DC Pandey Solutions: Electrostatics- 3 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


11. Net force on C = 0
F
Q
a
CB
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CD
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CA
= ×
- 1
4
2 2 1
2
0
2 2
2
pe
( )
F
Q
a
CO
= ×
- 1
4
2 2 2 1
0
2
2
pe
( )
Net force on C
F F F F F
CA CO CB CO
= + + ° + ° cos cos 45 45
=
- -
+
- é
ë
ê
ê
( ) ( ) ( ) 2 2 1 2 2 1
2
2 2 1
2
2
Q
a
Q Q
+
-
+
ù
û
ú
ú
=
( ) 2 2 1
2
2
1
0
Q q
q
Q
= -
7
4
12. E =
s
e
0
F eF
e
= =
s
e
0
Acceleration of proton
a
F
m
e
m
= =
s
e
0
s ut
a
t = +
1
2
             u =0
  t
a
m
e
= =
25 25
0
e
s
=
´ ´ ´ ´ ´
´ ´ ´
- -
- -
2 10 10
10 10
27 12
9 19
0.1 1.67 8.8
2.21 1.6
  =2 2 ms
13. Data is not suf fi cient.
14. If the charges have op po site sign, elec tric
field is zero on the left of smaller charge.
15. Net field is only due to charge on C.
E
q
a
= ×
1
4 2
0
2
pe ( )
 =
q
a 16
0
2
pe
16. On touch ing two spheres, equal charge will
ap pear on both the spheres and for a given
to tal charge, force be tween two spheres is
max i mum if charges on them are equal.
17. Charge dis tri bu tion is shown in fig ure.
18. V
q
r
= ×
1
4
0
pe
If drops coalesce, total volume remains
conserved,
4
3
1000
4
3
3 3
p p R r = ´
R r =10
V
q
q
¢= × =
1
4
1000
10
10
0
pe
V
19. V
q
r
Q
R
A
= × +
é
ë
ê
ù
û
ú
1
4
0
pe
 48
r
q
A
R
B
A
q
B
q
q q
E D
F qC
C 
F
CD
F
CA
F
CB
F
CO
O
A B
D
+8Q
–2Q
+2Q
–4Q
+4Q
Page 2


11. Net force on C = 0
F
Q
a
CB
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CD
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CA
= ×
- 1
4
2 2 1
2
0
2 2
2
pe
( )
F
Q
a
CO
= ×
- 1
4
2 2 2 1
0
2
2
pe
( )
Net force on C
F F F F F
CA CO CB CO
= + + ° + ° cos cos 45 45
=
- -
+
- é
ë
ê
ê
( ) ( ) ( ) 2 2 1 2 2 1
2
2 2 1
2
2
Q
a
Q Q
+
-
+
ù
û
ú
ú
=
( ) 2 2 1
2
2
1
0
Q q
q
Q
= -
7
4
12. E =
s
e
0
F eF
e
= =
s
e
0
Acceleration of proton
a
F
m
e
m
= =
s
e
0
s ut
a
t = +
1
2
             u =0
  t
a
m
e
= =
25 25
0
e
s
=
´ ´ ´ ´ ´
´ ´ ´
- -
- -
2 10 10
10 10
27 12
9 19
0.1 1.67 8.8
2.21 1.6
  =2 2 ms
13. Data is not suf fi cient.
14. If the charges have op po site sign, elec tric
field is zero on the left of smaller charge.
15. Net field is only due to charge on C.
E
q
a
= ×
1
4 2
0
2
pe ( )
 =
q
a 16
0
2
pe
16. On touch ing two spheres, equal charge will
ap pear on both the spheres and for a given
to tal charge, force be tween two spheres is
max i mum if charges on them are equal.
17. Charge dis tri bu tion is shown in fig ure.
18. V
q
r
= ×
1
4
0
pe
If drops coalesce, total volume remains
conserved,
4
3
1000
4
3
3 3
p p R r = ´
R r =10
V
q
q
¢= × =
1
4
1000
10
10
0
pe
V
19. V
q
r
Q
R
A
= × +
é
ë
ê
ù
û
ú
1
4
0
pe
 48
r
q
A
R
B
A
q
B
q
q q
E D
F qC
C 
F
CD
F
CA
F
CB
F
CO
O
A B
D
+8Q
–2Q
+2Q
–4Q
+4Q
V
q Q
R
B
= ×
+
é
ë
ê
ù
û
ú
1
4
0
pe
           
