò
l
pe 4
0
2 2
 32
L
x
d
P
2l
x
d
d
P
O d–x
Page 4


In tro duc tory Ex er cise 21.1
1. No, be cause charged body can at tract an
un charged by in duc ing charge on it.
2. Yes.
3. On clear ing, a pho no graph re cord be comes
charged by fric tion.
4. No. of electrons in 3 g mole of hy dro gen atom
     = ´ ´ 3 10
23
6.022
\ q ne = = ´ ´ ´ ´
-
3 6.022 1.6 10 10
23 19
     = ´ 2.9 10
5
 C
In tro duc tory Ex er cise 21.2
1. F
q q
r
e
r
e
= × = ×
1
4
1
4
0
1 2
2
0
2
2
pe pe
  F
Gm m
r
g
=
1 2
2
 
F
F
e
Gm m
e
g
=
×
2
0 1 2
4pe
     =
´ ´ ´
´ ´ ´ ´ ´
-
- - -
9 10 10
10 10 10
9 19 2
11 31 2
( ) 1.6
6.67 9.11 1.67
7
   = ´ 2.27 10
39
2.     F
q q
r
= ×
1
4
0
1 2
2
pe
     e
p
0
1 2
2
4
=
q q
Fr
   [ ]
[ ]
[ ][ ]
e
0
2
2
=
q
F r
e
       =
-
[ ] IT
[MLT ][L]
2
2
2
       =
- -
[ ] M L T I
1 3 4 2
SI units of e
0
=
- -
C N m
2 1 2
.
3. Let us find net force on charge at A.
   F
q
a
AB
= ×
1
4
0
2
2
pe
  F
q
a
AC
= ×
1
4
0
2
2
pe
Net force on charge at A
F F F
A AB AC
= ° + ° cos cos 30 30
         =
×
3
4
2
0
2
q
a pe
A
q
B
q
C
q
Electrostatics
21
60°
F
AC
F
AB
A
F sin 30°
AB 
F sin 30°
AC 
F cos 30°
AB 
F cos 30°
AC 
4. F F
® ®
= OA OC
and F F
® ®
= - OB OD
Hence, net force on charge at centre is zero.
5. No. In case of in duc tion while charge co mes
closer and like charge moves fur ther from
the source.
The cause of attraction is more attractive
force due to small distance. But if
electrostatic force becomes independent of
distance, attractive force will become equal
to repulsive force, hence net force becomes
zero.
6. When the charged glass rod is brought near
the metal sphere, neg a tive in duces on the
por tion of sphere near the charge, hence it
get at tracted. But when the sphere touches
the rod it be comes pos i tively charged due to
con duc tion and gets re pelled by the rod.
7. Yes as q e
min
= 
F
e
r
min
= ×
1
4
0
2
2
pe
8. No. Elec tro stati c force is in de pend ent of
pres ence or ab sence of other charges.
9. F F
21 12
4 3
® ®
= - = - + ( )
^ ^
i j N.
Introductory Exercise 21.3
1. False.  E
q
r
= ×
1
4
0
2
pe
2. V V
A B
> as elec tric lines of force move from
higher po ten tial to lower po ten tial.
3. False. Pos i tively charged par ti cle moves in
the di rec tion of elec tric field while neg a tively 
charged par ti cle moves op po site to the
di rec tion of elec tric field.
4. False. Di rec tion of mo tion can be dif fer ent
from di rec tion of force.
5. E =
s
e
0
 Þ s e = = ´ ´
-
E
0
12
10 3.0 8.85
         = ´
-
2.655 10
11
 C/m
2
6. q
1
 and q
3
 are pos i tively charged as lines of
