DC Pandey Solutions: Electrostatics- 2 Notes | Study DC Pandey Solutions for JEE Physics - JEE

 Page 1


8. As ball are in equi lib rium
  F T
e
= sin a
mg T = cos a
F mg
e
= tan a
    q r
2
0
2
4 = pe a tan
Here,      r l =2 sina
q l
2
0
2 2
16 = pe a a sin tan
          q = ´
-
3.3 10
8
 C.
9. Same as Q.7. In tro duc tory Ex er cise 21.3.
10. See Q.7. In tro duc tory Ex er cise 21.3.
11.    E
q
r
= ×
® 1
4
1
0
3
pe
r
   =
´ ´ - ´
+
-
-
9 10 10
9 9
2 232
( )
(( ) ( ))
( )
/
^ ^
8.0
1.2 1.6
1.2 1.6 i j
   =- - 18 2( )
^ ^
1.2 1.6 i j N/C.
12. Con sider an el e men tary por tion on the ring
of length dl sub t end ing an gle df at cen tre ‘O’
of the ring.
Charge on this portion,
dq dl Rd = = f l l
\ dE
dq
R
d
R
= × =
f 1
4
1
4
0
2
0
pe pe
l
Here, dE sin f components of field will cancel 
each other.
Hence, Net field at O
E dE
R
d = f = × f f
ò ò
-
cos cos
/
/ 1
4
0
2
2
pe
l
p
p
= ×
1
4
2
0
pe
l
R
13. Con sider el e men tary por tion of the rod of
length dl at a dis tance l from the cen tre O of
the rod.
Charge on this portion
dq dl
Q
L
dl = = l    
\ dE
dq
a
= ×
f
1
4
0
2
pe ( ) sec
        = ×
f
1
4
0
2 2
pe
Q dl
La sec
Now,
l a = f tan  
Þ dl a d = f f sec
2
\ dE
Qd
La
= ×
f 1
4
0
pe
 
Net Electric field at P.
E dE = f
ò
cos
[ dEsin f components will cancel each other
as rod in symmetrical about P.]
= × f f
-
ò
1
4
0
pe
q
q Q
La
d cos
         = ×
1
4
2
0
pe
q Q
La
sin
But sinq=
+
æ
è
ç
ö
ø
÷
=
+
L
a
L
L
a L
2
2
4
2
2 2 2
\ E
Q
a a L
= ×
+
1
4
2
4 0
2 2
pe
 38
dE
dE cos f
dE cos f
dE
dE sin f dE sin f
O
f
R
df
dl dl
dE
dE cos f
dE cos f
dE
dE sin f
dE sin f
df
Q
P
dl
l
dl
f
O
a
O 
a
F
e F
e
mg
T sin a
mg
r
T cos a
T T
Page 2


8. As ball are in equi lib rium
  F T
e
= sin a
mg T = cos a
F mg
e
= tan a
    q r
2
0
2
4 = pe a tan
Here,      r l =2 sina
q l
2
0
2 2
16 = pe a a sin tan
          q = ´
-
3.3 10
8
 C.
9. Same as Q.7. In tro duc tory Ex er cise 21.3.
10. See Q.7. In tro duc tory Ex er cise 21.3.
11.    E
q
r
= ×
® 1
4
1
0
3
pe
r
   =
´ ´ - ´
+
-
-
9 10 10
9 9
2 232
( )
(( ) ( ))
( )
/
^ ^
8.0
1.2 1.6
1.2 1.6 i j
   =- - 18 2( )
^ ^
1.2 1.6 i j N/C.
12. Con sider an el e men tary por tion on the ring
of length dl sub t end ing an gle df at cen tre ‘O’
of the ring.
Charge on this portion,
dq dl Rd = = f l l
\ dE
dq
R
d
R
= × =
f 1
4
1
4
0
2
0
pe pe
l
Here, dE sin f components of field will cancel 
each other.
Hence, Net field at O
E dE
R
d = f = × f f
ò ò
-
cos cos
/
/ 1
4
0
2
2
pe
l
p
p
= ×
1
4
2
0
pe
l
R
13. Con sider el e men tary por tion of the rod of
length dl at a dis tance l from the cen tre O of
the rod.
Charge on this portion
dq dl
Q
L
dl = = l    
\ dE
dq
a
= ×
f
1
4
0
2
pe ( ) sec
        = ×
f
1
4
0
2 2
pe
Q dl
La sec
Now,
l a = f tan  
Þ dl a d = f f sec
2
\ dE
Qd
La
= ×
f 1
4
0
pe
 