    V V
q
r R
A B
- = × -
é
ë
ê
ù
û
ú
4
1 1
0
pe
\ V V q
A B
- µ
If q is doubled, V V
A B
- will become double.
20. Charge dis tri bu tion is shown in fig ure.
21. f =
®
×
®
= + × = E S i j i ( ) ( )
^ ^ ^
5 2 5 V-m.
22. F F
Qq
a
DA DC
= = ×
1
4
0
2
pe
F
q
a
DB
= ×
1
4 2
0
2
2
pe
Net force on charge at D
F
0
45 45 0
®
= + ° + ° = F F F
DB DA DB
cos cos
Þ 
1
4
4
2 2 2
0
0 2
pe × + +
é
ë
ê
ù
û
ú
=
q
a
q Q Q
q Q = - 2 2
23. As V
B
= 0, To tal charge in side B must be zero 
and hence charge on its outer sur face is zero
and on its in ner sur face is - q.
24. V
V
p
=
1
0
1
4
1
2
1
4
3
2
0 0
pe pe
×
+
= × ×
q
R r
q
R
1 3
4 R r R +
=
°
      
4 3 3 R R r = +
r
R
=
3
    
25. Net charge on any di pole is zero.
26. For net force to be zero.
T mg cos q = Þ T
mg
=
cos q
or T qE sin q = Þ T
qE
=
sin q
27.  E
q
a
V
a
1
0
1
1
4
= × =
pe
   E
q
b
V
b
2
0
2
2
1
4
= =
pe
But     E E
1 2
=
       
V
a
V
b
1 2
=
Þ      
V
V
a
b
1
2
=
28. Elec tric field on equa to rial lines of di pole is
op po site to di pole mo ment.
29. Po ten tial dif fer ence be tween two con cen tric
spheres is in de pend ent of charge on outer
sphere.
30. E
q
r
= ×
1
4
0
2
pe
49  
2Q
–3Q
3Q
–Q
Q
F
DC
D C
B A
Q
F
DB
F
DA
q
Q
q
Q
P
O
R r
q
q
T T cos q
7 sin q
mg
q E
Page 3


11. Net force on C = 0
F
Q
a
CB
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CD
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CA
= ×
- 1
4
2 2 1
2
0
2 2
2
pe
( )
F
Q
a
CO
= ×
- 1
4
2 2 2 1
0
2
2
pe
( )
Net force on C
F F F F F
CA CO CB CO
= + + ° + ° cos cos 45 45
=
- -
+
- é
ë
ê
ê
( ) ( ) ( ) 2 2 1 2 2 1
2
2 2 1
2
2
Q
a
Q Q
+
-
+
ù
û
ú
ú
=
( ) 2 2 1
2
2
1
0
Q q
q
Q
= -
7
4
12. E =
s
e
0
F eF
e
= =
s
e
0
Acceleration of proton
a
F
m
e
m
= =
s
e
0
s ut
a
t = +
1
2
             u =0
  t
a
m
e
= =
25 25
0
e
s
=
´ ´ ´ ´ ´
´ ´ ´
- -
- -
2 10 10
10 10
27 12
9 19
0.1 1.67 8.8
2.21 1.6
  =2 2 ms
13. Data is not suf fi cient.
14. If the charges have op po site sign, elec tric
field is zero on the left of smaller charge.
15. Net field is only due to charge on C.
E
q
a
= ×
1
4 2
0
2
pe ( )
 =
q
a 16
0
2
pe
16. On touch ing two spheres, equal charge will
ap pear on both the spheres and for a given
to tal charge, force be tween two spheres is
max i mum if charges on them are equal.
17. Charge dis tri bu tion is shown in fig ure.
18. V
q
r
= ×
1
4
0
pe
If drops coalesce, total volume remains
conserved,
4
3
1000
4
3
3 3
p p R r = ´
R r =10
V
q
q
¢= × =
1
4
1000
10
10
0
pe
V
19. V
q
r
Q
R
A
= × +
é
ë
ê
ù
û
ú
1
4
0
pe
 48
r
q
A
R
B
A
q
B
q
q q
E D
F qC
C 
F
CD
F
CA
F
CB
F
CO
O
A B
D
+8Q
–2Q
+2Q
–4Q
+4Q
V
q Q
R
B
= ×
+
é
ë
ê
ù
û
ú
1
4
0
pe
           
