Chapter 21 - Electrostatics (Part - 4) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 21 - Electrostatics (Part - 4) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


27. q a
A
= s p ( ) 4
2
, q b
B
= - s p ( ) 4
2
, q c
C
= s p ( ) 4
2
Given, V V
A C
= 
1
4
1
4
0 0
pe pe
× + +
æ
è
ç
ö
ø
÷
= ×
+ +
æ
è
ç
ö
ø
÷
q
a
q
b
q
c
q q q
c
A B C A B C
Þ a b c
a b
c
c - + =
-
+
2 2
Þ a b c + =
28. Po ten tial at min i mum at mid-point in the
re gion be tween two charges, and is al ways
pos i tive.
29. U
q
r
U
i
= × =
1
4
2
p e
0
U
q
r
U
f
= × ´ =
1
4
3 3
2
pe
0
\ W U U U
f i
= - = 2
30. Loss of KE = Gain in PE
1
2
1
4
2
0
mv
qQ
r
= ×
pe
r
v
µ
1
2
31. When the spheres are in air
T mg cos q =
T F
e
sin q =
\ F mg
e
= tan q …(i)
When the spheres are immersed in liquid
F T
1
¢ = ¢sin q
g F T
B
- = ¢cos q    
\        F mg F
e B
¢= - ( )tanq …(ii)
On dividing Eq. (ii) by Eq. (i),
F
F
mg F
mg
e
e
B
¢
=
-
1
1 1
1
2 K
F
mg
B
= - = - =
0.8
1.6
K = 2
32. V
q
a
q
a b
P
= × ´ -
+
´
é
ë
ê
ê
ù
û
ú
ú
1
4
2 2
0
2 2
pe
= ×
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2
2 2
q
a b a
a a b
pe
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
=
2
4
1
0
2
2
1 2
2 2
q
a
b
a
a
a a b
pe
/
2
4
0
2
3
q b
a pe
× 
[As b a << ]
33. In any case elec tric field at or i gin is 
1
4
5
0
2
pe
×
q
r
along x-axis and 
1
4
5
0
2
pe
×
q
r
 along y-axis.
34. < > = = ×
æ
è
ç
ç
ö
ø
÷
÷
u E
q
R
1
2
1
2
1
4
0
2
0
0
2
2
e e
pe
         =
´ ´ ´
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
-
1
2
9 10
1
9
10
12
0
9 9
e
=
e
0
2
 J/m
3
    
35. If Q is ini tial charge on B  
then, V V
Q
b
V
A B
- = × =
1
4
0
pe
Now, if A is earthed, let charge q moves on A
from ground, then
V
q
a
Q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
pe
          q
a
b
Q = -
55  
Tcosq
T
O
q
Tsinq
F
e
F
e
mg
mg
q
T'cosq
T'
O
q
Tsinq
F'
e
mg
mg
q
T'
F'
e
F
B
Q
q
Page 2


27. q a
A
= s p ( ) 4
2
, q b
B
= - s p ( ) 4
2
, q c
C
= s p ( ) 4
2
Given, V V
A C
= 
1
4
1
4
0 0
pe pe
× + +
æ
è
ç
ö
ø
÷
= ×
+ +
æ
è
ç
ö
ø
÷
q
a
q
b
q
c
q q q
c
A B C A B C
Þ a b c
a b
c
c - + =
-
+
2 2
Þ a b c + =
28. Po ten tial at min i mum at mid-point in the
re gion be tween two charges, and is al ways
pos i tive.
29. U
q
r
U
i
= × =
1
4
2
p e
0
U
q
r
U
f
= × ´ =
1
4
3 3
2
pe
0
\ W U U U
f i
= - = 2
30. Loss of KE = Gain in PE
1
2
1
4
2
0
mv
qQ
r
= ×
pe
r
v
µ
1
2
31. When the spheres are in air
T mg cos q =
T F
e
sin q =
\ F mg
e
= tan q …(i)
When the spheres are immersed in liquid
F T
1
¢ = ¢sin q
g F T
B
- = ¢cos q    
\        F mg F
e B
¢= - ( )tanq …(ii)
On dividing Eq. (ii) by Eq. (i),
F
F
mg F
mg
e
e
B
¢
=
-
1
1 1
1
2 K
F
mg
B
= - = - =
0.8
1.6
K = 2
32. V
q
a
q
a b
P
= × ´ -
+
´
é
ë
ê
ê
ù
û
ú
ú
1
4
2 2
0
2 2
pe
= ×
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2
2 2
q
a b a
a a b
pe
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
=
2
4
1
0
2
2
1 2
2 2
q
a
b
a
a
a a b
pe
/
2
4
0
2
3
q b
a pe
× 
[As b a << ]
33. In any case elec tric field at or i gin is 
1
4
5
0
2
pe
×
q
r
along x-axis and 
1
4
5
0
2
pe
×
q
r
 along y-axis.
34. < > = = ×
æ
è
ç
ç
ö
ø
÷
÷
u E
q
R
1
2
1
2
1
4
0
2
0
0
2
2
e e
pe
         =
´ ´ ´
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
-
1
2
9 10
1
9
10
12
0
9 9
e
=
e
0
2
 J/m
3
    
