DPP for NEET: Daily Practice Problems, Ch 35: Electrostatics- 4 (Solutions)

# Electrostatics- 4 Practice Questions - DPP for NEET

``` Page 1

1. (a) By using
/ 1/ 1/
0
40 50 4/5
t CR CR CR
VVe ee
- --
= Þ= Þ=
Potential difference after 2 sec
2
2/ 1/2
0
4
' 50( ) 50 32
5
CR CR
VVeeV
-- æö
= = ==
ç÷
èø
Fraction of energy after 1 sec
2
2
2
1
()
40 16
2
1
50 25
()
2
æö
= ==
ç÷
èø
f
i
CV
CV
2. (c) The given circuit can be redrawn as follows. All
capacitors are identical and each having capacitance
0
A
C
d
e
=
V
+ –
1 2
3
5
2
4
4
| Charge on each capacitor | = | Charge on each plate |
0
A
V
d
e
=
Plate 1 is connected with positive terminal of battery
so charge on it will be
0
.
A
V
d
e
+
Plate 4 comes twice and it is connected with negative
terminal of battery , so charge on plate 4 will be
0
2 A
V
d
e
-
3. (c) Initially potential difference across both the capacitor
is same hence energy of the system is
2 22
1
11
22
U CV CV CV =+= .....(i)
In the second case when key K is opened and dielectric
medium is filled between the plates, capacitance of both
the capacitors becomes 3C, while potential difference
across A is V and potential difference across B is
3
V
hence energy of the system now is
2
22
2
1 1 10
(3) (3)
2 2 36
V
U C V C CV
æö
= +=
ç÷
èø
......(ii)
So,
1
2
3
5
U
U
=
4. (c) Plane conducting surfaces facing each other must have
equal and opposite charge densities. Here as the plate
areas are equal,
23
QQ =- .
The charge on a capacitor means the charge on the
inner surface of the positive plate (here it is
2
Q )
Potential difference between the plates
charge
capacitance
=
22
2
2
QQ
CC
==
23 22
()
22
QQ QQ
CC
- --
==
5. (b) While drawing the dielectric plate outside, the
capacitance decreases till the entire plate comes out
and then becomes constant. So,
V
increases and then
becomes constant.
6. (b) Given circuit can be reduced as follows
A B
3C 3C
(
C
= capacitance of each capacitor)
The capacitor 3 ,3 CC shown in figure can with stand
maximum 200
V
.
\
So maximum voltage that can be applied across
A and B equally shared. Hence maximum voltage applied
cross A and B be equally shared. Hence Maz. voltage
applied across A and B will be (200 + 200) = 400 volt.
7. (c) Common potential
1 21
12 12
0
'.
CVCC
VV
C C CC
+´
==
++
8. (b)
1 1 22
12
6 12 3 12
4 volt
36
- ´ -´
= ==
++
CV CV
V
CC
9. (d)
10
10
1
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
Page 2

1. (a) By using
/ 1/ 1/
0
40 50 4/5
t CR CR CR
VVe ee
- --
= Þ= Þ=
Potential difference after 2 sec
2
2/ 1/2
0
4
' 50( ) 50 32
5
CR CR
VVeeV
-- æö
= = ==
ç÷
èø
Fraction of energy after 1 sec
2
2
2
1
()
40 16
2
1
50 25
()
2
æö
= ==
ç÷
èø
f
i
CV
CV
2. (c) The given circuit can be redrawn as follows. All
capacitors are identical and each having capacitance
0
A
C
d
e
=
V
+ –
1 2
3
5
2
4
4
| Charge on each capacitor | = | Charge on each plate |
0
A
V
d
e
=
Plate 1 is connected with positive terminal of battery
so charge on it will be
0
.
