Electrostatics- 2 Practice Questions - DPP for NEET

 Page 1


1. (c) ABCDE is an equipotential surface, on equipotential
surface no work is done in shifting a charge from one
place to another.
2. (c) Potential at centre O of the square
Q Q
Q Q
O
a
2
a
0
0
4 ( / 2)
æö
=ç÷
ç÷
pe
èø
Q
V
a
Work done in shifting (– Q ) charge from centre to
infinity
00
() W Q V V QV
¥
=- -=
     
2
00
422
.
4
QQ
Q
aa
==
pe pe
3. (b) Using
21
42
= Þµ Þ = ==
AA
BB
vQ QVq
v vQ
M v Qq
4. (a) Work done in moving a charge from P to L , P to M
and P to N is zero while it is q (V
P
 – V
k
) > 0 for motion
from P to k.
5. (a)
12
( ) 2 (70 50) 40 KE q V V eV = - = ´ - W=
6. (a) The electric potential 
2
(, ,)4 Vxyzx = volt
Now 
$ $
VVV
E i jk
x yz
æö ¶ ¶¶
=- ++
ç÷
¶ ¶¶
èø
ur
$
Now 
8,0
VV
x
xy
¶¶
==
¶¶
and 
0
V
z
¶
=
¶
Hence 
8, =- Ei
ur
$
so at point (1 ,0,2) mm
8 E xi =-
ur
$
volt/meter or 8 along negative X - axis.
7. (b) Electric fields due to electrons on same line passing
through centre cancel each other while electric potential
due to each electron is negative at centre C. Therefore,
at centre 0,0 EV
ur
=±
8. (a) By using ( .) W QEr =D
urr
$ $ $
1 2 3 12
[( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+
$$
9. ( b) Potential at A= Potential due to (+q) charge
+ Potential due to (– q) charge
2 2 22
00
1 1 ()
.0
44
qq
a b ab
-
= +=
pe pe
++
10. (c) Point P will lie near the charge which is smaller in
magnitude i.e. 6 C -m . Hence potential at P
P
6 C - m 12 C m
x
20 cm
66
00
1 ( 6 10 ) 1 (12 10 )
0 0.2
4 4 (0.2)
V xm
xx
--
-´´
= + =Þ=
pe pe +
11. (a) Work done 
6
( );
AB
W q VV
-
=- where
6
3 10 coulomb q
-
=´ where
66
106
22
( 5 10 ) 2 10 1
10 10
15
1510 510
A
V
--
--
éù
-´´
= + =´ êú
´´ êú
ëû
volt
and  
66
10
22
(2 10 ) 5 10
10
15 10 5 10
B
V
--
--
éù
´´
=- êú
´´ êú
ëû
                = –
6
13
10
15
´
 volt
6 66
1 13
3 10 10 10
15 15
-
éù æö
\ =´ ´ -´
ç÷ êú
èø ëû
W
         = 2.8 J
12. (c)
A
B C
l l
l
+q
+q – 2q
60°
p
p
net
p
22
2 cos 60 3 3 ( ) =+ + == \=
net
P p p pp p ql p ql
o
Page 2


1. (c) ABCDE is an equipotential surface, on equipotential
surface no work is done in shifting a charge from one
place to another.
2. (c) Potential at centre O of the square
Q Q
Q Q
O
a
2
a
0
0
4 ( / 2)
æö
=ç÷
ç÷
pe
èø
Q
V
a
Work done in shifting (– Q ) charge from centre to
infinity
00
() W Q V V QV
¥
=- -=
     