Net Electric field at P.
E dE = f
ò
cos
[ dEsin f components will cancel each other
as rod in symmetrical about P.]
= × f f
-
ò
1
4
0
pe
q
q Q
La
d cos
         = ×
1
4
2
0
pe
q Q
La
sin
But sinq=
+
æ
è
ç
ö
ø
÷
=
+
L
a
L
L
a L
2
2
4
2
2 2 2
\ E
Q
a a L
= ×
+
1
4
2
4 0
2 2
pe
 38
dE
dE cos f
dE cos f
dE
dE sin f dE sin f
O
f
R
df
dl dl
dE
dE cos f
dE cos f
dE
dE sin f
dE sin f
df
Q
P
dl
l
dl
f
O
a
O 
a
F
e F
e
mg
T sin a
mg
r
T cos a
T T
14. (a) As shown in fig ure, di rec tion of elec tric
   field at P will be along + ve y-axis.
(b) Positive x-axis.
(c) Positive y-axis.
15. Let E
q
R
1
0
2
1
4
= ×
pe
Resultant fields of two opposite charges can
be shown as given in figure.
Clearly resultant field is along angle bisector 
of field towards 9 and 10.
Hence time shown by clock in the direction
of electric field is 9 : 30.
16. (a)  a
F
m
eE
m
= =
-
     =
- ´ ´ ´
´
-
-
1.6
9.1
10 1 10
10
19 3
31
     = - ´ 1.76 10
14
 ms
-2
u = ´ 5.00 10
8
 cm/s = ´ 5 10
6
 ms
-1
v = 0    
v u as
2 2
2 - =
       s =
´
´ ´
= ´ =
-
( ) 5 10
2 10
1 10
6 2
14
2
1.7
.4 1.4 cm
(b)  v u at = +
t =
´
´
= ´ =
-
5 10
10
10 28
6
14
8
1.76
2.8 ns.
(c) Dk = work done by electric field.
     = × = - F x eEx
= - ´ ´ ´ ´ ´
- -
1.6 10 1 10 8 10
19 3 3
     = - ´
-
1.28 J 10
18
Loss of KE = ´
-
1.28 J 10
18
17. Here, u u
x
= ° = cos45
25
2
 ms
-1
u u
y
= ° = sin45
25
2
 ms
-1
     a qE
x
= = ´ ´ ´
-
2 10 2 10
6 7
           =40 ms
-1
         a
y
= -10 ms
-1
          y u
a
t
yt
y
= +
1
2
          y t t = -
25
2
5
2
39  
E
1
E
E
2
P
x 
y
–Q
Q
y
+Q
+Q
E
2
E
1
E
x
P
–Q
+Q
E
1 E
2
E
P
12
E
12
2
3
4
5
6
7
8
9
10
11
E
1
E
2
E
3
E
4
E
5
E
6
E
7
E
8
E
9
E
10
E
11
1
6E
1
6E
1
6E
1
6E
1
E
6E
1
6E
1
q
u
E
Page 3


8. As ball are in equi lib rium
  F T
e
= sin a
mg T = cos a
F mg
e
= tan a
    q r
2
0
2
4 = pe a tan
Here,      r l =2 sina
q l
2
0
2 2
16 = pe a a sin tan
          q = ´
-
3.3 10
8
 C.
9. Same as Q.7. In tro duc tory Ex er cise 21.3.
10. See Q.7. In tro duc tory Ex er cise 21.3.
11.    E
q
r
= ×
® 1
4
1
0
3
pe
r
   =
´ ´ - ´
+
-
-
9 10 10
9 9
2 232
( )
(( ) ( ))
( )
/
^ ^
8.0
1.2 1.6
1.2 1.6 i j
   =- - 18 2( )
^ ^
1.2 1.6 i j N/C.
12. Con sider an el e men tary por tion on the ring
of length dl sub t end ing an gle df at cen tre ‘O’
of the ring.
Charge on this portion,
dq dl Rd = = f l l
\ dE
dq
R
d
R
= × =
f 1
4
1
4
0
2
0
pe pe
l
Here, dE sin f components of field will cancel 
each other.
Hence, Net field at O
E dE
R
d = f = × f f
ò ò
-
cos cos
/
/ 1
4
0
2
2
pe
l
p
p
= ×
1
4
2
0
pe
l
R
13. Con sider el e men tary por tion of the rod of
length dl at a dis tance l from the cen tre O of
the rod.
Charge on this portion
dq dl
Q
L
dl = = l    
\ dE
dq
a
= ×
f
1
4
0
2
pe ( ) sec
        = ×
f
1
4
0
2 2
pe
Q dl
La sec
Now,
l a = f tan  
Þ dl a d = f f sec
2
\ dE
Qd
La
= ×
f 1
4
0
pe
 