    V V
q
r R
A B
- = × -
é
ë
ê
ù
û
ú
4
1 1
0
pe
\ V V q
A B
- µ
If q is doubled, V V
A B
- will become double.
20. Charge dis tri bu tion is shown in fig ure.
21. f =
®
×
®
= + × = E S i j i ( ) ( )
^ ^ ^
5 2 5 V-m.
22. F F
Qq
a
DA DC
= = ×
1
4
0
2
pe
F
q
a
DB
= ×
1
4 2
0
2
2
pe
Net force on charge at D
F
0
45 45 0
®
= + ° + ° = F F F
DB DA DB
cos cos
Þ 
1
4
4
2 2 2
0
0 2
pe × + +
é
ë
ê
ù
û
ú
=
q
a
q Q Q
q Q = - 2 2
23. As V
B
= 0, To tal charge in side B must be zero 
and hence charge on its outer sur face is zero
and on its in ner sur face is - q.
24. V
V
p
=
1
0
1
4
1
2
1
4
3
2
0 0
pe pe
×
+
= × ×
q
R r
q
R
1 3
4 R r R +
=
°
      
4 3 3 R R r = +
r
R
=
3
    
25. Net charge on any di pole is zero.
26. For net force to be zero.
T mg cos q = Þ T
mg
=
cos q
or T qE sin q = Þ T
qE
=
sin q
27.  E
q
a
V
a
1
0
1
1
4
= × =
pe
   E
q
b
V
b
2
0
2
2
1
4
= =
pe
But     E E
1 2
=
       
V
a
V
b
1 2
=
Þ      
V
V
a
b
1
2
=
28. Elec tric field on equa to rial lines of di pole is
op po site to di pole mo ment.
29. Po ten tial dif fer ence be tween two con cen tric
spheres is in de pend ent of charge on outer
sphere.
30. E
q
r
= ×
1
4
0
2
pe
49  
2Q
–3Q
3Q
–Q
Q
F
DC
D C
B A
Q
F
DB
F
DA
q
Q
q
Q
P
O
R r
q
q
T T cos q
7 sin q
mg
q E
      V
q
r
Er = × =
1
4
0
pe
       r
V
E
=
½
½
½
½
½
½
= =
3000
500
6 m
       q rV = =
´ -
´
= - 4
6 3000
9 10
2
0 9
pe m
( )
C
31. F F
1 2
=
1
4
1
4
0
1 2
1
2
0
1 2
2
2
pe p e
× = ×
q q
r K
q q
r
   r
r
K
2
1
50
5
10 5 = = = m
   »22.3 m
32. Elec tric field at a dis tance r from in fi nite line 
charge
E
r
=
l
pe 2
0
dV E dr = -  
dV Edr
V
V
a
b
1
2
ò ò
= -    
Þ V V
2 1
0
2
1
2
- = ×
l
pe
ln   
          W q V V
q
= - = ( ) ln
2 1
0
2
1
2
l
pe
33. As neg a tive charge is at less dis tance from
the line charge, it is at tracted to wards the
line charge.
34. r = - + - + - ( ) ( ) ( ) 4 1 2 2 0 4
2 2 2
 = 5 m
V
q
r
= × =
´ ´ ´
=
-
1
4
9 10 2 10
5
36
0
9 8
pe
 V
(b) and (c) are wrong.
35. V
q
R
= ×
1
4
0
pe
At a distance r from the centre,
E
q
r
VR
r
= × =
1
4
0
2 2
pe
36. When outer sphere is earthed field be tween
the re gion of two spheres in non-zero and is
zero in all other re gions.
37. W =
®
×
®
F s = qEscos q 
E
W
qs
= =
´ ´ °
=
cos cos q
4
2 60
20
0.2
 N/C
38. V
Q
R
Q
d R
1
0
2 2
1
4
= × -
+
é
ë
ê
ê
ù
û
ú
ú pe
V
Q
R
Q
d R
2
0
2 2
1
4
= - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
Q
d R
1 2
0
2 2
1
4
2 2
- = - -
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
d R
1 2
0
2 2
4
1 1
- = - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
39. Elec tric field in side a hol low sphere is
al way s zero .
40. W =
®
×
®
F r =
®
×
®
q E r
  = + - + q E E a b ( ) ( )
^ ^ ^ ^
1 2
i j i j
        = + q aE bE ( )
1 2
JEE Corner
As s er tion and Rea s on
1. Neg a tive charge al ways moved to wards
in cr eas ing po ten tial.
On moving from A to B potential energy of
negative charge decreases hence its KE
increases.
2. U
qq
r
= ×
1
4
0
1 2
pe
If q
1
 and q
2
 have opposite sign, U decreases
with decrease in r.
F
dU
dr
= - Þ work done by conservative force
always decreases PE.
3. E
dV
dr
= - = - = ( ) 10 10 V/m along x-axis.
4. V
q
R
= ×
1
4
0
pe
 50
r
Q –Q
1 2
d
R R
Page 4