35. If Q is ini tial charge on B  
then, V V
Q
b
V
A B
- = × =
1
4
0
pe
Now, if A is earthed, let charge q moves on A
from ground, then
V
q
a
Q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
pe
          q
a
b
Q = -
55  
Tcosq
T
O
q
Tsinq
F
e
F
e
mg
mg
q
T'cosq
T'
O
q
Tsinq
F'
e
mg
mg
q
T'
F'
e
F
B
Q
q
         V
q Q
b
B
= ×
+ 1
4
0
pe
           = × -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1
4
1 1
0
pe
Q
b
a
b
V
a
b
36. E i j k
®
= -
¶
¶
+
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
v
z
^ ^ ^
  = -
-
+
-
+
-
æ
è
ç
ö
ø
÷
2
1
2
1
2
1
i j k
^ ^ ^
  = + + 2( )
^ ^ ^
i j k N/C
If V
P
 is potential at P, then
V V
P
- = -
®
×
®
0
E r
V
P
- = - + + × + + = - 10 2 6 ( ) ( )
^ ^ ^ ^ ^ ^
i j k i j k
V
P
= 4 V
37. On touch ing two spheres, charge is equally
di vided among them, then due to in duc tion a 
charge -
æ
è
ç
ö
ø
÷
q
2
 ap pears on the earthed sphere.
38. Neg a tive charge will in duce on the con duc tor 
near P.
39.
0 for
for
for
r r
kQ
r
r r r
k Q Q
r
r r
A
P
P B
A B
B
<
< <
-
>
ì
í
ï
ï
ï
î
ï
ï
ï
( )
As | | | | Q Q
B A
> 
E is –ve for r r
B
> .
40. E i j
®
= -
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
^ ^
  = + k y x ( )
^ ^
i j
| | E
®
= + = k y x kr
2 2
41. Let charge on outer shell be comes q.
V
Q q
r
B
= ×
+ æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
q Q = -
42. Let charge q¢ flows through the switch to the
ground, then
1
4 2
0
0
pe
×
- ¢
-
é
ë
ê
ê
ù
û
ú
ú
=
Q q
r
Q
r
q Q ¢ =
1
2
43. Af ter n steps
    q Q
n
¢=
1
2
 and q Q
n
=
-
-
1
2
1
\   V
q
r
q
r
A
= ×
¢
+
æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
   V
q q
r
B
= ×
¢+ 1
4 2
0
pe
      =
é
ë
ê
ê
ù
û
ú
ú
+
1
2 4
1
0
n
Q
r pe
44. Con side r a spher i cal Gaussi an sur face of
ra diu s r R ( ) < and con cen tric with the sphere,
Charge on a small sphere of radius r
dq dV r dr = = d p d 4
2
     = -
æ
è
ç
ç
ö
ø
÷
÷
r r
r
R
dr pd
0
2
3
Total charge inside the Gaussian surface,
q r
r
R
dr
r
= -
æ
è
ç
ç
ö
ø
÷
÷
ò
4
0
2
3
0
p d
= -
é
ë
ê
ê
ù
û
ú
ú
4
3 4
0
3 4
pd
r r
R
\ E
q
r
r r
R
= × = -
é
ë
ê
ê
ù
û
ú
ú
1
4 3 4
0
2
0
0
2
pe
d
e
45. To tal charge in side the sur face.
Q
R R
r
R = -
é
ë
ê
ê
ù
û
ú
ú
= 4
3
1
3
0
3 3
0
3
pd pd
E
Q
r
R
r
= × =
1
4 12
0
2
0
3
0
2
pe
d
e
46. E
r r
R
= -
é
ë
ê
ê
ù
û
ú
ú
d
e
0
0
2
3 4
For maximum intensity of electric field
 56
B
S
2
S
1
2r
A
r
R O
r
Page 3


27. q a
A
= s p ( ) 4
2
, q b
B
= - s p ( ) 4
2
, q c
C
= s p ( ) 4
2
Given, V V
A C
= 
1
4
1
4
0 0
pe pe
× + +
æ
è
ç
ö
ø
÷
= ×
+ +
æ
è
ç
ö
ø
÷
q
a
q
b
q
c
q q q
c
A B C A B C
Þ a b c
a b
c
c - + =
-
+
2 2
Þ a b c + =
28. Po ten tial at min i mum at mid-point in the
re gion be tween two charges, and is al ways
pos i tive.
29. U
q
r
U
i
= × =
1
4
2
p e
0
U
q
r
U
f
= × ´ =
1
4
3 3
2
pe
0
\ W U U U
f i
= - = 2
30. Loss of KE = Gain in PE
1
2
1
4
2
0
mv
qQ
r
= ×
pe
r
v
µ
1
2
31. When the spheres are in air
T mg cos q =
T F
e
sin q =
\ F mg
e
= tan q …(i)
When the spheres are immersed in liquid
F T
1
¢ = ¢sin q
g F T
B
- = ¢cos q    
\        F mg F
e B
¢= - ( )tanq …(ii)
On dividing Eq. (ii) by Eq. (i),
F
F
mg F
mg
e
e
B
¢
=
-
1
1 1
1
2 K
F
mg
B
= - = - =
0.8
1.6
K = 2
32. V
q
a
q
a b
P
= × ´ -
+
´
é
ë
ê
ê
ù
û
ú
ú
1
4
2 2
0
2 2
pe
= ×
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2
2 2
q
a b a
a a b
pe
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
=
2
4
1
0
2
2
1 2
2 2
q
a
b
a
a
a a b
pe
/
2
4
0
2
3
q b
a pe
× 
[As b a << ]
33. In any case elec tric field at or i gin is 
1
4
5
0
2
pe
×
q
r
along x-axis and 
1
4
5
0
2
pe
×
q
r
 along y-axis.
34. < > = = ×
æ
è
ç
ç
ö
ø
÷
÷
u E
q
R
1
2
1
2
1
4
0
2
0
0
2
2
e e
pe
         =
´ ´ ´
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
-
1
2
9 10
1
9
10
12
0
9 9
e
=
e
0
2
 J/m
3
    