Net Electric field at P.
E dE = f
ò
cos
[ dEsin f components will cancel each other
as rod in symmetrical about P.]
= × f f
-
ò
1
4
0
pe
q
q Q
La
d cos
         = ×
1
4
2
0
pe
q Q
La
sin
But sinq=
+
æ
è
ç
ö
ø
÷
=
+
L
a
L
L
a L
2
2
4
2
2 2 2
\ E
Q
a a L
= ×
+
1
4
2
4 0
2 2
pe
 38
dE
dE cos f
dE cos f
dE
dE sin f dE sin f
O
f
R
df
dl dl
dE
dE cos f
dE cos f
dE
dE sin f
dE sin f
df
Q
P
dl
l
dl
f
O
a
O 
a
F
e F
e
mg
T sin a
mg
r
T cos a
T T
14. (a) As shown in fig ure, di rec tion of elec tric
   field at P will be along + ve y-axis.
(b) Positive x-axis.
(c) Positive y-axis.
15. Let E
q
R
1
0
2
1
4
= ×
pe
Resultant fields of two opposite charges can
be shown as given in figure.
Clearly resultant field is along angle bisector 
of field towards 9 and 10.
Hence time shown by clock in the direction
of electric field is 9 : 30.
16. (a)  a
F
m
eE
m
= =
-
     =
- ´ ´ ´
´
-
-
1.6
9.1
10 1 10
10
19 3
31
     = - ´ 1.76 10
14
 ms
-2
u = ´ 5.00 10
8
 cm/s = ´ 5 10
6
 ms
-1
v = 0    
v u as
2 2
2 - =
       s =
´
´ ´
= ´ =
-
( ) 5 10
2 10
1 10
6 2
14
2
1.7
.4 1.4 cm
(b)  v u at = +
t =
´
´
= ´ =
-
5 10
10
10 28
6
14
8
1.76
2.8 ns.
(c) Dk = work done by electric field.
     = × = - F x eEx
= - ´ ´ ´ ´ ´
- -
1.6 10 1 10 8 10
19 3 3
     = - ´
-
1.28 J 10
18
Loss of KE = ´
-
1.28 J 10
18
17. Here, u u
x
= ° = cos45
25
2
 ms
-1
u u
y
= ° = sin45
25
2
 ms
-1
     a qE
x
= = ´ ´ ´
-
2 10 2 10
6 7
           =40 ms
-1
         a
y
= -10 ms
-1
          y u
a
t
yt
y
= +
1
2
          y t t = -
25
2
5
2
39  
E
1
E
E
2
P
x 
y
–Q
Q
y
+Q
+Q
E
2
E
1
E
x
P
–Q
+Q
E
1 E
2
E
P
12
E
12
2
3
4
5
6
7
8
9
10
11
E
1
E
2
E
3
E
4
E
5
E
6
E
7
E
8
E
9
E
10
E
11
1
6E
1
6E
1
6E
1
6E
1
E
6E
1
6E
1
q
u
E
at the end of motion,
t T = and y = 0
\ T =
5
2
 s
Also at the end of motion,
         x R =
\         x u t a t
x x
= +
1
2
2
  R= ´ + ´
æ
è
ç
ö
ø
÷
25
2
5
2
20
5
2
2
           =312.5 m
18. (a)    R
qE
=
m q
2
2 sin
   sin2
2
q=
qER
mu
        =
´ ´ ´ ´
´ ´ ´
- -
-
1.6 1.27
1.67 9.55
10 720 10
10 10
19 3
27 3 2
( )
        =0.96
      2 88 q= ° or 92°
       q= ° 44 or 46°
      T
mh
E
=
2
2
sinq
 
        =
´ ´ ´ ´ ´
´ ´
-
-
2 10
1
2
10
10 720
3 31
19
9.55 1.67
1.6
        = ´
-
1.95 10
11
 s
19. (a) a
E
j
®
®
-
-
= - = -
´ ´
´
e
m
1.6
9.1
10 120
10
19
31
^
= - ´ 2.1 10
13
i
^
 m/s
(b) t
x
u
x
= =
´
´
= ´
-
-
D 2 10
10
4
3
10
2
5
7
1.5
 s
v u a t
y y y
= +
       = ´ ´ ´ ´ ´
-
3.0 2.1 10 10
4
3
10
6 13 7
        = ´ 0.2 10
6
 m/s
v i j
®
= ´ + ´ ( ) ( )
^ ^
1.5 0.2 10 10
5 6
20. Ab so lute po ten tial can be zero at two points
on the x-axis. One in be tween the charges
and other on the left of charge a
1
 (smaller in
mag ni tu de).
Case I.
In between two charges : let potential is zero
at a distance x from q
1
 towards q
2
.
V
q
x
q
x
= × + ×
-
=
1
4
1
4 100
0
0
1
0
2
pe pe
        = ×
´
- ×
´
-
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
Þ 200 2 3 - = x x
        x =20 cm
Case II.
Consider the potential is zero at a distance x
from charge q, on its left.
\   V
q
x
q
x
= × + ×
+
=
1
4
1
4 100
0
0
1
0
2
pe pe
    = ×
´
+ ×
- ´
+
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
200 2 3 + = x x
          x =200 cm
21. Let us first find the po ten tial at a point on
the per pen dic u lar bi sec tor of a line charge.
Consider a line of carrying a line charge
density l having length L.
Consider an elementary portion of length dl
on the rod. 
Charge on this portion
           dq dl = l
\ dV
dl
r
= ×
f
1
4
0
pe
l
sec
  
Now,         l r = f tan
            dl r d = f f sec
2
 40
O
100cm
q
1
q
2
X
q
1
q
2
x 100–x
q
1
q
2
x 100 cm
f
q
L
r
l
dl
Page 4