11. Net force on C = 0
F
Q
a
CB
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CD
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CA
= ×
- 1
4
2 2 1
2
0
2 2
2
pe
( )
F
Q
a
CO
= ×
- 1
4
2 2 2 1
0
2
2
pe
( )
Net force on C
F F F F F
CA CO CB CO
= + + ° + ° cos cos 45 45
=
- -
+
- é
ë
ê
ê
( ) ( ) ( ) 2 2 1 2 2 1
2
2 2 1
2
2
Q
a
Q Q
+
-
+
ù
û
ú
ú
=
( ) 2 2 1
2
2
1
0
Q q
q
Q
= -
7
4
12. E =
s
e
0
F eF
e
= =
s
e
0
Acceleration of proton
a
F
m
e
m
= =
s
e
0
s ut
a
t = +
1
2
             u =0
  t
a
m
e
= =
25 25
0
e
s
=
´ ´ ´ ´ ´
´ ´ ´
- -
- -
2 10 10
10 10
27 12
9 19
0.1 1.67 8.8
2.21 1.6
  =2 2 ms
13. Data is not suf fi cient.
14. If the charges have op po site sign, elec tric
field is zero on the left of smaller charge.
15. Net field is only due to charge on C.
E
q
a
= ×
1
4 2
0
2
pe ( )
 =
q
a 16
0
2
pe
16. On touch ing two spheres, equal charge will
ap pear on both the spheres and for a given
to tal charge, force be tween two spheres is
max i mum if charges on them are equal.
17. Charge dis tri bu tion is shown in fig ure.
18. V
q
r
= ×
1
4
0
pe
If drops coalesce, total volume remains
conserved,
4
3
1000
4
3
3 3
p p R r = ´
R r =10
V
q
q
¢= × =
1
4
1000
10
10
0
pe
V
19. V
q
r
Q
R
A
= × +
é
ë
ê
ù
û
ú
1
4
0
pe
 48
r
q
A
R
B
A
q
B
q
q q
E D
F qC
C 
F
CD
F
CA
F
CB
F
CO
O
A B
D
+8Q
–2Q
+2Q
–4Q
+4Q
V
q Q
R
B
= ×
+
é
ë
ê
ù
û
ú
1
4
0
pe
           
    V V
q
r R
A B
- = × -
é
ë
ê
ù
û
ú
4
1 1
0
pe
\ V V q
A B
- µ
If q is doubled, V V
A B
- will become double.
20. Charge dis tri bu tion is shown in fig ure.
21. f =
®
×
®
= + × = E S i j i ( ) ( )
^ ^ ^
5 2 5 V-m.
22. F F
Qq
a
DA DC
= = ×
1
4
0
2
pe
F
q
a
DB
= ×
1
4 2
0
2
2
pe
Net force on charge at D
F
0
45 45 0
®
= + ° + ° = F F F
DB DA DB
cos cos
Þ 
1
4
4
2 2 2
0
0 2
pe × + +
é
ë
ê
ù
û
ú
=
q
a
q Q Q
q Q = - 2 2
23. As V
B
= 0, To tal charge in side B must be zero 
and hence charge on its outer sur face is zero
and on its in ner sur face is - q.
24. V
V
p
=
1
0
1
4
1
2
1
4
3
2
0 0
pe pe
×
+
= × ×
q
R r
q
R
1 3
4 R r R +
=
°
      
4 3 3 R R r = +
r
R
=
3
    
25. Net charge on any di pole is zero.
26. For net force to be zero.
T mg cos q = Þ T
mg
=
cos q
or T qE sin q = Þ T
qE
=
sin q
27.  E
q
a
V
a
1
0
1
1
4
= × =
pe
   E
q
b
V
b
2
0
2
2
1
4
= =
pe
But     E E
1 2
=
       