35. If Q is ini tial charge on B  
then, V V
Q
b
V
A B
- = × =
1
4
0
pe
Now, if A is earthed, let charge q moves on A
from ground, then
V
q
a
Q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
pe
          q
a
b
Q = -
55  
Tcosq
T
O
q
Tsinq
F
e
F
e
mg
mg
q
T'cosq
T'
O
q
Tsinq
F'
e
mg
mg
q
T'
F'
e
F
B
Q
q
         V
q Q
b
B
= ×
+ 1
4
0
pe
           = × -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1
4
1 1
0
pe
Q
b
a
b
V
a
b
36. E i j k
®
= -
¶
¶
+
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
v
z
^ ^ ^
  = -
-
+
-
+
-
æ
è
ç
ö
ø
÷
2
1
2
1
2
1
i j k
^ ^ ^
  = + + 2( )
^ ^ ^
i j k N/C
If V
P
 is potential at P, then
V V
P
- = -
®
×
®
0
E r
V
P
- = - + + × + + = - 10 2 6 ( ) ( )
^ ^ ^ ^ ^ ^
i j k i j k
V
P
= 4 V
37. On touch ing two spheres, charge is equally
di vided among them, then due to in duc tion a 
charge -
æ
è
ç
ö
ø
÷
q
2
 ap pears on the earthed sphere.
38. Neg a tive charge will in duce on the con duc tor 
near P.
39.
0 for
for
for
r r
kQ
r
r r r
k Q Q
r
r r
A
P
P B
A B
B
<
< <
-
>
ì
í
ï
ï
ï
î
ï
ï
ï
( )
As | | | | Q Q
B A
> 
E is –ve for r r
B
> .
40. E i j
®
= -
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
^ ^
  = + k y x ( )
^ ^
i j
| | E
®
= + = k y x kr
2 2
41. Let charge on outer shell be comes q.
V
Q q
r
B
= ×
+ æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
q Q = -
42. Let charge q¢ flows through the switch to the
ground, then
1
4 2
0
0
pe
×
- ¢
-
é
ë
ê
ê
ù
û
ú
ú
=
Q q
r
Q
r
q Q ¢ =
1
2
43. Af ter n steps
    q Q
n
¢=
1
2
 and q Q
n
=
-
-
1
2
1
\   V
q
r
q
r
A
= ×
¢
+
æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
   V
q q
r
B
= ×
¢+ 1
4 2
0
pe
      =
é
ë
ê
ê
ù
û
ú
ú
+
1
2 4
1
0
n
Q
r pe
44. Con side r a spher i cal Gaussi an sur face of
ra diu s r R ( ) < and con cen tric with the sphere,
Charge on a small sphere of radius r
dq dV r dr = = d p d 4
2
     = -
æ
è
ç
ç
ö
ø
÷
÷
r r
r
R
dr pd
0
2
3
Total charge inside the Gaussian surface,
q r
r
R
dr
r
= -
æ
è
ç
ç
ö
ø
÷
÷
ò
4
0
2
3
0
p d
= -
é
ë
ê
ê
ù
û
ú
ú
4
3 4
0
3 4
pd
r r
R
\ E
q
r
r r
R
= × = -
é
ë
ê
ê
ù
û
ú
ú
1
4 3 4
0
2
0
0
2
pe
d
e
45. To tal charge in side the sur face.
Q
R R
r
R = -
é
ë
ê
ê
ù
û
ú
ú
= 4
3
1
3
0
3 3
0
3
pd pd
E
Q
r
R
r
= × =
1
4 12
0
2
0
3
0
2
pe
d
e
46. E
r r
R
= -
é
ë
ê
ê
ù
û
ú
ú
d
e
0
0
2
3 4
For maximum intensity of electric field
 56
B
S
2
S
1
2r
A
r
R O
r
57  
dE
dr
r
R
= -
é
ë
ê
ù
û
ú
=
d
e
0
0
1
3 2
0
Þ             r R =
2
3
d E
dr R
2
2
0
0
2
= - =
d
e
- ve,
hence E is maximum at r R =
2
3
.
47. E
R
R
R
q
max
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
=
r
e
l
r
e
0
0
2
0
0
2
3
3
2
3
4
48. Po ten tial dif fer ence be tween two con cen tric
spheres do not de pend on the charge on
outer sphere.
49. When outer sphere B is earthed
V
Q q
b
B
= ×
+
æ
è
ç
ö
ø
÷ =
1
4
0
0
1
pe
q Q
1
= -
Now, if A is earthed
q
q
a
q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
2 1
pe
q
a
b
q
a
b
Q
2 1
= - =     
50. When con nected by con duct ing wires, en tire
charge from in ner sphere flows to the outer
sphere, ie,
q q q
a
b
Q
3 1 2
1 = + -
æ
è
ç
ö
ø
÷
=
- a b
b
Q
More than One Correct Options
1. Be fore eart hing the sur face B,
V
q
R
q
R
A
A B
= × +
æ
è
ç
ö
ø
÷
=
1
4 2
2
0
pe
 V
V
q q
R
B
A B
= ×
+
æ
è
ç
ö
ø
÷
=
1
4 2
3
2
0
pe
 V
Þ
q
q
A
B
=
1
2
On earthing the sphere B,
q q
A A
¢ =
V
q q
R
B
A B
= ×
¢ + ¢
=
1
4 2
0
0
pe
q q
B A
¢ = - ¢
Þ
q
q
A
B
¢
¢
= - 1
As potential difference does not depend on
charge on outer sphere,
V V V V
V
A B A B
¢ - ¢ = - =
2
V V
A
¢ =
1
2
  