8. As ball are in equi lib rium
  F T
e
= sin a
mg T = cos a
F mg
e
= tan a
    q r
2
0
2
4 = pe a tan
Here,      r l =2 sina
q l
2
0
2 2
16 = pe a a sin tan
          q = ´
-
3.3 10
8
 C.
9. Same as Q.7. In tro duc tory Ex er cise 21.3.
10. See Q.7. In tro duc tory Ex er cise 21.3.
11.    E
q
r
= ×
® 1
4
1
0
3
pe
r
   =
´ ´ - ´
+
-
-
9 10 10
9 9
2 232
( )
(( ) ( ))
( )
/
^ ^
8.0
1.2 1.6
1.2 1.6 i j
   =- - 18 2( )
^ ^
1.2 1.6 i j N/C.
12. Con sider an el e men tary por tion on the ring
of length dl sub t end ing an gle df at cen tre ‘O’
of the ring.
Charge on this portion,
dq dl Rd = = f l l
\ dE
dq
R
d
R
= × =
f 1
4
1
4
0
2
0
pe pe
l
Here, dE sin f components of field will cancel 
each other.
Hence, Net field at O
E dE
R
d = f = × f f
ò ò
-
cos cos
/
/ 1
4
0
2
2
pe
l
p
p
= ×
1
4
2
0
pe
l
R
13. Con sider el e men tary por tion of the rod of
length dl at a dis tance l from the cen tre O of
the rod.
Charge on this portion
dq dl
Q
L
dl = = l    
\ dE
dq
a
= ×
f
1
4
0
2
pe ( ) sec
        = ×
f
1
4
0
2 2
pe
Q dl
La sec
Now,
l a = f tan  
Þ dl a d = f f sec
2
\ dE
Qd
La
= ×
f 1
4
0
pe
 
Net Electric field at P.
E dE = f
ò
cos
[ dEsin f components will cancel each other
as rod in symmetrical about P.]
= × f f
-
ò
1
4
0
pe
q
q Q
La
d cos
         = ×
1
4
2
0
pe
q Q
La
sin
But sinq=
+
æ
è
ç
ö
ø
÷
=
+
L
a
L
L
a L
2
2
4
2
2 2 2
\ E
Q
a a L
= ×
+
1
4
2
4 0
2 2
pe
 38
dE
dE cos f
dE cos f
dE
dE sin f dE sin f
O
f
R
df
dl dl
dE
dE cos f
dE cos f
dE
dE sin f
dE sin f
df
Q
P
dl
l
dl
f
O
a
O 
a
F
e F
e
mg
T sin a
mg
r
T cos a
T T
14. (a) As shown in fig ure, di rec tion of elec tric
   field at P will be along + ve y-axis.
(b) Positive x-axis.
(c) Positive y-axis.
15. Let E
q
R
1
0
2
1
4
= ×
pe
Resultant fields of two opposite charges can
be shown as given in figure.
Clearly resultant field is along angle bisector 
of field towards 9 and 10.
Hence time shown by clock in the direction
of electric field is 9 : 30.
16. (a)  a
F
m
eE
m
= =
-
     =
- ´ ´ ´
´
-
-
1.6
9.1
10 1 10
10
19 3
31
     = - ´ 1.76 10
14
 ms
-2
u = ´ 5.00 10
8
 cm/s = ´ 5 10
6
 ms
-1
v = 0    
v u as
2 2
2 - =
       s =
´
´ ´
= ´ =
-
( ) 5 10
2 10
1 10
6 2
14
2
1.7
.4 1.4 cm
(b)  v u at = +
t =
´
´
= ´ =
-
5 10
10
10 28
6
14
8
1.76
2.8 ns.
(c) Dk = work done by electric field.
     = × = - F x eEx
= - ´ ´ ´ ´ ´
- -
1.6 10 1 10 8 10
19 3 3
     = - ´
-
1.28 J 10
18
Loss of KE = ´
-
1.28 J 10
18
17. Here, u u
x
= ° = cos45
25
2
 ms
-1
u u
y
= ° = sin45
25
2
 ms
-1
     a qE
x
= = ´ ´ ´
-
2 10 2 10
6 7
           =40 ms
-1
         a
y
= -10 ms
-1
          y u
a
t
yt
y
= +
1
2
          y t t = -
25
2
5
2
39  
E
1
E
E
2
P
x 
y
–Q
Q
y
+Q
+Q
E
2
E
1
E
x
P
–Q
+Q
E
1 E
2
E
P
12
E
12
2
3
4
5
6
7
8
9
10
11
E
1
E
2
E
3
E
4
E
5
E
6
E
7
E
8
E
9
E
10
E
11
1
6E
1
6E
1
6E
1
6E
1
E
6E
1
6E
1
q
u
E
at the end of motion,
t T = and y = 0
\ T =
5
2
 s
Also at the end of motion,
         x R =
\         x u t a t
x x
= +
1
2
2
  R= ´ + ´
æ
è
ç
ö
ø
÷
25
2
5
2
20
5
2
2
           =312.5 m
18. (a)    R
qE
=
m q
2
2 sin
   sin2
2
q=
qER
mu
        =
´ ´ ´ ´
´ ´ ´
- -
-
1.6 1.27
1.67 9.55
10 720 10
10 10
19 3
27 3 2
( )
        =0.96
      2 88 q= ° or 92°
       q= ° 44 or 46°
      T
mh
E
=
2
2
sinq
 
