DC Pandey Solutions: Capacitors- 2

# DC Pandey Solutions: Capacitors- 2 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

31.
(a) Consider the circuit as combination of two
cells of emf E and OV.
E
E R E R
R R
V
e
=
+
+
=
1 2 2 1
1 2
2
R R
R R
e
= + =
2
3
2

q q e
t
= -
-
0
1 ( )
/t

q
CE
0
2
=
t =
3
2
RC

\ q
CE
e
t RC
= -
-
2
1
2 3
( )
/
(b) I
dq
dt
E
R
e
t RC
1
2 3
3
= =
- /
In loop EDBA
q
C
I R I R + - =
1 2
0
I
q
RC
I
2 1
= +
= - +
- -
E
R
e
E
R
e
t RC t RC
2
1
3
2 3 2 3
( )
/ /
= -
-
E
R
e
t RC
6
3
2 3
( )
/
Objective Questions (Level 1)
1. F
Q
A
=
2
0
2e
is independent of d.
2. C
q
V
=
On connecting the plates V becomes zero.
3. The system can assumed to a parallel
combination of two spherical conductors.
C C C a b = + = +
1 2 0 0
4 4 pe pe
= + 4
0
pe ( ) a b
4. V
q
C
=
On connecting in series
q q ¢ = = Charge on any capacitor
C
C
n
¢ =
\ V
nq
C
nV ¢ = =
5. In cor rect di a gram.
6. Charge on capacitor of capacitance
C C
V
2 2
30 = = C
V
q
C
MN
= = 30 V
7. For equilibrium,
qE mg =
q
V
d
r g =
4
3
3
p r
V
r
q
µ
3
71
A B
I
I
1
I
E
R
F
V
R
F D
C
I
2
R
C
3R/2
V/2
I
60V C C
2C
2C
A
B
M
N
60V C
A
B
C
—
2
E
I
V
C
R
B
G
C
R
A F
I
1
q
I–I
1
R
qE
mg
Page 2

31.
(a) Consider the circuit as combination of two
cells of emf E and OV.
E
E R E R
R R
V
e
=
+
+
=
1 2 2 1
1 2
2
R R
R R
e
= + =
2
3
2

q q e
t
= -
-
0
1 ( )
/t

q
CE
0
2
=
t =
3
2
RC

\ q
CE
e
t RC
= -
-
2
1
2 3
( )
/
(b) I
dq
dt
E
R
e
t RC
1
2 3
3
= =
- /
In loop EDBA
q
C
I R I R + - =
1 2
0
I
q
RC
I
2 1
= +
= - +
- -
E
R
e
E
R
e
t RC t RC
2
1
3
2 3 2 3
( )
/ /
= -
-
E
R
e
t RC
6
3
2 3
( )
/
Objective Questions (Level 1)
1. F
Q
A
=
2
0
2e
is independent of d.
2. C
q
V
=
On connecting the plates V becomes zero.
3. The system can assumed to a parallel
combination of two spherical conductors.
C C C a b = + = +
1 2 0 0
4 4 pe pe
= + 4
0
pe ( ) a b
4. V
q
C
=
On connecting in series
q q ¢ = = Charge on any capacitor
C
C
n
¢ =
\ V
nq
C
nV ¢ = =
5. In cor rect di a gram.
6. Charge on capacitor of capacitance
C C
V
2 2
30 = = C
V
q
C
MN
= = 30 V
7. For equilibrium,
qE mg =
q
V
d
r g =
4
3
3
p r
V
r
q
µ
3
71
A B
I
I
1
I
E
R
F
V
R
F D
C
I
2
R
C
3R/2
V/2
I
60V C C
2C
2C
A
B
M
N
60V C
A
B
C
—
2
E
I
V
C
R
B
G
C
R
A F
I
1
q
I–I
1
R
qE
mg
V
V
r
r
q
q
2
1
2
1
3
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
´
Þ           V
2
4 = V
8. Electric field between the plates is uniform
but in all other regions it is zero.
9. Initially the capacitor offers zero resistance.
i =
+
=
12
1 2
4 A
10. q CV CE = =
11. In the steady state, capacitor behaves as
open circuit. the equivalent diagram is given
by
I
E
R R
E
R
=
+
=
3
4
4
7
But potential difference across capacitor,
V IR =
Þ 10
4
7
=
E
R
R
Þ E =17.5 V
12. As all the capacitors are connected in series
potential difference across each capacitor is
V
E
= = =
4
10
4
2.5 V
V V
A N
- = = 3V 7.5 V
V
A
= 7.5 V
V V
N B
- =2.5 V
V
B
= -2.5 V
13. Heat produced = Loss of energy
=
+
-
C C
C C
V V
1 2
1 2
1 2
2
2( )
( )
=
´ ´ ´
+ ´
-
- -
-
2 10 2 10
2 2 2 10
100 0
6 6
6
( )
( )
= ´ =
-
5 10 5
3
J mJ
14.  q qe
t
=
-
0
/h
I I e
t
=
-
0
/h
P I R I e R
t
= =
- 2
0
2 2 /h
=
-
Pe
t
0
2 /h
Þ     h
h
¢=
2
15. Common potential =
+
+
=
C V C V
C C
E
1 1 2 2
1 2
2
16. V V
A B
- = + ´ - + ´ = 6 3 2
9
1
3 3 12 V
72
I
R
3
—
4
E
R
R R
3R
E
R
R
E
R R R
1W
12V
6W
3W
12V
4A
1W
2W
Page 3