2. For the mo tion of par ti cle
u
x
= 0, v v
x
= , a
q E
m
x
= , a g
y
= - ,
x
0
0 = , y
0
0 =
x x u t a t
x x
= + +
0
2
1
2
           x
qE
m
t =
2
2
…(i)
     y y u t a t
y y
= + +
0
2
1
2
= - ut gt
1
2
2
…(ii)
At the end of motion
t T = , y = 0, x R =
q
1
Q
B
A
q
1
q
2
m
y
x
E
Page 4


27. q a
A
= s p ( ) 4
2
, q b
B
= - s p ( ) 4
2
, q c
C
= s p ( ) 4
2
Given, V V
A C
= 
1
4
1
4
0 0
pe pe
× + +
æ
è
ç
ö
ø
÷
= ×
+ +
æ
è
ç
ö
ø
÷
q
a
q
b
q
c
q q q
c
A B C A B C
Þ a b c
a b
c
c - + =
-
+
2 2
Þ a b c + =
28. Po ten tial at min i mum at mid-point in the
re gion be tween two charges, and is al ways
pos i tive.
29. U
q
r
U
i
= × =
1
4
2
p e
0
U
q
r
U
f
= × ´ =
1
4
3 3
2
pe
0
\ W U U U
f i
= - = 2
30. Loss of KE = Gain in PE
1
2
1
4
2
0
mv
qQ
r
= ×
pe
r
v
µ
1
2
31. When the spheres are in air
T mg cos q =
T F
e
sin q =
\ F mg
e
= tan q …(i)
When the spheres are immersed in liquid
F T
1
¢ = ¢sin q
g F T
B
- = ¢cos q    
\        F mg F
e B
¢= - ( )tanq …(ii)
On dividing Eq. (ii) by Eq. (i),
F
F
mg F
mg
e
e
B
¢
=
-
1
1 1
1
2 K
F
mg
B
= - = - =
0.8
1.6
K = 2
32. V
q
a
q
a b
P
= × ´ -
+
´
é
ë
ê
ê
ù
û
ú
ú
1
4
2 2
0
2 2
pe
= ×
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2
2 2
q
a b a
a a b
pe
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
=
2
4
1
0
2
2
1 2
2 2
q
a
b
a
a
a a b
pe
/
2
4
0
2
3
q b
a pe
× 
[As b a << ]
33. In any case elec tric field at or i gin is 
1
4
5
0
2
pe
×
q
r
along x-axis and 
1
4
5
0
2
pe
×
q
r
 along y-axis.
34. < > = = ×
æ
è
ç
ç
ö
ø
÷
÷
u E
q
R
1
2
1
2
1
4
0
2
0
0
2
2
e e
pe
         =
´ ´ ´
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
-
1
2
9 10
1
9
10
12
0
9 9
e
=
e
0
2
 J/m
3
    