V
a
V
b
1 2
=
Þ      
V
V
a
b
1
2
=
28. Elec tric field on equa to rial lines of di pole is
op po site to di pole mo ment.
29. Po ten tial dif fer ence be tween two con cen tric
spheres is in de pend ent of charge on outer
sphere.
30. E
q
r
= ×
1
4
0
2
pe
49  
2Q
–3Q
3Q
–Q
Q
F
DC
D C
B A
Q
F
DB
F
DA
q
Q
q
Q
P
O
R r
q
q
T T cos q
7 sin q
mg
q E
      V
q
r
Er = × =
1
4
0
pe
       r
V
E
=
½
½
½
½
½
½
= =
3000
500
6 m
       q rV = =
´ -
´
= - 4
6 3000
9 10
2
0 9
pe m
( )
C
31. F F
1 2
=
1
4
1
4
0
1 2
1
2
0
1 2
2
2
pe p e
× = ×
q q
r K
q q
r
   r
r
K
2
1
50
5
10 5 = = = m
   »22.3 m
32. Elec tric field at a dis tance r from in fi nite line 
charge
E
r
=
l
pe 2
0
dV E dr = -  
dV Edr
V
V
a
b
1
2
ò ò
= -    
Þ V V
2 1
0
2
1
2
- = ×
l
pe
ln   
          W q V V
q
= - = ( ) ln
2 1
0
2
1
2
l
pe
33. As neg a tive charge is at less dis tance from
the line charge, it is at tracted to wards the
line charge.
34. r = - + - + - ( ) ( ) ( ) 4 1 2 2 0 4
2 2 2
 = 5 m
V
q
r
= × =
´ ´ ´
=
-
1
4
9 10 2 10
5
36
0
9 8
pe
 V
(b) and (c) are wrong.
35. V
q
R
= ×
1
4
0
pe
At a distance r from the centre,
E
q
r
VR
r
= × =
1
4
0
2 2
pe
36. When outer sphere is earthed field be tween
the re gion of two spheres in non-zero and is
zero in all other re gions.
37. W =
®
×
®
F s = qEscos q 
E
W
qs
= =
´ ´ °
=
cos cos q
4
2 60
20
0.2
 N/C
38. V
Q
R
Q
d R
1
0
2 2
1
4
= × -
+
é
ë
ê
ê
ù
û
ú
ú pe
V
Q
R
Q
d R
2
0
2 2
1
4
= - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
Q
d R
1 2
0
2 2
1
4
2 2
- = - -
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
d R
1 2
0
2 2
4
1 1
- = - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
39. Elec tric field in side a hol low sphere is
al way s zero .
40. W =
®
×
®
F r =
®
×
®
q E r
  = + - + q E E a b ( ) ( )
^ ^ ^ ^
1 2
i j i j
        = + q aE bE ( )
1 2
JEE Corner
As s er tion and Rea s on
1. Neg a tive charge al ways moved to wards
in cr eas ing po ten tial.
On moving from A to B potential energy of
negative charge decreases hence its KE
increases.
2. U
qq
r
= ×
1
4
0
1 2
pe
If q
1
 and q
2
 have opposite sign, U decreases
with decrease in r.
F
dU
dr
= - Þ work done by conservative force
always decreases PE.
3. E
dV
dr
= - = - = ( ) 10 10 V/m along x-axis.
4. V
q
R
= ×
1
4
0
pe
 50
r
Q –Q
1 2
d
R R
Inside the solid sphere.
E
qr
R
= ×
1
4
0
3
pe
at r
R
=
2
 
E
q
R
V
R
= × =
1
4 2 2
0
2
pe
Assertion is correct.
Reason is false as electric field inside the
sphere is directly proportional to distance
from centre but not outside it.
5. Gauss the o rem is valid only for closed
sur face but elec tric flux can be ob tained for
any sur face.
6. Let V
0
= Po ten tial at or i gin,
   V
A
= - + × = - ( ) ( )
^ ^ ^
4 4 4 16 i j i V
    V
B
= - + × = - ( ) ( )
^ ^ ^
4 4 4 16 i j i V
     V V
A B
=
Hence, Assertion is false.
7. In the line go ing A and B, the en ergy of third 
charge is min i mum at cen tre .
8. Di pole has both neg a tive and pos i tive
charges hence work done is not pos i tive.
9. Charge out side a closed sur face can pro duce
elec tric field but can not pro duce flux.
10. E
qx
x a
= ×
+
1
4
0
2 2 3 2
pe ( )
/
 is max i mum at x
a
=
2
But V
q
a x
= ×
+
1
4
0
2 2
pe
 is maximum at x = 0.
Ob jec ti ve Ques tions (Level-2)
1. Elec tro static force al ways acts along the line
join ing the two charges, hence net torque on
charge + 2q is al ways zero.
As net torque is zero angular momentum of
charge remains conserved.
Initial angular momentum
L m V R
i
= ° ( sin ) 30
When the separation between the charges
become minimum, direction of motion of
charge +2q become perpendicular to the line 
joining the charges.
\ find angular momentum
L mv r
mvr
f
= ¢ =
2
By conservation of angular momentum
L L
i f
= Þ r R =
3
2
2. v j
1
®
= v
^
, v i j
2
2 30 2 30
®
= ° + ° v v cos sin
^ ^
= + 3 i j
^ ^
v
As velocity along y-axis is unchanged,
electric field along x-axis is zero.
For motion along x-axis,
v u a x x
x x x
2 2
0
2 - = - ( )
a
v
a
v
a
x
=
-
=
( ) 3 0
2
3
2
2 2
F ma
mv
a
x x
= =
3
2
2
Þ F i
®
=
3
2
2
mv
a
^
Also,          F E
®
= -
®
e 
E i
®
= -
3
2
2
mv
ea
^
Rate of work done by electric field at B
P =
®
×
®
F v =
æ
è
ç
ö
ø
÷ × +
3
2
3
mv
a
v v i i j
^ ^ ^
( )
=
3 3
2
3
mv
a
3. Elec tric field is al ways pos si ble, hence a
must be pos i tive and  b must be neg a tive.
51  
30°
+Q +2q
R
U
r
+1q
+2q
V'
–Q +q
a b
Page 5