        =
´ ´ ´ ´ ´
´ ´
-
-
2 10
1
2
10
10 720
3 31
19
9.55 1.67
1.6
        = ´
-
1.95 10
11
 s
19. (a) a
E
j
®
®
-
-
= - = -
´ ´
´
e
m
1.6
9.1
10 120
10
19
31
^
= - ´ 2.1 10
13
i
^
 m/s
(b) t
x
u
x
= =
´
´
= ´
-
-
D 2 10
10
4
3
10
2
5
7
1.5
 s
v u a t
y y y
= +
       = ´ ´ ´ ´ ´
-
3.0 2.1 10 10
4
3
10
6 13 7
        = ´ 0.2 10
6
 m/s
v i j
®
= ´ + ´ ( ) ( )
^ ^
1.5 0.2 10 10
5 6
20. Ab so lute po ten tial can be zero at two points
on the x-axis. One in be tween the charges
and other on the left of charge a
1
 (smaller in
mag ni tu de).
Case I.
In between two charges : let potential is zero
at a distance x from q
1
 towards q
2
.
V
q
x
q
x
= × + ×
-
=
1
4
1
4 100
0
0
1
0
2
pe pe
        = ×
´
- ×
´
-
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
Þ 200 2 3 - = x x
        x =20 cm
Case II.
Consider the potential is zero at a distance x
from charge q, on its left.
\   V
q
x
q
x
= × + ×
+
=
1
4
1
4 100
0
0
1
0
2
pe pe
    = ×
´
+ ×
- ´
+
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
200 2 3 + = x x
          x =200 cm
21. Let us first find the po ten tial at a point on
the per pen dic u lar bi sec tor of a line charge.
Consider a line of carrying a line charge
density l having length L.
Consider an elementary portion of length dl
on the rod. 
Charge on this portion
           dq dl = l
\ dV
dl
r
= ×
f
1
4
0
pe
l
sec
  
Now,         l r = f tan
            dl r d = f f sec
2
 40
O
100cm
q
1
q
2
X
q
1
q
2
x 100–x
q
1
q
2
x 100 cm
f
q
L
r
l
dl
\   dV
d
=
f f l
pe
sec
4
0
\     V dV d = = × f f
ò ò
-
l
pe
q
q
4
0
2
sec
       = +
-
l
pe
q q
q
q
4
0
[ln| tan |] sec
       =
+
-
½
½
½
½
½
½
é
ë
ê
ê
ù
û
ú
ú
l
pe
q q
q q 4
0
ln
tan
tan
sec
sec
       = +
2
4
0
l
pe
q q ln| tan | sec
In the given condition
q = ° 60
Potential due to one side
 V V V
1 2 3
0
2
4
60 60 = = = × ° + °
l
pe
ln| tan | sec
= × +
2
4
2 3
0
l
pe
ln| |
Total potential at O
V V = = × + 3
6
4
2 3
1
0
l
pe
ln| |
= × +
Q
a 2
2 3
0
pe
ln| |   
22. (a) V V
2 1
2
250 20 10 - = - × = - ´ ´
® ®
-
E d 
= - 50 V
     W V q V V = = - D ( )
2 1
= ´ ´ -
-
12 10 50
6
 = -0.6 mJ
(b) V V
2 1
50 - = - V
23. By work en ergy the o rem
       W K = D
q V V mv mv ( )
1 2 2
2
1
2
1
2
1
2
- = -
 - ´ -
-
5 10 20 800
6
( )
= ´ ´ -
-
1
2
2 10 5
4
2
2 2
( ( ) ) V
       v
2
2
55 =
       v
2
55 = = 7.42 ms
-1
When a particle is released in electric field it 
moves in such a way that, it decreases its PE 
and increases KE
Hence, particle at B is faster than that at A.
24. Cen tr e of cir cle is equi dis tant from ev ery
point on its pe riph ery,
Hence, V
q
R
0
0
1
4
= ×
pe
, 
where q Q Q SQ = + = -
1 2
\    V
Q
R
0
0
1
4
5
= - ×
pe
Similarly,      V
q
R Z
p
= ×
+
1
4
0
2 2
pe
          = - ×
+
1
4
0
2 2
pe
SQ
R Z
25. Ini tial PE
U
q q
r
i
= ×
1
4
0
1 2
1
p e
U
q q
r
f
= ×
1
4
0
1 2
2
p e
Work done by electric force
W U U U
f i
= - = - - D ( )
    = - × -
æ
è
ç
ç
ö
ø
÷
÷
1
4
1 1
0
1 2
2 1
pe
q q
r r
Þ  W = - ´ ´ ´ ´ - ´
- -
9 10 10 10
9 6 6
2.4 4.3 ( ) 
1 1
0.25 2 0.15
-
æ
è
ç
ö
ø
÷
  W = -0.356 mJ
26. (a) 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
     = ´
´ ´ - ´ é
ë
ê
ê
- -
9 10
4 10 3 10
9
9 9
( )
0.2
+
- ´ ´ ´
- -
( ) ( ) 3 10 2 10
9 9
0.1
+
´ ´ ´ ù
û
ú
ú
- -
4 10 2 10
9 9
0.1
U = ´ - - + = -
-
9 10 6 6 8 360
8
[ ] nJ
(b) Let the distance of q
3
 from q
1
 is x cm. Then
U
q q q q
x
q q
x
= × +
-
+
é
ë
ê
ê
ù
û
ú
ú
=
1
4
0
0
1 2 2 3 3 1
pe 0.2 0.2
Þ   9 10
4 10 3 10
20 10
9
9 9
0 2
´
´ ´ - ´
´
é
ë
ê
ê
- -
-
( )
+
- ´ ´ ´
- ´
- -
-
( )
( )
3 10 2 10
20 10
9 9
2
x
41  
O
Page 5