A
V
d
e
+
Plate 4 comes twice and it is connected with negative
terminal of battery , so charge on plate 4 will be
0
2 A
V
d
e
-
3. (c) Initially potential difference across both the capacitor
is same hence energy of the system is
2 22
1
11
22
U CV CV CV =+= .....(i)
In the second case when key K is opened and dielectric
medium is filled between the plates, capacitance of both
the capacitors becomes 3C, while potential difference
across A is V and potential difference across B is
3
V
hence energy of the system now is
2
22
2
1 1 10
(3) (3)
2 2 36
V
U C V C CV
æö
= +=
ç÷
èø
......(ii)
So,
1
2
3
5
U
U
=
4. (c) Plane conducting surfaces facing each other must have
equal and opposite charge densities. Here as the plate
areas are equal,
23
QQ =- .
The charge on a capacitor means the charge on the
inner surface of the positive plate (here it is
2
Q )
Potential difference between the plates
charge
capacitance
=
22
2
2
QQ
CC
==
23 22
()
22
QQ QQ
CC
- --
==
5. (b) While drawing the dielectric plate outside, the
capacitance decreases till the entire plate comes out
and then becomes constant. So,
V
increases and then
becomes constant.
6. (b) Given circuit can be reduced as follows
A B
3C 3C
(
C
= capacitance of each capacitor)
The capacitor 3 ,3 CC shown in figure can with stand
maximum 200
V
.
\
So maximum voltage that can be applied across
A and B equally shared. Hence maximum voltage applied
cross A and B be equally shared. Hence Maz. voltage
applied across A and B will be (200 + 200) = 400 volt.
7. (c) Common potential
1 21
12 12
0
'.
CVCC
VV
C C CC
+´
==
++
8. (b)
1 1 22
12
6 12 3 12
4 volt
36
- ´ -´
= ==
++
CV CV
V
CC
9. (d)
10
10
1
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
DPP/ P 35
97
20
20
2
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
and
3 0 30
3
22
KA KA
C
dd
ee
==
Now,
30 1 2 12
3
1 2 12
.
2
eq
KA CC KK
CC
C C K Kd
æöe
= + =+
ç÷
++
èø
10. (b) Given circuit is a balanced Wheatstone bridge.
11. (a)
P
Q
2C
2C
2C
C + C = C 2
C
C C
Þ
P
Q
2C
2C
2C
2C
22 C/ = C
C
Þ
P
Q
2C
2C
C + C = C 2
C C
Þ
P
Q
2C
2C
2C

Þ

3
PQ
CC =
12. (d) The two capacitors formed by the slabs may assumed
to be in series combination.
13. (d) In the given system, no current will flow through the
branch
CD
so it can be removed
A
C
B
10µF 10µF
10µF 10µF
D
5µF
5µF
Effective capacitance of the system 5 5 10 F = + =m
14. (c) V olume of 8 small drops = V olume of big drop
33
44
82
33
r R Rr ´ p = p Þ=
As capacity is proportional to r , hence capacity
becomes 2 times.
15. (c) Potential difference between the plates
air medium
V VV =+
00
() dtt
K
ss
= ´ -+´
ee
0
()
t
V dt
K
s
Þ = -+
e
0
()
Qt
dt
AK
= -+
e
K A
+
+
+
+
+
+
+
–
–
–
–
–
–
–
t
Hence capacitance
0
()
QQ
C
Qt
V
dt
AK
==
-+
e
00
1
() 1
AA
t
dt dt
K K
ee
==
æö
-+ --
ç÷
èø
16. (b)
0
1
A
C pF
d
e
==
and
0
' 24
2
KA
C pFK
d
e
= = \=
17. (a) Potential difference across the condenser
121122 12
1 0 20
VVVEtEt tt
KK
ss
=+= + =+
ee
1 2 12
0 1 2 0 12
tt tt Q
V
KK A KK
æö æö s
Þ= + =+
ç÷ ç÷
ee
èø èø
18. (d) When the battery is disconnected, the charge will
remain same in any case.