11. Net force on C = 0
F
Q
a
CB
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CD
= ×
- 1
4
2 2 1
0
2 2
2
pe
( )
F
Q
a
CA
= ×
- 1
4
2 2 1
2
0
2 2
2
pe
( )
F
Q
a
CO
= ×
- 1
4
2 2 2 1
0
2
2
pe
( )
Net force on C
F F F F F
CA CO CB CO
= + + ° + ° cos cos 45 45
=
- -
+
- é
ë
ê
ê
( ) ( ) ( ) 2 2 1 2 2 1
2
2 2 1
2
2
Q
a
Q Q
+
-
+
ù
û
ú
ú
=
( ) 2 2 1
2
2
1
0
Q q
q
Q
= -
7
4
12. E =
s
e
0
F eF
e
= =
s
e
0
Acceleration of proton
a
F
m
e
m
= =
s
e
0
s ut
a
t = +
1
2
             u =0
  t
a
m
e
= =
25 25
0
e
s
=
´ ´ ´ ´ ´
´ ´ ´
- -
- -
2 10 10
10 10
27 12
9 19
0.1 1.67 8.8
2.21 1.6
  =2 2 ms
13. Data is not suf fi cient.
14. If the charges have op po site sign, elec tric
field is zero on the left of smaller charge.
15. Net field is only due to charge on C.
E
q
a
= ×
1
4 2
0
2
pe ( )
 =
q
a 16
0
2
pe
16. On touch ing two spheres, equal charge will
ap pear on both the spheres and for a given
to tal charge, force be tween two spheres is
max i mum if charges on them are equal.
17. Charge dis tri bu tion is shown in fig ure.
18. V
q
r
= ×
1
4
0
pe
If drops coalesce, total volume remains
conserved,
4
3
1000
4
3
3 3
p p R r = ´
R r =10
V
q
q
¢= × =
1
4
1000
10
10
0
pe
V
19. V
q
r
Q
R
A
= × +
é
ë
ê
ù
û
ú
1
4
0
pe
 48
r
q
A
R
B
A
q
B
q
q q
E D
F qC
C 
F
CD
F
CA
F
CB
F
CO
O
A B
D
+8Q
–2Q
+2Q
–4Q
+4Q
V
q Q
R
B
= ×
+
é
ë
ê
ù
û
ú
1
4
0
pe
           
    V V
q
r R
A B
- = × -
é
ë
ê
ù
û
ú
4
1 1
0
pe
\ V V q
A B
- µ
If q is doubled, V V
A B
- will become double.
20. Charge dis tri bu tion is shown in fig ure.
21. f =
®
×
®
= + × = E S i j i ( ) ( )
^ ^ ^
5 2 5 V-m.
22. F F
Qq
a
DA DC
= = ×
1
4
0
2
pe
F
q
a
DB
= ×
1
4 2
0
2
2
pe
Net force on charge at D
F
0
45 45 0
®
= + ° + ° = F F F
DB DA DB
cos cos
Þ 
1
4
4
2 2 2
0
0 2
pe × + +
é
ë
ê
ù
û
ú
=
q
a
q Q Q
q Q = - 2 2
23. As V
B
= 0, To tal charge in side B must be zero 
and hence charge on its outer sur face is zero
and on its in ner sur face is - q.
24. V
V
p
=
1
0
1
4
1
2
1
4
3
2
0 0
pe pe
×
+
= × ×
q
R r
q
R
1 3
4 R r R +
=
°
      
