DPP for NEET: Daily Practice Problems, Ch 34: Electrostatics-3 (Solutions)

# DPP for NEET: Daily Practice Problems, Ch 34: Electrostatics-3 (Solutions) | Physics Class 12 PDF Download

Page 1

1. (b) Net electrostatic energy
2
0
2
kQq kq kQq
U
aa a
= + +=
2
0
2 22
kq Qq
QqQ
a
æö
Þ + + = Þ =-
ç÷
+ èø
2. (c) Electric field is perpendicular to the equipotential
surface and is zero every where inside the metal.
3. (c) ( ) 0
2
kqq
QqQQ
l
Þ + + = Þ =-
net potential energy
2
0
1
3.
4
net
q
U
l
=´
pe
4. (d) Length of the diagonal of a cube having each side
b
is
3. b
So distance of centre of cube from each vertex is
3
.
2
b
Hence potential energy of the given system of charge is
( )( )
2
0 0
14
8.
4 3/23
qq q
U
bb
ì -ü -
=´=
íý
pe pe
îþ
5. (a) Change in potential energy ( )
fi
U UU
D
=-
D
30 cm
40 cm
2
q
1
q
3
q
50 cm
10 cm
40 cm
1 3 23 1 3 23
0
1
4 0.4 0.1 0.4 0.5
qq qq qq qq
U
éù æöæö
Þ = + -+
êú ç÷ç÷
pe
èøèø ëû
V
3
232
00
1
[8 ] (8 )
44
q
U qqq Þ==
pe pe
V
2
8 kq \=
6. (c) For pair of charge
12
0
1
.
4
qq
U
r
=
pe
66
0
1 10 10 10 10
4 10 /100
System
U
--
é
´´
= ê
pe
ê
ë
66 66
10 10 10 10 10 10 10 10
10 /100 10/100
-- --
ù
´ ´´ ´ ´´
++ ú
ú
û
12
9
100 10 100
3 9 10 27
10
J
-
´´
=´´´=
7. (c)
0 00
1 ()(2 ) 1 (2 )() 1 ()()
4 4 42
System
q q qq qq
U
a aa
--
= ++
pe pe pe
2
0
7
8
=-
pe
System
q
U
a
8. (c) In the given condition angle between p
ur
and
E
ur
is zero.
Hence potential energy cos0 U pE pE =- =-= min.
Also in uniform electric field 0
net
F =
9. (b) U =
2
0
1 ( e)( e)
4
r
--
pÎ
As r decreases then U
increases and sign of U is '+ve' so, U increases.
10. (c)
2
1 111
.....
U 2kq
a 2a 3a 4a
éù
= - +- ++
êú
ëû

2
0
2q 1 11
.....
1
4 a 234
éù
=- - + -+
êú
pe
ëû

2
e
0
2q log 2
4a
=-
pe
11. (b) The initial energy of the system
q q
q
a
U
i
=
2
0
1q
U
4a
=
pe
The final energy of the system
U
f
=
2 22
0
1 q qq
3U
4 a aa
éù
+ += êú
pe
êú
ëû
Thus work done, W = U
f
– U
i
= 3U – U = 2U
12. (d)
12
=
kqq
U
r
13. (c) As potential at A and B is same, . ==
AB
kQ
VV
d
So,
work done in both cases will be same.
14. (b)
12
. =e
kqq
U
r
There will be 6 pairs, 4 on a side of square
and 2 as diagonal.
Page 2

1. (b) Net electrostatic energy
2
0
2
kQq kq kQq
U
aa a
= + +=
2
0
2 22
kq Qq
QqQ
a
æö
Þ + + = Þ =-
ç÷
+ èø
2. (c) Electric field is perpendicular to the equipotential
surface and is zero every where inside the metal.
3. (c) ( ) 0
2
kqq
QqQQ
l
Þ + + = Þ =-
net potential energy
2
0
1
3.
4
net
q
U
l
=´
pe
4. (d) Length of the diagonal of a cube having each side
b
is
3. b
So distance of centre of cube from each vertex is
3
.
2
b
Hence potential energy of the given system of charge is
( )( )
2
0 0
14
8.
4 3/23
qq q
U
bb
ì -ü -
=´=
íý
pe pe
îþ
5. (a) Change in potential energy ( )
fi
U UU
D
=-
D
30 cm
40 cm
2
q
1
q
3
q
50 cm
10 cm
40 cm
1 3 23 1 3 23
0
1
4 0.4 0.1 0.4 0.5
qq qq qq qq
U
éù æöæö
Þ = + -+
êú ç÷ç÷
pe
èøèø ëû
V
3
232
00
1
[8 ] (8 )
44
q
U qqq Þ==
pe pe
V
2
8 kq \=
6. (c) For pair of charge
12
0
1
.
4
qq
U
r
=
pe
66
0
1 10 10 10 10
4 10 /100
System
U
--
é
´´
= ê
pe
ê
ë
66 66
10 10 10 10 10 10 10 10
10 /100 10/100
-- --
ù
´ ´´ ´ ´´
++ ú
ú
û
12
9
100 10 100
3 9 10 27
10
J
-
´´
=´´´=
7. (c)
0 00
1 ()(2 ) 1 (2 )() 1 ()()
4 4 42
System
q q qq qq
U
a aa
--
= ++
pe pe pe
2
0
7
8
=-
pe
System
q
U
a
8. (c) In the given condition angle between p
ur
and
E
ur
is zero.
Hence potential energy cos0 U pE pE =- =-= min.
Also in uniform electric field 0
net
F =
9. (b) U =
2
0
1 ( e)( e)
4
r
--
pÎ
As r decreases then U
increases and sign of U is '+ve' so, U increases.
10. (c)
2
1 111
.....
U 2kq
a 2a 3a 4a
éù
= - +- ++
êú
ëû