Capacitance of a parallel plate capacitor is given by
0
A
C
d
e
=
Page 3

1. (a) By using
/ 1/ 1/
0
40 50 4/5
t CR CR CR
VVe ee
- --
= Þ= Þ=
Potential difference after 2 sec
2
2/ 1/2
0
4
' 50( ) 50 32
5
CR CR
VVeeV
-- æö
= = ==
ç÷
èø
Fraction of energy after 1 sec
2
2
2
1
()
40 16
2
1
50 25
()
2
æö
= ==
ç÷
èø
f
i
CV
CV
2. (c) The given circuit can be redrawn as follows. All
capacitors are identical and each having capacitance
0
A
C
d
e
=
V
+ –
1 2
3
5
2
4
4
| Charge on each capacitor | = | Charge on each plate |
0
A
V
d
e
=
Plate 1 is connected with positive terminal of battery
so charge on it will be
0
.
A
V
d
e
+
Plate 4 comes twice and it is connected with negative
terminal of battery , so charge on plate 4 will be
0
2 A
V
d
e
-
3. (c) Initially potential difference across both the capacitor
is same hence energy of the system is
2 22
1
11
22
U CV CV CV =+= .....(i)
In the second case when key K is opened and dielectric
medium is filled between the plates, capacitance of both
the capacitors becomes 3C, while potential difference
across A is V and potential difference across B is
3
V
hence energy of the system now is
2
22
2
1 1 10
(3) (3)
2 2 36
V
U C V C CV
æö
= +=
ç÷
èø
......(ii)
So,
1
2
3
5
U
U
=
4. (c) Plane conducting surfaces facing each other must have
equal and opposite charge densities. Here as the plate
areas are equal,
23
QQ =- .
The charge on a capacitor means the charge on the
inner surface of the positive plate (here it is
2
Q )
Potential difference between the plates
charge
capacitance
=
22
2
2
QQ
CC
==
23 22
()
22
QQ QQ
CC
- --
==
5. (b) While drawing the dielectric plate outside, the
capacitance decreases till the entire plate comes out
and then becomes constant. So,
V
increases and then
becomes constant.
6. (b) Given circuit can be reduced as follows
A B
3C 3C
(
C
= capacitance of each capacitor)
The capacitor 3 ,3 CC shown in figure can with stand
maximum 200
V
.
\
So maximum voltage that can be applied across
A and B equally shared. Hence maximum voltage applied
cross A and B be equally shared. Hence Maz. voltage
applied across A and B will be (200 + 200) = 400 volt.
7. (c) Common potential
1 21
12 12
0
'.
CVCC
VV
C C CC
+´
==
++
8. (b)
1 1 22
12
6 12 3 12
4 volt
36
- ´ -´
= ==
++
CV CV
V
CC
9. (d)
10
10
1
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
DPP/ P 35
97
20
20
2
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
and
3 0 30
3
22
KA KA
C
dd
ee
==
Now,
30 1 2 12
3
1 2 12
.
2
eq
KA CC KK
CC
C C K Kd
æöe
= + =+
ç÷
++
èø
10. (b) Given circuit is a balanced Wheatstone bridge.
11. (a)
P
Q
2C
2C
2C
C + C = C 2
C
C C
Þ
P
Q
2C
2C
2C
2C
22 C/ = C
C
Þ
P
Q
2C
2C
C + C = C 2
C C
Þ
P
Q
2C
2C
2C

Þ

3
PQ
CC =
12. (d) The two capacitors formed by the slabs may assumed
to be in series combination.
13. (d) In the given system, no current will flow through the
branch
CD
so it can be removed
A
C
B
10µF 10µF
10µF 10µF
D
5µF
5µF
Effective capacitance of the system 5 5 10 F = + =m
14. (c) V olume of 8 small drops = V olume of big drop
33
44
82
33
r R Rr ´ p = p Þ=
As capacity is proportional to r , hence capacity
becomes 2 times.