4 3 3 R R r = +
r
R
=
3
    
25. Net charge on any di pole is zero.
26. For net force to be zero.
T mg cos q = Þ T
mg
=
cos q
or T qE sin q = Þ T
qE
=
sin q
27.  E
q
a
V
a
1
0
1
1
4
= × =
pe
   E
q
b
V
b
2
0
2
2
1
4
= =
pe
But     E E
1 2
=
       
V
a
V
b
1 2
=
Þ      
V
V
a
b
1
2
=
28. Elec tric field on equa to rial lines of di pole is
op po site to di pole mo ment.
29. Po ten tial dif fer ence be tween two con cen tric
spheres is in de pend ent of charge on outer
sphere.
30. E
q
r
= ×
1
4
0
2
pe
49  
2Q
–3Q
3Q
–Q
Q
F
DC
D C
B A
Q
F
DB
F
DA
q
Q
q
Q
P
O
R r
q
q
T T cos q
7 sin q
mg
q E
      V
q
r
Er = × =
1
4
0
pe
       r
V
E
=
½
½
½
½
½
½
= =
3000
500
6 m
       q rV = =
´ -
´
= - 4
6 3000
9 10
2
0 9
pe m
( )
C
31. F F
1 2
=
1
4
1
4
0
1 2
1
2
0
1 2
2
2
pe p e
× = ×
q q
r K
q q
r
   r
r
K
2
1
50
5
10 5 = = = m
   »22.3 m
32. Elec tric field at a dis tance r from in fi nite line 
charge
E
r
=
l
pe 2
0
dV E dr = -  
dV Edr
V
V
a
b
1
2
ò ò
= -    
Þ V V
2 1
0
2
1
2
- = ×
l
pe
ln   
          W q V V
q
= - = ( ) ln
2 1
0
2
1
2
l
pe
33. As neg a tive charge is at less dis tance from
the line charge, it is at tracted to wards the
line charge.
34. r = - + - + - ( ) ( ) ( ) 4 1 2 2 0 4
2 2 2
 = 5 m
V
q
r
= × =
´ ´ ´
=
-
1
4
9 10 2 10
5
36
0
9 8
pe
 V
(b) and (c) are wrong.
35. V
q
R
= ×
1
4
0
pe
At a distance r from the centre,
E
q
r
VR
r
= × =
1
4
0
2 2
pe
36. When outer sphere is earthed field be tween
the re gion of two spheres in non-zero and is
zero in all other re gions.
37. W =
®
×
®
F s = qEscos q 
E
W
qs
= =
´ ´ °
=
cos cos q
4
2 60
20
0.2
 N/C
38. V
Q
R
Q
d R
1
0
2 2
1
4
= × -
+
é
ë
ê
ê
ù
û
ú
ú pe
V
Q
R
Q
d R
2
0
2 2
1
4
= - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
Q
d R
1 2
0
2 2
1
4
2 2
- = - -
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
V V
Q
R
d R
1 2
0
2 2
4
1 1
- = - +
+
é
ë
ê
ê
ù
û
ú
ú
pe
.
39. Elec tric field in side a hol low sphere is
al way s zero .
40. W =
®
×
®
F r =
®
×
®
q E r
  = + - + q E E a b ( ) ( )
^ ^ ^ ^
1 2
i j i j
        = + q aE bE ( )
1 2
JEE Corner
As s er tion and Rea s on
1. Neg a tive charge al ways moved to wards
in cr eas ing po ten tial.
On moving from A to B potential energy of
negative charge decreases hence its KE
increases.
2. U
qq
r
= ×
1
4
0
1 2
pe
If q
1
 and q
2
 have opposite sign, U decreases
with decrease in r.
F
dU
dr
= - Þ work done by conservative force
always decreases PE.
3. E
dV
dr
= - = - = ( ) 10 10 V/m along x-axis.
4. V
q
R
= ×
1
4
0
pe
 50
r
Q –Q
1 2
d
R R
Inside the solid sphere.
E
qr
R
= ×
1
4
0
3
pe
at r
R
=
2
 