force are di rected away from q
1
 and q
3
. q
2
 is
neg a tively charged be cause elec tric field
lines are to wards q
2
.
7. If a charge q is placed at A also net field at
cen tre will be zero.
Hence net field at O is same as produced by 
A done but in opposite direction,02 i.e.,
E
q
a
= ×
1
4
2
pe
8. Net field at the cen tre (O) of wire is zero. If a
small length of the wire is cut-off, net field
will be equal to the field
due to cut-off por tion, i.e.,
  dE
dq
R
= ×
1
4
0
2
pe
     = ×
1
4
2
0
2
pe
p
q
R
dl
R
     =
qdl
R 8
2
0
3
p e
31  
O
A
q
B
q
q
D
q
C
–q
+
+
+
+
–
–
–
–
+++++
A
E
B
D
C
q
q
q
O
q
O
R
9.             E r
®
®
= ×
1
4
0
3
pe
q
r
        =-
´ ´ ´
+
+
-
9 10 2 10
3 4
3 4
9 6
2 232
( )
( )
/
^ ^
i j =- + 1443 4 ( )
^ ^
i j N/C
Introductory Exercise 21.4
1. Gain in KE = loss of PE
1
2
1
4
1 1
2
0
1 2
1 2
mv q q
r r
= × -
æ
è
ç
ç
ö
ø
÷
÷
pe
1
2
10
4 2
´
-
v
= - ´ ´ ´ ´ ´ -
æ
è
ç
ö
ø
÷
- -
1 10 2 10 9 10
1
1
1
6 6 9
0.5
v
2
360 =
      v=6 10 ms
-1
2. W q V V
A B
= - ( )
= ´ ×
- ´
- ×
- ´ æ
è
ç
ç
ö
ø
÷
÷
-
- -
2 10
1
4
1 10
1
1
4
1 10
2
6
0
6
0
6
pe pe
=- ´
-
9 10
3
J 
=-9mJ
3. When ever work is done by elec tric force,
po ten tial en ergy is de creased.
W U = - D
U U W
2 1
8
1 0 = - = - ´
-
8 . 6 J
4. No. As U
q q
r
=
1 2
0
4pe
If there are three particles
U
q q
r
q q
r
q q
r
= × + +
é
ë
ê
ê
ù
û
ú
ú
1
4
0
1 2
12
2 3
23
3 1
31
pe
Here U may be zero.
In case of more than two particles PE of
systems may same as if they were separated
by infinite distance but not in case of two
particles.
Introductory Exercise 21.5
1. V
W
q
ba
a b
= = ´
®
12 10
2
 = 1200 V
2. l a = x
(a) SI Units of l = C/m
             a
l
=
x
Hence SI unit of a =
C/ m
m
 = C/m
2
.
(b) Consider an elementary portion of rod at
a distance x from origin having length
dx. Electric potential at P due to this
element.
dV
dx
x d
= ×
+
1
4
0
pe
l
Net electric potential at P
V
dx
x d
L
= ×
+
ò
1
4
0
0
pe
l
Þ
a
pe 4
0
0
×
+
ò
x dx
x d
L
      = × -
+
é
ë
ê
ê
ù
û
ú
ú
ò ò
a
pe 4
0
0 0
dx d
dx
x d
L L
      = × - +
a
pe 4
0
0 0
[[ ] [ln( )] ] x d x d
L L
      = × -
+
é
ë
ê
ù
û
ú
a
pe 4
0
L d
L d
d
ln
3. Con sider an el e men tary por tion of length d x
at a dis tance x fro my centre O of the rod.
Electric potential at P due to this element,
dV
dx
d x
= ×
+
1
4
0
2 2
pe
l
     V
dx
d x
l
l
= ×
+
-
ò
l
pe 4
0
2 2
 32
L
x
d
P
2l
x
d
d
P
O d–x
= ×
é
ë
ê
ù
û
ú
-
-
l
pe 4
0
1
sin
x
d
l
l
    =
×
´
-
q
l
x
d 4 2
2
0
1
pe
sin
          V
q
l
x
d
=
-
4
0
1
pe
sin
4. Con sider the cone to be made up of large
num ber of el e men tary rings.
 