15. (c) Potential difference between the plates
air medium
V VV =+
00
() dtt
K
ss
= ´ -+´
ee
0
()
t
V dt
K
s
Þ = -+
e
0
()
Qt
dt
AK
= -+
e
K A
+
+
+
+
+
+
+
–
–
–
–
–
–
–
t
Hence capacitance
0
()
QQ
C
Qt
V
dt
AK
==
-+
e
00
1
() 1
AA
t
dt dt
K K
ee
==
æö
-+ --
ç÷
èø
16. (b)
0
1
A
C pF
d
e
==
and
0
' 24
2
KA
C pFK
d
e
= = \=
17. (a) Potential difference across the condenser
121122 12
1 0 20
VVVEtEt tt
KK
ss
=+= + =+
ee
1 2 12
0 1 2 0 12
tt tt Q
V
KK A KK
æö æö s
Þ= + =+
ç÷ ç÷
ee
èø èø
18. (d) When the battery is disconnected, the charge will
remain same in any case.
Capacitance of a parallel plate capacitor is given by
0
A
C
d
e
=
98
DPP/ P 35
When
d
is increased, capacitance will decreases and
because the charge remains the same, so according to
, = q CV the voltage will increase. Hence the
electrostatics energy stored in the capacitor will
increase.
19. (c)
s –s
.C
.B A.
E
A
=
00
0
22
ss
-=
ee
E
C
= 0,
E
B
=
0 00
.
22
s ss
+=
e ee
20. (b) E =
00
V/kV V
d d kd
==
21. (c) C
eq
=
( ) ( )
( )( )
3 3 11
1
3 3 11
+ ´+
+
+ ++
=
62
1
62
´ æö
+
ç÷
+
èø
=
5
2
F m
\ Q  = C ´ V =
5
100
2
´ = 250 mC
Change in 6 mF branch – VC =
62
100
62
´ æö
ç÷
+
èø
= 150 mC
100 V
6mF 2mF
1mF
A
B
C
V
AB
=
150
6
= 25 V  and V
BC
= 100 – V
AB
= 75 V
22. (c) Capacitance will be increased when a dielectric is
introduced in the capacitor but potential difference will
remain the same because battery is still connected. So
according to q = CV , charge will increase i.e. Q > Q
0
and
00 00 00
11
UQV,U QVQQsoU U
22
= = Þ>>
23. (a) Electric field between the plates of a parallel plate
capacitor
0
0
.
s
==µ
ee
Q
E ieEd
A
24. (a) Capacitance of parallel plate condenser
0
A
d
e
=
25. (a)
C
1
C
2
+C
3
V
23
1
1 23
()
()
C
VCC
V
C CC
+
=
++
Initially C
3
= 0
So
2
1
12
6
C
VC
V
CC
==
+
........... (1)
Now, at
3
1
10,
C
VC = =¥
Þ
23
1 23
()
10
()
VCC
C CC
+
=
++
Þ
1
23
10 10
1
V
V
C
CC
= Þ=
æö
+
ç÷
+èø
.......... (2)
Eq. (1) and (2),
1
2 1
2
10
6531
1
C
C C
C
æö
=Þ=+
ç÷
æö èø
+
ç÷
èø
1
2
52
1
33
C
C
Þ = -=
26. (b) Now ,
23
1
1 23
10()
8
()
C
CC
V
C CC
+
==
++
3 2
11
3 2
11
10
8
1
C C
CC
C C
CC
æö
+
ç÷
èø
Þ=
æö
++
ç÷
èø
Þ  C
3
= 2.5C
1
27. (c)
3 2 11
1 11
C C CC
+»
+
(where C
3
®
¥
)
\
Total energy = energy in C
1
\ Required ratio = 1
Page 4

1. (a) By using
/ 1/ 1/
0
40 50 4/5
t CR CR CR
VVe ee
- --
= Þ= Þ=
Potential difference after 2 sec
2
2/ 1/2
0
4
' 50( ) 50 32
5
CR CR
VVeeV
-- æö
= = ==
ç÷
èø
Fraction of energy after 1 sec
2
2
2
1
()
40 16
2
1
50 25
()
2
æö
= ==
ç÷
èø
f
i
CV
CV
2. (c) The given circuit can be redrawn as follows. All
capacitors are identical and each having capacitance
0
A
C
d
e
=
V
+ –
1 2
3
5
2
4
4
| Charge on each capacitor | = | Charge on each plate |
0
A
V
d
e
=
Plate 1 is connected with positive terminal of battery
so charge on it will be
0
.