2
00
422
.
4
QQ
Q
aa
==
pe pe
3. (b) Using
21
42
= Þµ Þ = ==
AA
BB
vQ QVq
v vQ
M v Qq
4. (a) Work done in moving a charge from P to L , P to M
and P to N is zero while it is q (V
P
 – V
k
) > 0 for motion
from P to k.
5. (a)
12
( ) 2 (70 50) 40 KE q V V eV = - = ´ - W=
6. (a) The electric potential 
2
(, ,)4 Vxyzx = volt
Now 
$ $
VVV
E i jk
x yz
æö ¶ ¶¶
=- ++
ç÷
¶ ¶¶
èø
ur
$
Now 
8,0
VV
x
xy
¶¶
==
¶¶
and 
0
V
z
¶
=
¶
Hence 
8, =- Ei
ur
$
so at point (1 ,0,2) mm
8 E xi =-
ur
$
volt/meter or 8 along negative X - axis.
7. (b) Electric fields due to electrons on same line passing
through centre cancel each other while electric potential
due to each electron is negative at centre C. Therefore,
at centre 0,0 EV
ur
=±
8. (a) By using ( .) W QEr =D
urr
$ $ $
1 2 3 12
[( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+
$$
9. ( b) Potential at A= Potential due to (+q) charge
+ Potential due to (– q) charge
2 2 22
00
1 1 ()
.0
44
qq
a b ab
-
= +=
pe pe
++
10. (c) Point P will lie near the charge which is smaller in
magnitude i.e. 6 C -m . Hence potential at P
P
6 C - m 12 C m
x
20 cm
66
00
1 ( 6 10 ) 1 (12 10 )
0 0.2
4 4 (0.2)
V xm
xx
--
-´´
= + =Þ=
pe pe +
11. (a) Work done 
6
( );
AB
W q VV
-
=- where
6
3 10 coulomb q
-
=´ where
66
106
22
( 5 10 ) 2 10 1
10 10
15
1510 510
A
V
--
--
éù
-´´
= + =´ êú
´´ êú
ëû
volt
and  
66
10
22
(2 10 ) 5 10
10
15 10 5 10
B
V
--
--
éù
´´
=- êú
´´ êú
ëû
                = –
6
13
10
15
´
 volt
6 66
1 13
3 10 10 10
15 15
-
éù æö
\ =´ ´ -´
ç÷ êú
èø ëû
W
         = 2.8 J
12. (c)
A
B C
l l
l
+q
+q – 2q
60°
p
p
net
p
22
2 cos 60 3 3 ( ) =+ + == \=
net
P p p pp p ql p ql
o
92
DPP/ P 33
13. (d) According to figure, potential at 
A
and 
C
are both
equal to kQ. Hence work done in moving q - charge
from 
A
to 
C
()
AC
qVV =-- = 0
A
B C
l l
l
–q
+Q
14. (c)
19
9
10
( 1.6 10 )
9 10 27.2
0.53 10
Q
VkV
r
-
-
+´
=´=´´=
´
15. (c) Potential will be zero at two points
O
M N
x = 0 x = 4 x = 6 x = 12
1
2 q C =m
2
1 q C = -m
l ' l
6
At internal point (M) :
66
0
1 2 10 (110)
0
4 (6)ll
--
éù
´ -´
´ += êú
pe-
êú
ëû
2l Þ=
So distance of M from origin;
6 24 x=-=
At exterior point (N):
66
0
1 2 10 ( 1 10 )
0
4 (6 ')ll
--
éù
´ -´
´ += êú
¢ pe-
êú
ëû
'6l Þ=
So distance of 
N
from origin, 
6 6 12 x= +=
16. (a) V = V
AB
 + V
BC
 + V
CD
   
0 00
.5 .( 2 ) .(3 ) k Q k Q kQ
R RR
-
= ++
   
0
6kQ
R
=
   
0
0
3
2
Q
R
=
pÎ
17. (a)
9
2
9 10.
p
V
r
=´
19 10
9
102
(1.6 10 ) 1.28 10
9 10 0.13
(1210)
V
--
-
´ ´´
=´´=
´
18. (b)
02 01
() =- W qVV
where 
12
01
0 0
4 42
=+
pe pe
QQ
V
R R
and 
21
02
0 0
4
42
=+
pe pe
QQ
V
R R
21
02 01
0
() ( 2 1)
()
4 2
qQQ
W qVV
R
- -
Þ = -=
pe
19. (d)  
00 00
1 1 111
1 ... ...
4 3 5 4 2 46
qq
V
xx
é ù éù
= + + + - + ++
ê ú êú
pe pe ë û ëû
   