31.
(a) Consider the circuit as combination of two
cells of emf E and OV.
E
E R E R
R R
V
e
=
+
+
=
1 2 2 1
1 2
2
R R
R R
e
= + =
2
3
2

q q e
t
= -
-
0
1 ( )
/t

q
CE
0
2
=
t =
3
2
RC

\ q
CE
e
t RC
= -
-
2
1
2 3
( )
/
(b) I
dq
dt
E
R
e
t RC
1
2 3
3
= =
- /
In loop EDBA
q
C
I R I R + - =
1 2
0
I
q
RC
I
2 1
= +
= - +
- -
E
R
e
E
R
e
t RC t RC
2
1
3
2 3 2 3
( )
/ /
= -
-
E
R
e
t RC
6
3
2 3
( )
/
Objective Questions (Level 1)
1. F
Q
A
=
2
0
2e
is independent of d.
2. C
q
V
=
On connecting the plates V becomes zero.
3. The system can assumed to a parallel
combination of two spherical conductors.
C C C a b = + = +
1 2 0 0
4 4 pe pe
= + 4
0
pe ( ) a b
4. V
q
C
=
On connecting in series
q q ¢ = = Charge on any capacitor
C
C
n
¢ =
\ V
nq
C
nV ¢ = =
5. In cor rect di a gram.
6. Charge on capacitor of capacitance
C C
V
2 2
30 = = C
V
q
C
MN
= = 30 V
7. For equilibrium,
qE mg =
q
V
d
r g =
4
3
3
p r
V
r
q
µ
3
71
A B
I
I
1
I
E
R
F
V
R
F D
C
I
2
R
C
3R/2
V/2
I
60V C C
2C
2C
A
B
M
N
60V C
A
B
C
—
2
E
I
V
C
R
B
G
C
R
A F
I
1
q
I–I
1
R
qE
mg
V
V
r
r
q
q
2
1
2
1
3
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
´
Þ           V
2
4 = V
8. Electric field between the plates is uniform
but in all other regions it is zero.
9. Initially the capacitor offers zero resistance.
i =
+
=
12
1 2
4 A
10. q CV CE = =
11. In the steady state, capacitor behaves as
open circuit. the equivalent diagram is given
by
I
E
R R
E
R
=
+
=
3
4
4
7
But potential difference across capacitor,
V IR =
Þ 10
4
7
=
E
R
R
Þ E =17.5 V
12. As all the capacitors are connected in series
potential difference across each capacitor is
V
E
= = =
4
10
4
2.5 V
V V
A N
- = = 3V 7.5 V
V
A
= 7.5 V
V V
N B
- =2.5 V
V
B
= -2.5 V
13. Heat produced = Loss of energy
=
+
-
C C
C C
V V
1 2
1 2
1 2
2
2( )
( )
=
´ ´ ´
+ ´
-
- -
-
2 10 2 10
2 2 2 10
100 0
6 6
6
( )
( )
= ´ =
-
5 10 5
3
J mJ
14.  q qe
t
=
-
0
/h
I I e
t
=
-
0
/h
P I R I e R
t
= =
- 2
0
2 2 /h
=
-
Pe
t
0
2 /h
Þ     h
h
¢=
2
15. Common potential =
+
+
=
C V C V
C C
E
1 1 2 2
1 2
2
16. V V
A B
- = + ´ - + ´ = 6 3 2
9
1
3 3 12 V
72
I
R
3
—
4
E
R
R R
3R
E
R
R
E
R R R
1W
12V
6W
3W
12V
4A
1W
2W
17. In the steady state, current through battery
I =
+
=
12
6 2
3
2
A
Potential difference across the capacitor,
V
AB
= ´ = 6
3
2
9 N
\ q CV
AB
= = ´ = 2 9 18 mC
18. C
2
and C
3
are in parallel
Hence, V V
2 3
=
Again Kirchhoff’s junction rule
- + + = q q q
1 2 3
0
Þ q q q
1 2 3
= +
19. For the motion of electron
R
mu
eE
l = =
2
2 sin q
…(i)
and H
mu
eE
d = =
2 2
2
sin q
…(ii)
Dividing Eq. (ii) by Eq. (i),
tan q =
4d
l
20. V Ed = Þ d
V
E
V
= =
2
6
0
e
=
´ ´ ´
-
-
2 5 10
10
12
7
8.85
= ´ =
-
8.85 0.88 10
4
mm
21.
P and Q are at same potential, hence
capacitor connected between them have no
effect on equivalent capacitance.
22.
23. C
k A
d
k A
d
1
1 0 2 0
2 2
= +
e e