35. If Q is ini tial charge on B  
then, V V
Q
b
V
A B
- = × =
1
4
0
pe
Now, if A is earthed, let charge q moves on A
from ground, then
V
q
a
Q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
pe
          q
a
b
Q = -
55  
Tcosq
T
O
q
Tsinq
F
e
F
e
mg
mg
q
T'cosq
T'
O
q
Tsinq
F'
e
mg
mg
q
T'
F'
e
F
B
Q
q
         V
q Q
b
B
= ×
+ 1
4
0
pe
           = × -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1
4
1 1
0
pe
Q
b
a
b
V
a
b
36. E i j k
®
= -
¶
¶
+
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
v
z
^ ^ ^
  = -
-
+
-
+
-
æ
è
ç
ö
ø
÷
2
1
2
1
2
1
i j k
^ ^ ^
  = + + 2( )
^ ^ ^
i j k N/C
If V
P
 is potential at P, then
V V
P
- = -
®
×
®
0
E r
V
P
- = - + + × + + = - 10 2 6 ( ) ( )
^ ^ ^ ^ ^ ^
i j k i j k
V
P
= 4 V
37. On touch ing two spheres, charge is equally
di vided among them, then due to in duc tion a 
charge -
æ
è
ç
ö
ø
÷
q
2
 ap pears on the earthed sphere.
38. Neg a tive charge will in duce on the con duc tor 
near P.
39.
0 for
for
for
r r
kQ
r
r r r
k Q Q
r
r r
A
P
P B
A B
B
<
< <
-
>
ì
í
ï
ï
ï
î
ï
ï
ï
( )
As | | | | Q Q
B A
> 
E is –ve for r r
B
> .
40. E i j
®
= -
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
^ ^
  = + k y x ( )
^ ^
i j
| | E
®
= + = k y x kr
2 2
41. Let charge on outer shell be comes q.
V
Q q
r
B
= ×
+ æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
q Q = -
42. Let charge q¢ flows through the switch to the
ground, then
1
4 2
0
0
pe
×
- ¢
-
é
ë
ê
ê
ù
û
ú
ú
=
Q q
r
Q
r
q Q ¢ =
1
2
43. Af ter n steps
    q Q
n
¢=
1
2
 and q Q
n
=
-
-
1
2
1
\   V
q
r
q
r
A
= ×
¢
+
æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
   V
q q
r
B
= ×
¢+ 1
4 2
0
pe
      =
é
ë
ê
ê
ù
û
ú
ú
+
1
2 4
1
0
n
Q
r pe
44. Con side r a spher i cal Gaussi an sur face of
ra diu s r R ( ) < and con cen tric with the sphere,
Charge on a small sphere of radius r
dq dV r dr = = d p d 4
2
     = -
æ
è
ç
ç
ö
ø
÷
÷
r r
r
R
dr pd
0
2
3
Total charge inside the Gaussian surface,
q r
r
R
dr
r
= -
æ
è
ç
ç
ö
ø
÷
÷
ò
4
0
2
3
0
p d
= -
é
ë
ê
ê
ù
û
ú
ú
4
3 4
0
3 4
pd
r r
R
\ E
q
r
r r
R
= × = -
é
ë
ê
ê
ù
û
ú
ú
1
4 3 4
0
2
0
0
2
pe
d
e
45. To tal charge in side the sur face.
Q
R R
r
R = -
é
ë
ê
ê
ù
û
ú
ú
= 4
3
1
3
0
3 3
0
3
pd pd
E
Q
r
R
r
= × =
1
4 12
0
2
0
3
0
2
pe
d
e
46. E
r r
R
= -
é
ë
ê
ê
ù
û
ú
ú
d
e
0
0
2
3 4
For maximum intensity of electric field
 56
B
S
2
S
1
2r
A
r
R O
r
57  
dE
dr
r
R
= -
é
ë
ê
ù
û
ú
=
d
e
0
0
1
3 2
0
Þ             r R =
2
3
d E
dr R
2
2
0
0
2
= - =
d
e
- ve,
hence E is maximum at r R =
2
3
.
47. E
R
R
R
q
max
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
=
r
e
l
r
e
0
0
2
0
0
2
3
3
2
3
4
48. Po ten tial dif fer ence be tween two con cen tric
spheres do not de pend on the charge on
outer sphere.
49. When outer sphere B is earthed
V
Q q
b
B
= ×
+
æ
è
ç
ö
ø
÷ =
1
4
0
0
1
pe
q Q
1
= -
Now, if A is earthed
q
q
a
q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
2 1
pe
q
a
b
q
a
b
Q
2 1
= - =     
50. When con nected by con duct ing wires, en tire
charge from in ner sphere flows to the outer
sphere, ie,
q q q
a
b
Q
3 1 2
1 = + -
æ
è
ç
ö
ø
÷
=
- a b
b
Q
More than One Correct Options
1. Be fore eart hing the sur face B,
V
q
R
q
R
A
A B
= × +
æ
è
ç
ö
ø
÷
=
1
4 2
2
0
pe
 V
V
q q
R
B
A B
= ×
+
æ
è
ç
ö
ø
÷
=
1
4 2
3
2
0
pe
 V
Þ
q
q
A
B
=
1
2
On earthing the sphere B,
q q
A A
¢ =
V
q q
R
B
A B
= ×
¢ + ¢
=
1
4 2
0
0
pe
q q
B A
¢ = - ¢
Þ
q
q
A
B
¢
¢
= - 1
As potential difference does not depend on
charge on outer sphere,
V V V V
V
A B A B
¢ - ¢ = - =
2
V V
A
¢ =
1
2
  