E
q
R
V
R
= × =
1
4 2 2
0
2
pe
Assertion is correct.
Reason is false as electric field inside the
sphere is directly proportional to distance
from centre but not outside it.
5. Gauss the o rem is valid only for closed
sur face but elec tric flux can be ob tained for
any sur face.
6. Let V
0
= Po ten tial at or i gin,
   V
A
= - + × = - ( ) ( )
^ ^ ^
4 4 4 16 i j i V
    V
B
= - + × = - ( ) ( )
^ ^ ^
4 4 4 16 i j i V
     V V
A B
=
Hence, Assertion is false.
7. In the line go ing A and B, the en ergy of third 
charge is min i mum at cen tre .
8. Di pole has both neg a tive and pos i tive
charges hence work done is not pos i tive.
9. Charge out side a closed sur face can pro duce
elec tric field but can not pro duce flux.
10. E
qx
x a
= ×
+
1
4
0
2 2 3 2
pe ( )
/
 is max i mum at x
a
=
2
But V
q
a x
= ×
+
1
4
0
2 2
pe
 is maximum at x = 0.
Ob jec ti ve Ques tions (Level-2)
1. Elec tro static force al ways acts along the line
join ing the two charges, hence net torque on
charge + 2q is al ways zero.
As net torque is zero angular momentum of
charge remains conserved.
Initial angular momentum
L m V R
i
= ° ( sin ) 30
When the separation between the charges
become minimum, direction of motion of
charge +2q become perpendicular to the line 
joining the charges.
\ find angular momentum
L mv r
mvr
f
= ¢ =
2
By conservation of angular momentum
L L
i f
= Þ r R =
3
2
2. v j
1
®
= v
^
, v i j
2
2 30 2 30
®
= ° + ° v v cos sin
^ ^
= + 3 i j
^ ^
v
As velocity along y-axis is unchanged,
electric field along x-axis is zero.
For motion along x-axis,
v u a x x
x x x
2 2
0
2 - = - ( )
a
v
a
v
a
x
=
-
=
( ) 3 0
2
3
2
2 2
F ma
mv
a
x x
= =
3
2
2
Þ F i
®
=
3
2
2
mv
a
^
Also,          F E
®
= -
®
e 
E i
®
= -
3
2
2
mv
ea
^
Rate of work done by electric field at B
P =
®
×
®
F v =
æ
è
ç
ö
ø
÷ × +
3
2
3
mv
a
v v i i j
^ ^ ^
( )
=
3 3
2
3
mv
a
3. Elec tric field is al ways pos si ble, hence a
must be pos i tive and  b must be neg a tive.
51  
30°
+Q +2q
R
U
r
+1q
+2q
V'
–Q +q
a b
4. The sys tem can be as sumed as a
com bi na tion of three iden ti cal di poles as
shown in fig ure.
Here, P P P Q a
1 2 3
2 = = = ( )
Net dipole moment of the system
P P P P = + ° + °
1 2 3
60 60 cos cos
        = = 2 4 p Qa
Electric field on equatorial lines of short
dipole is given by
E
P
x
= ×
1
4
0
3
pe
 
          = × =
1
4
4
0
3
0
3
pe pe
Qa
x
Qa
x
5. Po ten tial at cen tre will be same as po ten tial
at the sur face of in ner shell i.e., 10 V.
6. Ini tial charge dis tri bu tion is shown in fig ure,
Initial energy of system
U
q
a
q
a
q
a
i
= × +
-
´
+
´
é
ë
ê
ê
1
4 2 2 2 2 3
0
2 2 2
pe
( )
+
-
+ +
-
ù
û
ú
q q
a
q q
a
q q
a
( ) ( ) ( )
2 3 3
        =
5
48
2
0
a
a pe
When switch S is closed, entire charge flows
to the outer surface of outer shell,
U
q
a
q
a
f
= ×
´
=
1
4 2 3 24
0
2 2
0
pe pe
Heat produced = = = U U
q
a
i f
2
0
8pe
=
kq
a
2
2
7. Let Q charge flows to C 
V
q Q
a
q
a
C
= ×
+
+
-
é
ë
ê
ù
û
ú
=
1
4 3 4
0
0
pe
( )
       Q
q
= -
4
\      V
q
a
Q
a
q
a
A
= × + +
-
é
ë
ê
ù
û
ú
1
4 2 3 4
0
pe
        =
é
ë
ê
ê
ù
û
ú
ú
1
4
0
pe s
q
a
  V V
q
a
A C
- = ×
1
4
0
pe s
        =
kq
a s
8. V
q
R
S
= ×
1
4
0
pe
  
and    V
q
R
C
= ×
1
4
3
2
0
pe
\  V V
q
R
C S
- = ×
1
4 2
0
pe
= × =
1
4
4
3
2
0
3
2
0
p
R
R
R
E e
p d
d
s
9. As par ti cle co mes to rest, force must be
re pul sive, hence it is pos i tively charged.
Again on moving down its KE first increases
than decreases, PE will first decrease than
increase.
 52
60°
60°
P
2
®
P
1
®
P
3
®
+q
–q
+q
S
2q
3a
a
D
–q
4a
q
Q
q
2a
A
B
3a
C