=
+ ( ) k k A
d
1 2 0
2
e
(Parallel grouping)
1
2 2
2 1 0 2 0
C
d
k A
d
k A
= +
e e
(Series grouping)
C
k k
k k
A
d
2
1 2
1 2
0
2
=
+
e
C
C
k k
k k
1
2
1 2
2
1 2
2
4
2 3
4 2 3
25
24
=
+
=
+
´ ´
=
( ) ( )
73
x
x
y
y
2mF 1mF
2mF
x y
8
—
3
mF
2mF
mF
2
—
3
A B
P Q R S
B
S
A
C C C C/2
R
C
C
B A
C C C 3C/2 C
3C/14
x y
1mF 1mF
1mF
2mF
I
A B
12V
2W
4W
2mF
6W
I
O
Page 4

31.
(a) Consider the circuit as combination of two
cells of emf E and OV.
E
E R E R
R R
V
e
=
+
+
=
1 2 2 1
1 2
2
R R
R R
e
= + =
2
3
2

q q e
t
= -
-
0
1 ( )
/t

q
CE
0
2
=
t =
3
2
RC

\ q
CE
e
t RC
= -
-
2
1
2 3
( )
/
(b) I
dq
dt
E
R
e
t RC
1
2 3
3
= =
- /
In loop EDBA
q
C
I R I R + - =
1 2
0
I
q
RC
I
2 1
= +
= - +
- -
E
R
e
E
R
e
t RC t RC
2
1
3
2 3 2 3
( )
/ /
= -
-
E
R
e
t RC
6
3
2 3
( )
/
Objective Questions (Level 1)
1. F
Q
A
=
2
0
2e
is independent of d.
2. C
q
V
=
On connecting the plates V becomes zero.
3. The system can assumed to a parallel
combination of two spherical conductors.
C C C a b = + = +
1 2 0 0
4 4 pe pe
= + 4
0
pe ( ) a b
4. V
q
C
=
On connecting in series
q q ¢ = = Charge on any capacitor
C
C
n
¢ =
\ V
nq
C
nV ¢ = =
5. In cor rect di a gram.
6. Charge on capacitor of capacitance
C C
V
2 2
30 = = C
V
q
C
MN
= = 30 V
7. For equilibrium,
qE mg =
q
V
d
r g =
4
3
3
p r
V
r
q
µ
3
71
A B
I
I
1
I
E
R
F
V
R
F D
C
I
2
R
C
3R/2
V/2
I
60V C C
2C
2C
A
B
M
N
60V C
A
B
C
—
2
E
I
V
C
R
B
G
C
R
A F
I
1
q
I–I
1
R
qE
mg
V
V
r
r
q
q
2
1
2
1
3
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
´
Þ           V
2
4 = V
8. Electric field between the plates is uniform
but in all other regions it is zero.
9. Initially the capacitor offers zero resistance.
i =
+
=
12
1 2
4 A
10. q CV CE = =
11. In the steady state, capacitor behaves as
open circuit. the equivalent diagram is given
by
I
E
R R
E
R
=
+
=
3
4
4
7
But potential difference across capacitor,
V IR =
Þ 10
4
7
=
E
R
R
Þ E =17.5 V
12. As all the capacitors are connected in series
potential difference across each capacitor is
V
E
= = =
4
10
4
2.5 V
V V
A N
- = = 3V 7.5 V
V
A
= 7.5 V
V V
N B
- =2.5 V
V
B
= -2.5 V
13. Heat produced = Loss of energy
=
+
-
C C
C C
V V
1 2
1 2
1 2
2
2( )
( )
=
´ ´ ´
+ ´
-
- -
-
2 10 2 10
2 2 2 10
100 0
6 6
6
( )
( )
= ´ =
-
5 10 5
3
J mJ
14.  q qe
t
=
-
0
/h
I I e
t
=
-
0
/h
P I R I e R
t
= =
- 2
0
2 2 /h
=
-
Pe
t
0
2 /h
Þ     h
h
¢=
2
15. Common potential =
+
+
=
C V C V
C C
E
1 1 2 2
1 2
2
16. V V
A B
- = + ´ - + ´ = 6 3 2
9
1
3 3 12 V
72
I
R
3
—
4
E
R
R R
3R
E
R
R
E
R R R
1W
12V
6W
3W
12V
4A
1W
2W
17. In the steady state, current through battery
I =
+
=
12
6 2
3
2
A
Potential difference across the capacitor,
V
AB
= ´ = 6
3
2
9 N
\ q CV
AB
= = ´ = 2 9 18 mC
18. C
2
and C
3
are in parallel
Hence, V V
2 3
=
Again Kirchhoff’s junction rule
- + + = q q q
1 2 3
0
Þ q q q
1 2 3
= +
19. For the motion of electron
R
mu
eE
l = =
2
2 sin q
…(i)
and H
mu
eE
d = =
2 2
2
sin q
…(ii)
Dividing Eq. (ii) by Eq. (i),
tan q =
4d
l
20. V Ed = Þ d
V
E
V
= =
2
6
0
e
=
´ ´ ´
-
-
2 5 10
10
12
7
8.85
= ´ =
-
8.85 0.88 10
4
mm
21.
P and Q are at same potential, hence
capacitor connected between them have no
effect on equivalent capacitance.
22.
23. C
k A
d
k A
d
1
1 0 2 0
2 2
= +
e e