2
0
2q 1 11
.....
1
4 a 234
éù
=- - + -+
êú
pe
ëû

2
e
0
2q log 2
4a
=-
pe
11. (b) The initial energy of the system
q q
q
a
U
i
=
2
0
1q
U
4a
=
pe
The final energy of the system
U
f
=
2 22
0
1 q qq
3U
4 a aa
éù
+ += êú
pe
êú
ëû
Thus work done, W = U
f
– U
i
= 3U – U = 2U
12. (d)
12
=
kqq
U
r
13. (c) As potential at A and B is same, . ==
AB
kQ
VV
d
So,
work done in both cases will be same.
14. (b)
12
. =e
kqq
U
r
There will be 6 pairs, 4 on a side of square
and 2 as diagonal.
DPP/ P 34
95
15. (c) Apply conservation of mechanical energy between
point a and b : (K.E. + P .E.)
a
= (K.E. + P .E.)
a
99
00
(3 10 ) (3 10 )
0
0.01 0.02
--
´´
Þ+-
k qkq
99
2 00
(3 10) (3 10) 1
2 0.02 0.01
--
´´
=+-
kqkq
mv
Put the values we get : v = 12 15 = 46 m/s
16. (b)
2
0 / 1/4
2
kqQ kqQ kq
U Qq
r rr
=- - + =Þ=
17. (b) Find potential at A and C due to charge at B, then
required work done is W = q (V
A
– V
C
)
18. (d) It depends whether both charges are of same or opposite
sign.
19. (a) Because work is to be done by an external agent in
moving a positive charge from low potential to high
potential and this work gets stored in the form of
potential energy of the system. Hence, it increases.
20. (a)
2
3 31
23
æö
= -+-
ç÷
èø
U kq
d dd
22
2 31 2 21
23 23
æöæö
+ -+- + -+-
ç÷ç÷
èøèø
kq kq
dd d d dd
22
1 2 1 21
232
æö æö
+ -+ - + -+
ç÷ ç÷
èø èø
kq kq
dd ddd
22
121
2
æö æö
+ -+ +-
ç÷ ç÷
èø èø
kq kq
dd d
2
12 12 4
23
æö
=- +-
ç÷
èø
U kq
d dd
2
12 11
1
2 33
æö
=- -+
ç÷
èø
kq
d
21. (c)
22
2 0
0
4
11
22
e
= Þ=e=
kQ kQ
V uE
r
r
4
µ Vu
22. q 4 B
A
C
AC = 5m,
96
kq 9 10 1 10
V
AC5
-
´ ´´
==
= 1.8 × 10
3
= 1.8 kV
V
B
= (V
B
)
due to q
+ (V
B
)
i
, where (V
B
)
i
= Potential at B
due to induced charge
\ 1.8 × 10
3
=
Bi
kq
(V)
AB
+
Þ 1.8 × 10
3
= 2.25 × 10
3
+ (V
B
)
i
Þ (V
B
)
i
= – 0.45 kV
23. (b) Force = eE
Work done = force × distance
Force and distance are in opposite direction, so work
is negative.
W = – eE × d
Here, distance increases so, potential energy increases.
24. (d) Under electrostatic condition, all points lying on the
conductor are in same potential. Therefore, potential
at A = potential at B.
From Gauss's theorem, total flux through the surface
of the cavity will be
0
q/e .
Note : Instead of an elliptical cavity, if it would and
been a spherical cavity then options (a) and (b) were
also correct.
25. (b) q
1
+ q
2
= 0
A
B
C
Q
q
1
q
2
12
24
A
kq kq kQ
V
R RR
= ++
12
4 44
C
kq kq kQ
V
R RR
= ++
V
A
= V
C
\
q
1
= – Q/3 and q
2
= Q/3
26. (b)
0
–
3 2 12 16
A
QQQQ
Vk
RRRR
éù
= + +=
êú
pe ëû
27. (c)
0
–5
6 2 12 48
B
QQQQ
Vk
RRRR
éù
= + +=
êú
pe ëû
28. (a) Inside electric field is zero but not outside.
29. (c) Earth also has some surface charge density due to which
it produces electric field in the surrounding space.
30. (d) Net potential at centre
0
/ 2 / 2 /2 /2
kq kq kq kq
aa aa
+ -+- =
and field is zero due to symmetry .
–q
a
a
a
+q –q
a
+q
If electric potential at a point is zero then the magnitude
of electric field at that point is not necessarily to be
zero.

## Physics Class 12

157 videos|453 docs|185 tests

## Physics Class 12

157 videos|453 docs|185 tests

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