2. For the mo tion of par ti cle
u
x
= 0, v v
x
= , a
q E
m
x
= , a g
y
= - ,
x
0
0 = , y
0
0 =
x x u t a t
x x
= + +
0
2
1
2
           x
qE
m
t =
2
2
…(i)
     y y u t a t
y y
= + +
0
2
1
2
= - ut gt
1
2
2
…(ii)
At the end of motion
t T = , y = 0, x R =
q
1
Q
B
A
q
1
q
2
m
y
x
E
From Eq. (ii),
0
1
2
= -
æ
è
ç
ö
ø
÷
u gT T
  T
u
g
= =
´
=
2 2 10
10
2 s
From Eq. (i),
R
qE
m
T =
2
2
=
´ ´
´
´
-
1 10 10
2
4
3 4
g
 = 10 m
Now, v u a y y
y y y
2 2
0
2 - = - ( )
At highest point (i.e., y H = ), v
y
= 0
0 10 2 10 0
2
- = - ´ - ( ) ( ) H
H = 5 m
3. Let R be the ra dius of the sphere
V
q
R r
1
0 1
1
4
= ×
+ pe
Þ
q q
R S
´ ´
+ ´
=
-
10
10
100
9
2
( )
…(i)
V
q
R r
2
0 2
1
4
= ×
+ pe
Þ 
9 10
10 10
75
9
2
´ ´
+ ´
=
-
q
R ( )
…(ii)
On solving,
R = 10 cm, 
q = ´
-
5
3
10
9
 C = ´
-
50
3
10
10
 C
Electric field on surface,
E
q
R
= × =
´ ´ ´
´
-
-
1
4
9 10
5
3
10
10 10
0
2
9 9
2
pe ( )
= 1500 V/m
Potential at surface,
V
q
R
= ×
´ ´ ´
´
-
-
1
4
9 10
5
3
10
10 10
0
9 9
2
pe
= 150 V
Potential at Centre
V V
C S
= =
3
2
225 V
4. For all charges to be in equi lib rium, force
ex pe ri enced by ei ther charge must be zero
ie., force due to other two charges must be
equal and op po site.
Hence all the charges must be collinear,
charges q
1
, and q
3
 must have same sign and 
q
2
 must have opposite sign, q
2
 must have
maximum magnitude.
Such on equilibrium is always unstable.
5. Flux through any closed sur face de pends
only on charge in side the sur face but elec tric 
field at any point on the sur face de pends on
charges in side as well as out side the sur face.
6. As net charge on an elec tric di pole is zero,
net flux through the sphere is zero.
But electric field at any point due to a dipole
cannot be zero.
7. Gauss’s law gives to tal elec tric field and flux
due to all charges.
8. If two con cen tric spheres carry equal and
op po site charges, Elec tric field is non-zero
only in the re gion be tween two sphere and
po ten tial is  is zero only out side both the
spheres.
9. As force on the rod due to elec tric field is
to wards right, force on the rod due to hinge
must be left.
The equilibrium is clearly neutral.
10. If moved along per pen dic u lar bi sec tor, for all 
iden ti cal charges, elec tro static po ten tial
en ergy is max i mum at mid point and if
moved along the line join ing the par ti cles,
elec tr o static po ten ti al en er gy is min i mum at
the mid-point.
 58
q
1
q
2
q
3
Q
1
Q
2
Page 5


27. q a
A
= s p ( ) 4
2
, q b
B
= - s p ( ) 4
2
, q c
C
= s p ( ) 4
2
Given, V V
A C
= 
1
4
1
4
0 0
pe pe
× + +
æ
è
ç
ö
ø
÷
= ×
+ +
æ
è
ç
ö
ø
÷
q
a
q
b
q
c
q q q
c
A B C A B C
Þ a b c
a b
c
c - + =
-
+
2 2
Þ a b c + =
28. Po ten tial at min i mum at mid-point in the
re gion be tween two charges, and is al ways
pos i tive.
29. U
q
r
U
i
= × =
1
4
2
p e
0
U
q
r
U
f
= × ´ =
1
4
3 3
2
pe
0
\ W U U U
f i
= - = 2
30. Loss of KE = Gain in PE
1
2
1
4
2
0
mv
qQ
r
= ×
pe
r
v
µ
1
2
31. When the spheres are in air
T mg cos q =
T F
e
sin q =
\ F mg
e
= tan q …(i)
When the spheres are immersed in liquid
F T
1
¢ = ¢sin q
g F T
B
- = ¢cos q    
\        F mg F
e B
¢= - ( )tanq …(ii)
On dividing Eq. (ii) by Eq. (i),
F
F
mg F
mg
e
e
B
¢
=
-
1
1 1
1
2 K
F
mg
B
= - = - =
0.8
1.6
K = 2
32. V
q
a
q
a b
P
= × ´ -
+
´
é
ë
ê
ê
ù
û
ú
ú
1
4
2 2
0
2 2
pe
= ×
+ -
+
é
ë
ê
ê
ù
û
ú
ú
2
4
0
2 2
2 2
q
a b a
a a b
pe
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
+
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
=
2
4
1
0
2
2
1 2
2 2
q
a
b
a
a
a a b
pe
/
2
4
0
2
3
q b
a pe
× 
[As b a << ]
33. In any case elec tric field at or i gin is 
1
4
5
0
2
pe
×
q
r
along x-axis and 
1
4
5
0
2
pe
×
q
r
 along y-axis.
34. < > = = ×
æ
è
ç
ç
ö
ø
÷
÷
u E
q
R
1
2
1
2
1
4
0
2
0
0
2
2
e e
pe
         =
´ ´ ´
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
-
1
2
9 10
1
9
10
12
0
9 9
e
=
e
0
2
 J/m
3
    