8. As ball are in equi lib rium
  F T
e
= sin a
mg T = cos a
F mg
e
= tan a
    q r
2
0
2
4 = pe a tan
Here,      r l =2 sina
q l
2
0
2 2
16 = pe a a sin tan
          q = ´
-
3.3 10
8
 C.
9. Same as Q.7. In tro duc tory Ex er cise 21.3.
10. See Q.7. In tro duc tory Ex er cise 21.3.
11.    E
q
r
= ×
® 1
4
1
0
3
pe
r
   =
´ ´ - ´
+
-
-
9 10 10
9 9
2 232
( )
(( ) ( ))
( )
/
^ ^
8.0
1.2 1.6
1.2 1.6 i j
   =- - 18 2( )
^ ^
1.2 1.6 i j N/C.
12. Con sider an el e men tary por tion on the ring
of length dl sub t end ing an gle df at cen tre ‘O’
of the ring.
Charge on this portion,
dq dl Rd = = f l l
\ dE
dq
R
d
R
= × =
f 1
4
1
4
0
2
0
pe pe
l
Here, dE sin f components of field will cancel 
each other.
Hence, Net field at O
E dE
R
d = f = × f f
ò ò
-
cos cos
/
/ 1
4
0
2
2
pe
l
p
p
= ×
1
4
2
0
pe
l
R
13. Con sider el e men tary por tion of the rod of
length dl at a dis tance l from the cen tre O of
the rod.
Charge on this portion
dq dl
Q
L
dl = = l    
\ dE
dq
a
= ×
f
1
4
0
2
pe ( ) sec
        = ×
f
1
4
0
2 2
pe
Q dl
La sec
Now,
l a = f tan  
Þ dl a d = f f sec
2
\ dE
Qd
La
= ×
f 1
4
0
pe
 
Net Electric field at P.
E dE = f
ò
cos
[ dEsin f components will cancel each other
as rod in symmetrical about P.]
= × f f
-
ò
1
4
0
pe
q
q Q
La
d cos
         = ×
1
4
2
0
pe
q Q
La
sin
But sinq=
+
æ
è
ç
ö
ø
÷
=
+
L
a
L
L
a L
2
2
4
2
2 2 2
\ E
Q
a a L
= ×
+
1
4
2
4 0
2 2
pe
 38
dE
dE cos f
dE cos f
dE
dE sin f dE sin f
O
f
R
df
dl dl
dE
dE cos f
dE cos f
dE
dE sin f
dE sin f
df
Q
P
dl
l
dl
f
O
a
O 
a
F
e F
e
mg
T sin a
mg
r
T cos a
T T
14. (a) As shown in fig ure, di rec tion of elec tric
   field at P will be along + ve y-axis.
(b) Positive x-axis.
(c) Positive y-axis.
15. Let E
q
R
1
0
2
1
4
= ×
pe
Resultant fields of two opposite charges can
be shown as given in figure.
Clearly resultant field is along angle bisector 
of field towards 9 and 10.
Hence time shown by clock in the direction
of electric field is 9 : 30.
16. (a)  a
F
m
eE
m
= =
-
     =
- ´ ´ ´
´
-
-
1.6
9.1
10 1 10
10
19 3
31
     = - ´ 1.76 10
14
 ms
-2
u = ´ 5.00 10
8
 cm/s = ´ 5 10
6
 ms
-1
v = 0    
v u as
2 2
2 - =
       s =
´
´ ´
= ´ =
-
( ) 5 10
2 10
1 10
6 2
14
2
1.7
.4 1.4 cm
(b)  v u at = +
t =
´
´
= ´ =
-
5 10
10
10 28
6
14
8
1.76
2.8 ns.
(c) Dk = work done by electric field.
     = × = - F x eEx
= - ´ ´ ´ ´ ´
- -
1.6 10 1 10 8 10
19 3 3
     = - ´
-
1.28 J 10
18
Loss of KE = ´
-
1.28 J 10
18
17. Here, u u
x
= ° = cos45
25
2
 ms
-1
u u
y
= ° = sin45
25
2
 ms
-1
     a qE
x
= = ´ ´ ´
-
2 10 2 10
6 7
           =40 ms
-1
         a
y
= -10 ms
-1
          y u
a
t
yt
y
= +
1
2
          y t t = -
25
2
5
2
39  
E
1
E
E
2
P
x 
y
–Q
Q
y
+Q
+Q
E
2
E
1
E
x
P
–Q
+Q
E
1 E
2
E
P
12
E
12
2
3
4
5
6
7
8
9
10
11
E
1
E
2
E
3
E
4
E
5
E
6
E
7
E
8
E
9
E
10
E
11
1
6E
1
6E
1
6E
1
6E
1
E
6E
1
6E
1
q
u
E
at the end of motion,
t T = and y = 0
\ T =
5
2
 s
Also at the end of motion,
         x R =
\         x u t a t
x x
= +
1
2
2
  R= ´ + ´
æ
è
ç
ö
ø
÷
25
2
5
2
20
5
2
2
           =312.5 m
18. (a)    R
qE
=
m q
2
2 sin
   sin2
2
q=
qER
mu
        =
´ ´ ´ ´
´ ´ ´
- -
-
1.6 1.27
1.67 9.55
10 720 10
10 10
19 3
27 3 2
( )
        =0.96
      2 88 q= ° or 92°
       q= ° 44 or 46°
      T
mh
E
=
2
2
sinq
 