A
V
d
e
+
Plate 4 comes twice and it is connected with negative
terminal of battery , so charge on plate 4 will be
0
2 A
V
d
e
-
3. (c) Initially potential difference across both the capacitor
is same hence energy of the system is
2 22
1
11
22
U CV CV CV =+= .....(i)
In the second case when key K is opened and dielectric
medium is filled between the plates, capacitance of both
the capacitors becomes 3C, while potential difference
across A is V and potential difference across B is
3
V
hence energy of the system now is
2
22
2
1 1 10
(3) (3)
2 2 36
V
U C V C CV
æö
= +=
ç÷
èø
......(ii)
So,
1
2
3
5
U
U
=
4. (c) Plane conducting surfaces facing each other must have
equal and opposite charge densities. Here as the plate
areas are equal,
23
QQ =- .
The charge on a capacitor means the charge on the
inner surface of the positive plate (here it is
2
Q )
Potential difference between the plates
charge
capacitance
=
22
2
2
QQ
CC
==
23 22
()
22
QQ QQ
CC
- --
==
5. (b) While drawing the dielectric plate outside, the
capacitance decreases till the entire plate comes out
and then becomes constant. So,
V
increases and then
becomes constant.
6. (b) Given circuit can be reduced as follows
A B
3C 3C
(
C
= capacitance of each capacitor)
The capacitor 3 ,3 CC shown in figure can with stand
maximum 200
V
.
\
So maximum voltage that can be applied across
A and B equally shared. Hence maximum voltage applied
cross A and B be equally shared. Hence Maz. voltage
applied across A and B will be (200 + 200) = 400 volt.
7. (c) Common potential
1 21
12 12
0
'.
CVCC
VV
C C CC
+´
==
++
8. (b)
1 1 22
12
6 12 3 12
4 volt
36
- ´ -´
= ==
++
CV CV
V
CC
9. (d)
10
10
1
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
DPP/ P 35
97
20
20
2
2
2
A
K
KA
C
d d
e
e
==
æö
ç÷
èø
and
3 0 30
3
22
KA KA
C
dd
ee
==
Now,
30 1 2 12
3
1 2 12
.
2
eq
KA CC KK
CC
C C K Kd
æöe
= + =+
ç÷
++
èø
10. (b) Given circuit is a balanced Wheatstone bridge.
11. (a)
P
Q
2C
2C
2C
C + C = C 2
C
C C
Þ
P
Q
2C
2C
2C
2C
22 C/ = C
C
Þ
P
Q
2C
2C
C + C = C 2
C C
Þ
P
Q
2C
2C
2C

Þ

3
PQ
CC =
12. (d) The two capacitors formed by the slabs may assumed
to be in series combination.
13. (d) In the given system, no current will flow through the
branch
CD
so it can be removed
A
C
B
10µF 10µF
10µF 10µF
D
5µF
5µF
Effective capacitance of the system 5 5 10 F = + =m
14. (c) V olume of 8 small drops = V olume of big drop
33
44
82
33
r R Rr ´ p = p Þ=
As capacity is proportional to r , hence capacity
becomes 2 times.
15. (c) Potential difference between the plates
air medium
V VV =+
00
() dtt
K
ss
= ´ -+´
ee
0
()
t
V dt
K
s
Þ = -+
e
0
()
Qt
dt
AK
= -+
e
K A
+
+
+
+
+
+
+
–
–
–
–
–
–
–
t
Hence capacitance
0
()
QQ
C
Qt
V
dt
AK
==
-+
e
00
1
() 1
AA
t
dt dt
K K
ee
==
æö
-+ --
ç÷
èø
16. (b)
0
1
A
C pF
d
e
==
and
0
' 24
2
KA
C pFK
d
e
= = \=
17. (a) Potential difference across the condenser
121122 12
1 0 20
VVVEtEt tt
KK
ss
=+= + =+
ee
1 2 12
0 1 2 0 12
tt tt Q
V
KK A KK
æö æö s
Þ= + =+
ç÷ ç÷
ee
èø èø
18. (d) When the battery is disconnected, the charge will
remain same in any case.