35. If Q is ini tial charge on B  
then, V V
Q
b
V
A B
- = × =
1
4
0
pe
Now, if A is earthed, let charge q moves on A
from ground, then
V
q
a
Q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
pe
          q
a
b
Q = -
55  
Tcosq
T
O
q
Tsinq
F
e
F
e
mg
mg
q
T'cosq
T'
O
q
Tsinq
F'
e
mg
mg
q
T'
F'
e
F
B
Q
q
         V
q Q
b
B
= ×
+ 1
4
0
pe
           = × -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1
4
1 1
0
pe
Q
b
a
b
V
a
b
36. E i j k
®
= -
¶
¶
+
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
v
z
^ ^ ^
  = -
-
+
-
+
-
æ
è
ç
ö
ø
÷
2
1
2
1
2
1
i j k
^ ^ ^
  = + + 2( )
^ ^ ^
i j k N/C
If V
P
 is potential at P, then
V V
P
- = -
®
×
®
0
E r
V
P
- = - + + × + + = - 10 2 6 ( ) ( )
^ ^ ^ ^ ^ ^
i j k i j k
V
P
= 4 V
37. On touch ing two spheres, charge is equally
di vided among them, then due to in duc tion a 
charge -
æ
è
ç
ö
ø
÷
q
2
 ap pears on the earthed sphere.
38. Neg a tive charge will in duce on the con duc tor 
near P.
39.
0 for
for
for
r r
kQ
r
r r r
k Q Q
r
r r
A
P
P B
A B
B
<
< <
-
>
ì
í
ï
ï
ï
î
ï
ï
ï
( )
As | | | | Q Q
B A
> 
E is –ve for r r
B
> .
40. E i j
®
= -
¶
¶
+
¶
¶
æ
è
ç
ç
ö
ø
÷
÷
v
x
v
y
^ ^
  = + k y x ( )
^ ^
i j
| | E
®
= + = k y x kr
2 2
41. Let charge on outer shell be comes q.
V
Q q
r
B
= ×
+ æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
q Q = -
42. Let charge q¢ flows through the switch to the
ground, then
1
4 2
0
0
pe
×
- ¢
-
é
ë
ê
ê
ù
û
ú
ú
=
Q q
r
Q
r
q Q ¢ =
1
2
43. Af ter n steps
    q Q
n
¢=
1
2
 and q Q
n
=
-
-
1
2
1
\   V
q
r
q
r
A
= ×
¢
+
æ
è
ç
ç
ö
ø
÷
÷
=
1
4 2
0
0
pe
   V
q q
r
B
= ×
¢+ 1
4 2
0
pe
      =
é
ë
ê
ê
ù
û
ú
ú
+
1
2 4
1
0
n
Q
r pe
44. Con side r a spher i cal Gaussi an sur face of
ra diu s r R ( ) < and con cen tric with the sphere,
Charge on a small sphere of radius r
dq dV r dr = = d p d 4
2
     = -
æ
è
ç
ç
ö
ø
÷
÷
r r
r
R
dr pd
0
2
3
Total charge inside the Gaussian surface,
q r
r
R
dr
r
= -
æ
è
ç
ç
ö
ø
÷
÷
ò
4
0
2
3
0
p d
= -
é
ë
ê
ê
ù
û
ú
ú
4
3 4
0
3 4
pd
r r
R
\ E
q
r
r r
R
= × = -
é
ë
ê
ê
ù
û
ú
ú
1
4 3 4
0
2
0
0
2
pe
d
e
45. To tal charge in side the sur face.
Q
R R
r
R = -
é
ë
ê
ê
ù
û
ú
ú
= 4
3
1
3
0
3 3
0
3
pd pd
E
Q
r
R
r
= × =
1
4 12
0
2
0
3
0
2
pe
d
e
46. E
r r
R
= -
é
ë
ê
ê
ù
û
ú
ú
d
e
0
0
2
3 4
For maximum intensity of electric field
 56
B
S
2
S
1
2r
A
r
R O
r
57  
dE
dr
r
R
= -
é
ë
ê
ù
û
ú
=
d
e
0
0
1
3 2
0
Þ             r R =
2
3
d E
dr R
2
2
0
0
2
= - =
d
e
- ve,
hence E is maximum at r R =
2
3
.
47. E
R
R
R
q
max
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
=
r
e
l
r
e
0
0
2
0
0
2
3
3
2
3
4
48. Po ten tial dif fer ence be tween two con cen tric
spheres do not de pend on the charge on
outer sphere.
49. When outer sphere B is earthed
V
Q q
b
B
= ×
+
æ
è
ç
ö
ø
÷ =
1
4
0
0
1
pe
q Q
1
= -
Now, if A is earthed
q
q
a
q
b
A
= × +
æ
è
ç
ö
ø
÷
=
1
4
0
0
2 1
pe
q
a
b
q
a
b
Q
2 1
= - =     
50. When con nected by con duct ing wires, en tire
charge from in ner sphere flows to the outer
sphere, ie,
q q q
a
b
Q
3 1 2
1 = + -
æ
è
ç
ö
ø
÷
=
- a b
b
Q
More than One Correct Options
1. Be fore eart hing the sur face B,
V
q
R
q
R
A
A B
= × +
æ
è
ç
ö
ø
÷
=
1
4 2
2
0
pe
 V
V
q q
R
B
A B
= ×
+
æ
è
ç
ö
ø
÷
=
1
4 2
3
2
0
pe
 V
Þ
q
q
A
B
=
1
2
On earthing the sphere B,
q q
A A
¢ =
V
q q
R
B
A B
= ×
¢ + ¢
=
1
4 2
0
0
pe
q q
B A
¢ = - ¢
Þ
q
q
A
B
¢
¢
= - 1
As potential difference does not depend on
charge on outer sphere,
V V V V
V
A B A B
¢ - ¢ = - =
2
V V
A
¢ =
1
2
  