        =
´ ´ ´ ´ ´
´ ´
-
-
2 10
1
2
10
10 720
3 31
19
9.55 1.67
1.6
        = ´
-
1.95 10
11
 s
19. (a) a
E
j
®
®
-
-
= - = -
´ ´
´
e
m
1.6
9.1
10 120
10
19
31
^
= - ´ 2.1 10
13
i
^
 m/s
(b) t
x
u
x
= =
´
´
= ´
-
-
D 2 10
10
4
3
10
2
5
7
1.5
 s
v u a t
y y y
= +
       = ´ ´ ´ ´ ´
-
3.0 2.1 10 10
4
3
10
6 13 7
        = ´ 0.2 10
6
 m/s
v i j
®
= ´ + ´ ( ) ( )
^ ^
1.5 0.2 10 10
5 6
20. Ab so lute po ten tial can be zero at two points
on the x-axis. One in be tween the charges
and other on the left of charge a
1
 (smaller in
mag ni tu de).
Case I.
In between two charges : let potential is zero
at a distance x from q
1
 towards q
2
.
V
q
x
q
x
= × + ×
-
=
1
4
1
4 100
0
0
1
0
2
pe pe
        = ×
´
- ×
´
-
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
Þ 200 2 3 - = x x
        x =20 cm
Case II.
Consider the potential is zero at a distance x
from charge q, on its left.
\   V
q
x
q
x
= × + ×
+
=
1
4
1
4 100
0
0
1
0
2
pe pe
    = ×
´
+ ×
- ´
+
=
- -
1
4
2 10 1
4
3 10
100
0
0
6
0
6
pe pe x x
200 2 3 + = x x
          x =200 cm
21. Let us first find the po ten tial at a point on
the per pen dic u lar bi sec tor of a line charge.
Consider a line of carrying a line charge
density l having length L.
Consider an elementary portion of length dl
on the rod. 
Charge on this portion
           dq dl = l
\ dV
dl
r
= ×
f
1
4
0
pe
l
sec
  
Now,         l r = f tan
            dl r d = f f sec
2
 40
O
100cm
q
1
q
2
X
q
1
q
2
x 100–x
q
1
q
2
x 100 cm
f
q
L
r
l
dl
\   dV
d
=
f f l
pe
sec
4
0
\     V dV d = = × f f
ò ò
-
l
pe
q
q
4
0
2
sec
       = +
-
l
pe
q q
q
q
4
0
[ln| tan |] sec
       =
+
-
½
½
½
½
½
½
é
ë
ê
ê
ù
û
ú
ú
l
pe
q q
q q 4
0
ln
tan
tan
sec
sec
       = +
2
4
0
l
pe
q q ln| tan | sec
In the given condition
q = ° 60
Potential due to one side
 V V V
1 2 3
0
2
4
60 60 = = = × ° + °
l
pe
ln| tan | sec
= × +
2
4
2 3
0
l
pe
ln| |
Total potential at O
V V = = × + 3
6
4
2 3
1
0
l
pe
ln| |
= × +
Q
a 2
2 3
0
pe
ln| |   
22. (a) V V
2 1
2
250 20 10 - = - × = - ´ ´
® ®
-
E d 
= - 50 V
     W V q V V = = - D ( )
2 1
= ´ ´ -
-
12 10 50
6
 = -0.6 mJ
(b) V V
2 1
50 - = - V
23. By work en ergy the o rem
       W K = D
q V V mv mv ( )
1 2 2
2
1
2
1
2
1
2
- = -
 - ´ -
-
5 10 20 800
6
( )
= ´ ´ -
-
1
2
2 10 5
4
2
2 2
( ( ) ) V
       v
2
2
55 =
       v
2
55 = = 7.42 ms
-1
When a particle is released in electric field it 
moves in such a way that, it decreases its PE 
and increases KE
Hence, particle at B is faster than that at A.
24. Cen tr e of cir cle is equi dis tant from ev ery
point on its pe riph ery,
Hence, V
q
R
0
0
1
4
= ×
pe
, 
where q Q Q SQ = + = -
1 2
\    V
Q
R
0
0
1
4
5
= - ×
pe
Similarly,      V
q
R Z
p
= ×
+
1
4
0
2 2
pe
          = - ×
+
1
4
0
2 2
pe
SQ
R Z
25. Ini tial PE
U
q q
r
i
= ×
1
4
0
1 2
1
p e
U
q q
r
f
= ×
1
4
0
1 2
2
p e
Work done by electric force
W U U U
f i
= - = - - D ( )
    = - × -
æ
è
ç
ç
ö
ø
÷
÷
1
4
1 1
0
1 2
2 1
pe
q q
r r
Þ  W = - ´ ´ ´ ´ - ´
- -
9 10 10 10
9 6 6
2.4 4.3 ( ) 
1 1
0.25 2 0.15
-
æ
è
ç
ö
ø
÷
  W = -0.356 mJ
26. (a) 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
     = ´
´ ´ - ´ é
ë
ê
ê
- -
9 10
4 10 3 10
9
9 9
( )
0.2
+
- ´ ´ ´
- -
( ) ( ) 3 10 2 10
9 9
0.1
+
´ ´ ´ ù
û
ú
ú
- -
4 10 2 10
9 9
0.1
U = ´ - - + = -
-
9 10 6 6 8 360
8
[ ] nJ
(b) Let the distance of q
3
 from q
1
 is x cm. Then
U
q q q q
x
q q
x
= × +
-
+
é
ë
ê
ê
ù
û
ú
ú
=
1
4
0
0
1 2 2 3 3 1
pe 0.2 0.2
Þ   9 10
4 10 3 10
20 10
9
9 9
0 2
´
´ ´ - ´
´
é
ë
ê
ê
- -
-
( )
+
- ´ ´ ´
- ´
- -
-
( )
( )
3 10 2 10
20 10
9 9
2
x
41  
O
+
´ ´ ´
´
ù
û
ú
ú
=
- -
-
2 10 4 10
10
0
9 9
2
x
 