Capacitance of a parallel plate capacitor is given by
0
A
C
d
e
=
98
DPP/ P 35
When
d
is increased, capacitance will decreases and
because the charge remains the same, so according to
, = q CV the voltage will increase. Hence the
electrostatics energy stored in the capacitor will
increase.
19. (c)
s –s
.C
.B A.
E
A
=
00
0
22
ss
-=
ee
E
C
= 0,
E
B
=
0 00
.
22
s ss
+=
e ee
20. (b) E =
00
V/kV V
d d kd
==
21. (c) C
eq
=
( ) ( )
( )( )
3 3 11
1
3 3 11
+ ´+
+
+ ++
=
62
1
62
´ æö
+
ç÷
+
èø
=
5
2
F m
\ Q  = C ´ V =
5
100
2
´ = 250 mC
Change in 6 mF branch – VC =
62
100
62
´ æö
ç÷
+
èø
= 150 mC
100 V
6mF 2mF
1mF
A
B
C
V
AB
=
150
6
= 25 V  and V
BC
= 100 – V
AB
= 75 V
22. (c) Capacitance will be increased when a dielectric is
introduced in the capacitor but potential difference will
remain the same because battery is still connected. So
according to q = CV , charge will increase i.e. Q > Q
0
and
00 00 00
11
UQV,U QVQQsoU U
22
= = Þ>>
23. (a) Electric field between the plates of a parallel plate
capacitor
0
0
.
s
==µ
ee
Q
E ieEd
A
24. (a) Capacitance of parallel plate condenser
0
A
d
e
=
25. (a)
C
1
C
2
+C
3
V
23
1
1 23
()
()
C
VCC
V
C CC
+
=
++
Initially C
3
= 0
So
2
1
12
6
C
VC
V
CC
==
+
........... (1)
Now, at
3
1
10,
C
VC = =¥
Þ
23
1 23
()
10
()
VCC
C CC
+
=
++
Þ
1
23
10 10
1
V
V
C
CC
= Þ=
æö
+
ç÷
+èø
.......... (2)
Eq. (1) and (2),
1
2 1
2
10
6531
1
C
C C
C
æö
=Þ=+
ç÷
æö èø
+
ç÷
èø
1
2
52
1
33
C
C
Þ = -=
26. (b) Now ,
23
1
1 23
10()
8
()
C
CC
V
C CC
+
==
++
3 2
11
3 2
11
10
8
1
C C
CC
C C
CC
æö
+
ç÷
èø
Þ=
æö
++
ç÷
èø
Þ  C
3
= 2.5C
1
27. (c)
3 2 11
1 11
C C CC
+»
+
(where C
3
®
¥
)
\
Total energy = energy in C
1
\ Required ratio = 1
DPP/ P 35
99
28. (b) The electric field due to one charged plate at the
location of the other is
0
2
E
s
=
e
and the force per unit
area is
2
0
.
2
FE
s
=s=
e
29. (d) A charged capacitor, after removing the battery, does
not discharge itself. If this capacitor is touched by
someone, he may feel shock due to large charge still
present on the capacitor. Hence it should be handled
cautiously otherwise this may cause a severe shock.
30. (b) By the formula capacitance of a capacitor
10
KAK
C
dd
=e´µ
Hence,
1 1 21
2 1 22
/21
36
C K dK d
C d K KK
= ´=´ =
or C
2
= 6C
1
Again for capacity of a capacitor
Q
C
V
=
Therefore, capacity of a capacitor does not depend
upon the nature of the material of the capacitor.
```

## Physics Class 12

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## Physics Class 12

105 videos|414 docs|114 tests

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