2. For the mo tion of par ti cle
u
x
= 0, v v
x
= , a
q E
m
x
= , a g
y
= - ,
x
0
0 = , y
0
0 =
x x u t a t
x x
= + +
0
2
1
2
           x
qE
m
t =
2
2
…(i)
     y y u t a t
y y
= + +
0
2
1
2
= - ut gt
1
2
2
…(ii)
At the end of motion
t T = , y = 0, x R =
q
1
Q
B
A
q
1
q
2
m
y
x
E
From Eq. (ii),
0
1
2
= -
æ
è
ç
ö
ø
÷
u gT T
  T
u
g
= =
´
=
2 2 10
10
2 s
From Eq. (i),
R
qE
m
T =
2
2
=
´ ´
´
´
-
1 10 10
2
4
3 4
g
 = 10 m
Now, v u a y y
y y y
2 2
0
2 - = - ( )
At highest point (i.e., y H = ), v
y
= 0
0 10 2 10 0
2
- = - ´ - ( ) ( ) H
H = 5 m
3. Let R be the ra dius of the sphere
V
q
R r
1
0 1
1
4
= ×
+ pe
Þ
q q
R S
´ ´
+ ´
=
-
10
10
100
9
2
( )
…(i)
V
q
R r
2
0 2
1
4
= ×
+ pe
Þ 
9 10
10 10
75
9
2
´ ´
+ ´
=
-
q
R ( )
…(ii)
On solving,
R = 10 cm, 
q = ´
-
5
3
10
9
 C = ´
-
50
3
10
10
 C
Electric field on surface,
E
q
R
= × =
´ ´ ´
´
-
-
1
4
9 10
5
3
10
10 10
0
2
9 9
2
pe ( )
= 1500 V/m
Potential at surface,
V
q
R
= ×
´ ´ ´
´
-
-
1
4
9 10
5
3
10
10 10
0
9 9
2
pe
= 150 V
Potential at Centre
V V
C S
= =
3
2
225 V
4. For all charges to be in equi lib rium, force
ex pe ri enced by ei ther charge must be zero
ie., force due to other two charges must be
equal and op po site.
Hence all the charges must be collinear,
charges q
1
, and q
3
 must have same sign and 
q
2
 must have opposite sign, q
2
 must have
maximum magnitude.
Such on equilibrium is always unstable.
5. Flux through any closed sur face de pends
only on charge in side the sur face but elec tric 
field at any point on the sur face de pends on
charges in side as well as out side the sur face.
6. As net charge on an elec tric di pole is zero,
net flux through the sphere is zero.
But electric field at any point due to a dipole
cannot be zero.
7. Gauss’s law gives to tal elec tric field and flux
due to all charges.
8. If two con cen tric spheres carry equal and
op po site charges, Elec tric field is non-zero
only in the re gion be tween two sphere and
po ten tial is  is zero only out side both the
spheres.
9. As force on the rod due to elec tric field is
to wards right, force on the rod due to hinge
must be left.
The equilibrium is clearly neutral.
10. If moved along per pen dic u lar bi sec tor, for all 
iden ti cal charges, elec tro static po ten tial
en ergy is max i mum at mid point and if
moved along the line join ing the par ti cles,
elec tr o static po ten ti al en er gy is min i mum at
the mid-point.
 58
q
1
q
2
q
3
Q
1
Q
2
Match the Col umns
1. (a ® s), (b ® q), (c ® r), (d ® p).
If charge at B is removed
E E E
D E net
= ° + ° cos cos 30 30
= 3E                        
If charge at C is removed 
E E E
D f net
= ° + ° cos cos 60 60
= E                          
If charge at D is removed
E
net
®
= 0 and E E
B E
®
= -
®
and E E
E E
®
= -
®
If charge at B and C both are removed,
E E E E
E D F net
= + ° + ° cos cos 60 60
         =2E
2. (a ® q), (b ® p), (c ® s), (d ® r).
V = -
®
×
®
E r
If r i
®
= 4
^
, V = - 8 V, 
If r i
®
= - 4
^
 , V = 8 V
If r j
®
= 4
^
, V = - 16 V,
If r j
®
= - 4
^
, V = 16 V
3. For a solid sphere
V
q
R
R r
in
= × -
1
4 2
3
0
3
2 2
pe
( )
at        r
R
=
2
     V
q
R
R
R
1
0
3
2
2
1
4 2
3
4
= × -
æ
è
ç
ç
ö
ø
÷
÷
pe
 
  =
11
8
V
V
q
r
out
= ×
1
4
0
pe
at r R = 2
V
q
R
V
2
0
1
4 2 2
= × =
pe
E
in
= ×
1
4
0
3
pe
qr
R
at r
R
=
2
E
q
R
V
R
V
1
0
2
1
4 2 2 2
= × = =
pe
(Q R =1 m)
E
out
= ×
1
4
0
2
pe
q
r
at r R = 2
E
q
R
V
R
V
2
0
2
1
4 2 4 4
= × = =
pe ( )
\ (a ® s), (b ® q), (c ® q), (d ® p).
4. (a ® r), (b ® q), (c ® q), (d ® s)
5. (a ® p), (b ® q), (c ® r), (d ® s)
For a spherical shell,
E
r R
Kq
r
r R
=
<
³
ì
í
ï
î
ï
0
2
for
for
V
Kq
R
r R
Kq
r
r R
=
£
³
ì
í
ï
ï
î
ï
ï
for
for
For a solid sphere,
E
Kqr
R
r R
Kq
r
r R
=
£
³
ì
í
ï
ï
î
ï
ï
3
2
for
for
                         
V
Kq
R
R r r R
Kq
r
r R
=
- £
³
ì
í
ï
ï
î
ï
ï
2
2
2
2 2
2
( ) for
for
   
59  
B C
F E
A D
®
E ®
E
®
E
®
E
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