Consider one such ring of radius x and
thickness dl. Let q be the semi-vertical angle
of cone and R be the radius of cone.
Charge on the elementary ring;
dQ dA
Q
RL
x dl = = × s
p
p 2
or       dQ
Ql
RL
dl =
2 sinq
Potential at O due to this ring
dV
dQ
l
= ×
1
4
0
pe
    =
Q
RL
dl
sinq
pe 2
0
Total potential at O
V
Q
RL
dl
QL
RL
L
= =
ò
sin sin q
pe
q
pe 2 2
0
0
0
         =
Q
L 2
0
pe
 [L R sin q = ]
       U qV =
         =
Qq
L 2
0
pe
Introductory Exercise 21.6
1. (a) V a x y = - ( )
2 2
   E
v
x
v
y
ax y = -
¶
¶
=
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
= - + i j i j
^ ^ ^ ^
2 2
(b) V axy =
   E
v
x
v
y
a y x = -
¶
¶
=
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
= - + i j i j
^ ^ ^ ^
( )
2. From x = -2 to x = 0 & x = 2 to x = 4 
V is increasing uniformly.
Hence, E is uniform and negative
From x = 0 to x = 2
V is constant hence E is zero.
For x > 4
V is decreasing at constant rate, hence E is
positive.
3. E
dV
dr
= - = -
-
-
=
( ) 50 100
5 0
10 V/m
True.
4. (a) V V
P D
- = × =
® ®
E l 0
(b) V V
P C
- = × = ´ ´ °
® ®
E l 20 1 0 cos
= 20 V
(c) V V
B D
- = - ´ = - 20 1 20 V
(d) V V
C D
- = - ´ = - 20 1 20 V
33  
–2 O 2 4
x
q
x
R
L
l
dl
O
A B
D C
1m
1m
E = 20V/m
®
Page 5