Þ - -
-
+ =
6
10
6
20
8
0
x x
Þ x = 6.43 cm
27. Let Q be the third charge
U
q
d
qQ
d
qQ
d
= × + +
é
ë
ê
ê
ù
û
ú
ú
=
1
4
0
0
2
pe
       Q
q
= -
2
28. V = - ×
® ®
E r
(a) r k
®
=5
^
V = - - - = ( ) ( )
^ ^ ^
5 3 5 0 i j k
(b) r i k
®
= + 4 3
^ ^
V = - - - + ( ) ( )
^ ^ ^ ^
5 3 4 3 i j i j
          = -20 kV
29. E
®
 = 400 j
^
 V/m
(a) r j
®
= 20
^
 cm = ( )
^
0.2 j m
V = - × = -
® ®
E r 80 V
(b) r j
®
= - ( )
^
0.3 m 
V = - × =
® ®
E r 120 V
(c) r k
®
= ( )
^
0.15
V = 0
30. E i
®
= 20
^
 N/C
(a) r i j
®
= + ( )
^ ^
4 2 m
V = - × = -
® ®
E r 80 V
(b) r i j
®
= + ( )
^ ^
2 3 m
V = - × = -
® ®
E r 40 V
31. (a) [
[ ]
[ ]
A
V
xy yz zx
] =
+ +
 =
- -
[ ]
[ ]
ML T I
L
2 3 1
2
                    =
- -
[ ] ML T I
0 3 1
(b) E V
v
x
v
y
v
z
= - Ñ = -
¶
¶
+
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
®
i j k
^ ^ ^
   = - + + + + + A y z z x x y [( ) ( ) ( ) ]
^ ^ ^
i j k
(c) at (1m, 1m, 1m)
E = - + + 10 2 2 2 ( )
^ ^ ^
i j k
= - + + 20( )
^ ^ ^
i j k  
32. V V
B
- = - ×
® ®
0
E r
Þ V - =- + 0 40 60 ( )
Þ V =-100  
33. (a) E
v
x
Ay Bx
x
= -
¶
¶
= - - ( ) 2
E
V
y
Ax C
y
-
¶
¶
= - + ( )
E
V
Z
z
= -
¶
¶
=0
(b) For E = 0
E
x
= 0 and E
y
= 0
Hence, E
y
= 0
Ax C + = 0
x
C
A
= -
E
x
= 0
Ay B
C
A
- -
æ
è
ç
ö
ø
÷
= 2 0
y
BC
A
= -
2
2
Hence, E is zero at - -
æ
è
ç
ö
ø
÷
C
A
BC
A
,
2
2
.
34. f =
q
e
0
     q = = ´ ´
-
e
0
12
10 360 f 8.8
           = ´
-
3.18 10
9
 C
           =3.186 nC
36. (a) f= = -
´
´
-
-
q
e
0
6
12
10
10
3.60
8.85
        = ´ 4.07 10
5
 V-m.
(b) f =
q
e
0
 Þ q = e
0
f
= ´ ´ = ´
- -
8.85 6.903 10 780 10
12 9
   q =6.903 nC
(c) No.
Net flux through a closed surface does
not depend on position of charge.
36. E i j
®
= +
æ
è
ç
ö
ø
÷
3
5
4
5
0 0
E E
^ ^
S j
®
= 0.2
^
 m
2
 =
1
5
j
^
 m
2
\ f = × =
® ®
E S
4
25
 Nm
2
/C
 42