0 0 00
111
1 ... log2
4 2344
e
qq
xx
éù
= - + - +=
êú
pe pe
ëû
20. (b) Potential decreases in the direction of electric field.
Dotted lines are equipotential surfaces
AC
VV \= and 
AB
VV >
E
y
C
A B
21. (d)
3
..
equatorial
kp
E ieE p
r
=µ and 
3 -
µ Er
22. (a) Suppose neutral point N lies at a distance x from dipole
of moment p or at a distance x
2
 from dipole of 64 p.
2
1
x
1
25 cm
p
64 
N
p
®
p
®
At  N |E.F . due to dipole 1 | = |E.F. due to dipole 2 |
Þ
3
0
12
.
4
p
x
pe
 = 
( )
( )
3
0
2 64 1
.
4
25
p
x
pe
-
Þ
3
1
x
 = 
( )
3
64
25 x -
  Þ  x = 5 cm.
23. (a). BC = 2R sin 
120
3R
2
æö
=
ç÷
èø
Electric field  at O = 
22
0
0
1 2q/3q
4R
R 6R
æö
=
ç÷
èø pe
pe
along negative X-axis.
Page 3


1. (c) ABCDE is an equipotential surface, on equipotential
surface no work is done in shifting a charge from one
place to another.
2. (c) Potential at centre O of the square
Q Q
Q Q
O
a
2
a
0
0
4 ( / 2)
æö
=ç÷
ç÷
pe
èø
Q
V
a
Work done in shifting (– Q ) charge from centre to
infinity
00
() W Q V V QV
¥
=- -=
     
2
00
422
.
4
QQ
Q
aa
==
pe pe
3. (b) Using
21
42
= Þµ Þ = ==
AA
BB
vQ QVq
v vQ
M v Qq
4. (a) Work done in moving a charge from P to L , P to M
and P to N is zero while it is q (V
P
 – V
k
) > 0 for motion
from P to k.
5. (a)
12
( ) 2 (70 50) 40 KE q V V eV = - = ´ - W=
6. (a) The electric potential 
2
(, ,)4 Vxyzx = volt
Now 
$ $
VVV
E i jk
x yz
æö ¶ ¶¶
=- ++
ç÷
¶ ¶¶
èø
ur
$
Now 
8,0
VV
x
xy
¶¶
==
¶¶
and 
0
V
z
¶
=
¶
Hence 
8, =- Ei
ur
$
so at point (1 ,0,2) mm
8 E xi =-
ur
$
volt/meter or 8 along negative X - axis.
7. (b) Electric fields due to electrons on same line passing
through centre cancel each other while electric potential
due to each electron is negative at centre C. Therefore,
at centre 0,0 EV
ur
=±
8. (a) By using ( .) W QEr =D
urr
$ $ $
1 2 3 12
[( ).( )] ( ) W Q e i e j e k ai bj Qea eb Þ= + + + =+
$$
9. ( b) Potential at A= Potential due to (+q) charge
+ Potential due to (– q) charge
2 2 22
00
1 1 ()
.0
44
qq
a b ab
-
= +=
pe pe
++
10. (c) Point P will lie near the charge which is smaller in
magnitude i.e. 6 C -m . Hence potential at P
P
6 C - m 12 C m
x
20 cm
66
00
1 ( 6 10 ) 1 (12 10 )
0 0.2
4 4 (0.2)
V xm
xx
--
-´´
= + =Þ=
pe pe +
11. (a) Work done 
6
( );
AB
W q VV
-
=- where
6
3 10 coulomb q
-
=´ where
66
106
22
( 5 10 ) 2 10 1
10 10
15
1510 510
A
V
--
--
éù
-´´
= + =´ êú
´´ êú
ëû
volt
and  
66
10
22
(2 10 ) 5 10
10
15 10 5 10
B
V
--
--
éù
´´
=- êú
´´ êú
ëû
                = –
6
13
10
15
´
 volt
6 66
1 13
3 10 10 10
15 15
-
éù æö
\ =´ ´ -´
ç÷ êú
èø ëû
W
         = 2.8 J
12. (c)
A
B C
l l
l
+q
+q – 2q
60°
p
p
net
p
22
2 cos 60 3 3 ( ) =+ + == \=
net
P p p pp p ql p ql
o
92
DPP/ P 33
13. (d) According to figure, potential at 
A
and 
C
are both
equal to kQ. Hence work done in moving q - charge
from 
A
to 
C
()
AC
qVV =-- = 0
A
B C
l l
l
–q
+Q
14. (c)
19
9
10
( 1.6 10 )
9 10 27.2
0.53 10
Q
VkV
r
-
-
+´
=´=´´=
´
15. (c) Potential will be zero at two points
O
M N
x = 0 x = 4 x = 6 x = 12
1
2 q C =m
2
1 q C = -m
l ' l
6
At internal point (M) :
66
0
1 2 10 (110)
0
4 (6)ll
--
éù
´ -´
´ += êú
pe-
êú
ëû
2l Þ=
So distance of M from origin;
6 24 x=-=
At exterior point (N):
66
0
1 2 10 ( 1 10 )
0
4 (6 ')ll
--
éù
´ -´
´ += êú
¢ pe-
êú
ëû
'6l Þ=
So distance of 
N
from origin, 
6 6 12 x= +=
16. (a) V = V
AB
 + V
BC
 + V
CD
   