In tro duc tory Ex er cise 21.1
1. No, be cause charged body can at tract an
un charged by in duc ing charge on it.
2. Yes.
3. On clear ing, a pho no graph re cord be comes
charged by fric tion.
4. No. of electrons in 3 g mole of hy dro gen atom
     = ´ ´ 3 10
23
6.022
\ q ne = = ´ ´ ´ ´
-
3 6.022 1.6 10 10
23 19
     = ´ 2.9 10
5
 C
In tro duc tory Ex er cise 21.2
1. F
q q
r
e
r
e
= × = ×
1
4
1
4
0
1 2
2
0
2
2
pe pe
  F
Gm m
r
g
=
1 2
2
 
F
F
e
Gm m
e
g
=
×
2
0 1 2
4pe
     =
´ ´ ´
´ ´ ´ ´ ´
-
- - -
9 10 10
10 10 10
9 19 2
11 31 2
( ) 1.6
6.67 9.11 1.67
7
   = ´ 2.27 10
39
2.     F
q q
r
= ×
1
4
0
1 2
2
pe
     e
p
0
1 2
2
4
=
q q
Fr
   [ ]
[ ]
[ ][ ]
e
0
2
2
=
q
F r
e
       =
-
[ ] IT
[MLT ][L]
2
2
2
       =
- -
[ ] M L T I
1 3 4 2
SI units of e
0
=
- -
C N m
2 1 2
.
3. Let us find net force on charge at A.
   F
q
a
AB
= ×
1
4
0
2
2
pe
  F
q
a
AC
= ×
1
4
0
2
2
pe
Net force on charge at A
F F F
A AB AC
= ° + ° cos cos 30 30
         =
×
3
4
2
0
2
q
a pe
A
q
B
q
C
q
Electrostatics
21
60°
F
AC
F
AB
A
F sin 30°
AB 
F sin 30°
AC 
F cos 30°
AB 
F cos 30°
AC 
4. F F
® ®
= OA OC
and F F
® ®
= - OB OD
Hence, net force on charge at centre is zero.
5. No. In case of in duc tion while charge co mes
closer and like charge moves fur ther from
the source.
The cause of attraction is more attractive
force due to small distance. But if
electrostatic force becomes independent of
distance, attractive force will become equal
to repulsive force, hence net force becomes
zero.
6. When the charged glass rod is brought near
the metal sphere, neg a tive in duces on the
por tion of sphere near the charge, hence it
get at tracted. But when the sphere touches
the rod it be comes pos i tively charged due to
con duc tion and gets re pelled by the rod.
7. Yes as q e
min
= 
F
e
r
min
= ×
1
4
0
2
2
pe
8. No. Elec tro stati c force is in de pend ent of
pres ence or ab sence of other charges.
9. F F
21 12
4 3
® ®
= - = - + ( )
^ ^
i j N.
Introductory Exercise 21.3
1. False.  E
q
r
= ×
1
4
0
2
pe
2. V V
A B
> as elec tric lines of force move from
higher po ten tial to lower po ten tial.
3. False. Pos i tively charged par ti cle moves in
the di rec tion of elec tric field while neg a tively 
charged par ti cle moves op po site to the
di rec tion of elec tric field.
4. False. Di rec tion of mo tion can be dif fer ent
from di rec tion of force.
5. E =
s
e
0
 Þ s e = = ´ ´
-
E
0
12
10 3.0 8.85
         = ´
-
2.655 10
11
 C/m
2
6. q
1
 and q
3
 are pos i tively charged as lines of
force are di rected away from q
1
 and q
3
. q
2
 is
neg a tively charged be cause elec tric field
lines are to wards q
2
.
7. If a charge q is placed at A also net field at
cen tre will be zero.
Hence net field at O is same as produced by 
A done but in opposite direction,02 i.e.,
E
q
a
= ×
1
4
2
pe
8. Net field at the cen tre (O) of wire is zero. If a
small length of the wire is cut-off, net field
will be equal to the field
due to cut-off por tion, i.e.,
  dE
dq
R
= ×
1
4
0
2
pe
     = ×
1
4
2
0
2
pe
p
q
R
dl
R
     =
qdl
R 8
2
0
3
p e
31  
O
A
q
B
q
q
D
q
C
–q
+
+
+
+
–
–
–
–
+++++
A
E
B
D
C
q
q
q
O
q
O
R
9.             E r
®
®
= ×
1
4
0
3
pe
q
r
        =-
´ ´ ´
+
+
-
9 10 2 10
3 4
3 4
9 6
2 232
( )
( )
/
^ ^
i j =- + 1443 4 ( )
^ ^
i j N/C
Introductory Exercise 21.4
1. Gain in KE = loss of PE
1
2
1
4
1 1
2
0
1 2
1 2
mv q q
r r
= × -
æ
è
ç
ç
ö
ø
÷
÷
pe
1
2
10
4 2
´
-
v
= - ´ ´ ´ ´ ´ -
æ
è
ç
ö
ø
÷
- -
1 10 2 10 9 10
1
1
1
6 6 9
0.5
v
2
360 =
      v=6 10 ms
-1
2. W q V V
A B
= - ( )
= ´ ×
- ´
- ×
- ´ æ
è
ç
ç
ö
ø
÷
÷
-
- -
2 10
1
4
1 10
1
1
4
1 10
2
6
0
6
0
6
pe pe
=- ´
-
9 10
3
J 
=-9mJ
3. When ever work is done by elec tric force,
po ten tial en ergy is de creased.
W U = - D
U U W
2 1
8
1 0 = - = - ´
-
8 . 6 J
4. No. As U
q q
r
=
1 2
0
4pe
If there are three particles
U
q q
r
q q
r
q q
r
= × + +
é
ë
ê
ê
ù
û
ú
ú
1
4
0
1 2
12
2 3
23
3 1
31
pe
Here U may be zero.
In case of more than two particles PE of
systems may same as if they were separated
by infinite distance but not in case of two
particles.
Introductory Exercise 21.5
1. V
W
q
ba
a b
= = ´
®
12 10
2
 = 1200 V
2. l a = x
(a) SI Units of l = C/m
             a
l
=
x
Hence SI unit of a =
C/ m
m
 = C/m
2
.
(b) Consider an elementary portion of rod at
a distance x from origin having length
dx. Electric potential at P due to this
element.
dV
dx
x d
= ×
+
1
4
0
pe
l
Net electric potential at P
V
dx
x d
L
= ×
+
ò
1
4
0
0
pe
l
Þ
a
pe 4
0
0
×
+
ò
x dx
x d
L
      = × -
+
é
ë
ê
ê
ù
û
ú
ú
ò ò
a
pe 4
0
0 0
dx d
dx
x d
L L
      = × - +
a
pe 4
0
0 0
[[ ] [ln( )] ] x d x d
L L
      = × -
+
é
ë
ê
ù
û
ú
a
pe 4
0
L d
L d
d
ln
3. Con sider an el e men tary por tion of length d x
at a dis tance x fro my centre O of the rod.
Electric potential at P due to this element,
dV
dx
d x
= ×
+
1
4
0
2 2
pe
l
     V
dx
d x
l
l
= ×
+
-
ò
l
pe 4
0
2 2
 32
L
x
d
P
2l
x
d
d
P
O d–x
= ×
é
ë
ê
ù
û
ú
-
-
l
pe 4
0
1
sin
x
d
l
l
    =
×
´
-
q
l
x
d 4 2
2
0
1
pe
sin
          V
q
l
x
d
=
-
4
0
1
pe
sin
4. Con sider the cone to be made up of large
num ber of el e men tary rings.
 