0 00
.5 .( 2 ) .(3 ) k Q k Q kQ
R RR
-
= ++
   
0
6kQ
R
=
   
0
0
3
2
Q
R
=
pÎ
17. (a)
9
2
9 10.
p
V
r
=´
19 10
9
102
(1.6 10 ) 1.28 10
9 10 0.13
(1210)
V
--
-
´ ´´
=´´=
´
18. (b)
02 01
() =- W qVV
where 
12
01
0 0
4 42
=+
pe pe
QQ
V
R R
and 
21
02
0 0
4
42
=+
pe pe
QQ
V
R R
21
02 01
0
() ( 2 1)
()
4 2
qQQ
W qVV
R
- -
Þ = -=
pe
19. (d)  
00 00
1 1 111
1 ... ...
4 3 5 4 2 46
qq
V
xx
é ù éù
= + + + - + ++
ê ú êú
pe pe ë û ëû
   
0 0 00
111
1 ... log2
4 2344
e
qq
xx
éù
= - + - +=
êú
pe pe
ëû
20. (b) Potential decreases in the direction of electric field.
Dotted lines are equipotential surfaces
AC
VV \= and 
AB
VV >
E
y
C
A B
21. (d)
3
..
equatorial
kp
E ieE p
r
=µ and 
3 -
µ Er
22. (a) Suppose neutral point N lies at a distance x from dipole
of moment p or at a distance x
2
 from dipole of 64 p.
2
1
x
1
25 cm
p
64 
N
p
®
p
®
At  N |E.F . due to dipole 1 | = |E.F. due to dipole 2 |
Þ
3
0
12
.
4
p
x
pe
 = 
( )
( )
3
0
2 64 1
.
4
25
p
x
pe
-
Þ
3
1
x
 = 
( )
3
64
25 x -
  Þ  x = 5 cm.
23. (a). BC = 2R sin 
120
3R
2
æö
=
ç÷
èø
Electric field  at O = 
22
0
0
1 2q/3q
4R
R 6R
æö
=
ç÷
èø pe
pe
along negative X-axis.
DPP/ P 33
93
C
A
B
60°
O
60°
60°
60°
30°
30°
120°
q/3
q/3
–2q/3
The potential energy of the system is non zero
Force between B & C
= 
2
22
0
0
1 (q / 3) ( 2q / 3) q
4
( 3R) 54 R
-
=
pe
pe
Potential at O =
0
1 q q 2q
0
4 3 33
æö
+-=
ç÷
èø pe
24. (d) The given graph is of charged conducting sphere of
radius R
0
. The whole charge q distributes on the sur-
face of sphere
25 (b), 26 (b), 27 c
+
–
F
+
F
–
(F
+
 > F
–
 as E
+
 > E
–
)
Net force F : 
F
–
F
+
F
y
x
Net torque immediately after it is released Þ clockwise
A body cannot exert force on itself.
28. (d) When the bird perches on a single high power line, no
current passes through its body because its body is at
equipotential surface i.e., there is no potential
difference. While when man touches the same line,
standing bare foot on ground the electrical circuit is
completed through the ground. The hands of man are
at high potential and his feet’s are at low potential.
Hence large amount of current flows through the body
of the man and person therefore gets a fatal shock.
29. (a) Electron has negative charge, in electric field negative
charge moves from lower potential to higher potential.
30. (b) Potential is constant on the surface of a sphere so it
behaves as an equipotential surface. Free charges
(electrons) are available in conductor. The two
statements are independent.