Consider one such ring of radius x and
thickness dl. Let q be the semi-vertical angle
of cone and R be the radius of cone.
Charge on the elementary ring;
dQ dA
Q
RL
x dl = = × s
p
p 2
or       dQ
Ql
RL
dl =
2 sinq
Potential at O due to this ring
dV
dQ
l
= ×
1
4
0
pe
    =
Q
RL
dl
sinq
pe 2
0
Total potential at O
V
Q
RL
dl
QL
RL
L
= =
ò
sin sin q
pe
q
pe 2 2
0
0
0
         =
Q
L 2
0
pe
 [L R sin q = ]
       U qV =
         =
Qq
L 2
0
pe
Introductory Exercise 21.6
1. (a) V a x y = - ( )
2 2
   E
v
x
v
y
ax y = -
¶
¶
=
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
= - + i j i j
^ ^ ^ ^
2 2
(b) V axy =
   E
v
x
v
y
a y x = -
¶
¶
=
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
= - + i j i j
^ ^ ^ ^
( )
2. From x = -2 to x = 0 & x = 2 to x = 4 
V is increasing uniformly.
Hence, E is uniform and negative
From x = 0 to x = 2
V is constant hence E is zero.
For x > 4
V is decreasing at constant rate, hence E is
positive.
3. E
dV
dr
= - = -
-
-
=
( ) 50 100
5 0
10 V/m
True.
4. (a) V V
P D
- = × =
® ®
E l 0
(b) V V
P C
- = × = ´ ´ °
® ®
E l 20 1 0 cos
= 20 V
(c) V V
B D
- = - ´ = - 20 1 20 V
(d) V V
C D
- = - ´ = - 20 1 20 V
33  
–2 O 2 4
x
q
x
R
L
l
dl
O
A B
D C
1m
1m
E = 20V/m
®
Introductory Exercise 21.7
1. F qE
1
= to wards right
F qE
2
= towards left
Net torque about q,
t q q = - + qE l x qEx ( ) sin sin 2
        = = q l E pE ( ) sin sin 2 q q
      t
® ®
®
= ´ p E
2.     E
q
y a
1
0
2 2 2
1
4
= ×
+
pe
( )
     E
q
y a
2
0
2 2 2
1
4
= ×
+
pe
( )
     E
q
y
3
0
2
1
4
2
= ×
pe
Net field at P
E j
®
= - - - ( cos cos )
^
E E E
3 1 2
q q                
=
-
-
+
-
+
é
ë
ê
ê
ù
û
ú
ú
q
y y a y a 4
2
0
2 2 2 2 2
pe
q q cos cos
^
j
= - -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
1
0
2 2 2 3 2
q
y
y
y a pe ( )
/
^
j        
= -
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2 3 2 3
2 2 2 3 2
q y a y
y y a pe
( )
( )
/
/
^
j         
= -
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
2
4
1
0
3
2
2
3 2
3
2 2 2 3 2
q
y
q
y
y
y y a pe
/
/
( )
ù
û
ú
ú
ú
ú
ú
ú
j
^
     
As y a >>
E
q
y
q
y
y
y
= - ×
+
æ
è
ç
ç
ö
ø
÷
÷
-
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
2
4
1
3
2
0
3
2
2
3
5
pe
j
^
            
E
qa
y
= -
3
4
2
0
4
pe
j
^
                                              
Introductory Exercise 21.8
1. (a) Charge q is com pletely the hemi sphere
    hence flux through hemi sphere is zero.
(b) Charge inside the sphere is q hence flux
through hemisphere 
f =
q
e
0
(c) As charge q is at the surface, net flux
through hemisphere
f =
q
2
0
e
2. When charge is at any of the ver tex, net flux
through the cube,
f =
q
8
0
e
If charge q is at D,
flux through three faces containing D is zero
and the flux f is divided equal among other
three faces, hence
f =
f
=
EFGH
1
2
0
q
pe
and f =
AEHD
0             
3. True. As elec tric field is uni form, flux
en ter ing the cube will be equal to flux
leav ing it.
\ f =
net
0 Þ f =
net
q
e
0
   Þ q =0
4. (a) As net charge in side hemi sphere is zero,
f + f =
1 2
0
But E is parallel to surface 2.
 34
E 
A
F
1
2l
B+q
Q
q
F
29
x
Q
2l
x
®
E 
1
2