=
+ ( ) k k A
d
1 2 0
2
e
(Parallel grouping)
1
2 2
2 1 0 2 0
C
d
k A
d
k A
= +
e e
(Series grouping)
C
k k
k k
A
d
2
1 2
1 2
0
2
=
+
e
C
C
k k
k k
1
2
1 2
2
1 2
2
4
2 3
4 2 3
25
24
=
+
=
+
´ ´
=
( ) ( )
73
x
x
y
y
2mF 1mF
2mF
x y
8
—
3
mF
2mF
mF
2
—
3
A B
P Q R S
B
S
A
C C C C/2
R
C
C
B A
C C C 3C/2 C
3C/14
x y
1mF 1mF
1mF
2mF
I
A B
12V
2W
4W
2mF
6W
I
O
24.
25. Cases (a), (b) and (c) are balanced
Wheatstone bridge.
26. The given arrangement can be considered as
the combination of three capacitors as shown
in figure.
Hence, C
k A
d
1
1 0
2
=
e
C
k
A
d
k A
d
2
2 0
2 0 2
2
= =
e
e
/
C
k
A
d
k A
d
3
3 0
3 0 2
2
= =
e
e
/
Effective capacitance,
C C
C C
C C
A
d
k k k
k k
= +
+
= +
+
é
ë
ê
ù
û
ú 1
2 3
1 2
0 1 2 3
2 3
2
e
27. Here, plate 1 is connected to plate 5 and
plate 3 is connected to plate 6.
Capacitance of all other capacitance is same,
i.e., C
A
d
=
e
0
but that of formed by plates 4
and 5 is
C
2
as distance between these two
plates is 2d.
The equivalent circuit is shown in figure.
\ C C
A
d
eq
F F = = = ´ =
11
7
11
7
11
7
7 11
0
e
m m
74
A B
4C
—
7
A B
4C
—
3 C
C C
11C
—
7
A
2 1 5 4 4 3
C
—
2
C
5 6
C 2
3
C
—
3
C
C
C
B
A B
C
C
1
2
3
4
5
6
d
d
d
2d
d
A
B
C
1
C
2
C
3
B
A
C C C
C C
C
P
A
B
Q
A
B
C C
Q
P
C
C C
A
B
C C
C
C C
A
B
C
C
—
2
C
—
2
A
B
2C
C
Page 5

31.
(a) Consider the circuit as combination of two
cells of emf E and OV.
E
E R E R
R R
V
e
=
+
+
=
1 2 2 1
1 2
2
R R
R R
e
= + =
2
3
2

q q e
t
= -
-
0
1 ( )
/t

q
CE
0
2
=
t =
3
2
RC

\ q
CE
e
t RC
= -
-
2
1
2 3
( )
/
(b) I
dq
dt
E
R
e
t RC
1
2 3
3
= =
- /
In loop EDBA
q
C
I R I R + - =
1 2
0
I
q
RC
I
2 1
= +
= - +
- -
E
R
e
E
R
e
t RC t RC
2
1
3
2 3 2 3
( )
/ /
= -
-
E
R
e
t RC
6
3
2 3
( )
/
Objective Questions (Level 1)
1. F
Q
A
=
2
0
2e
is independent of d.
2. C
q
V
=
On connecting the plates V becomes zero.
3. The system can assumed to a parallel
combination of two spherical conductors.
C C C a b = + = +
1 2 0 0
4 4 pe pe
= + 4
0
pe ( ) a b
4. V
q
C
=
On connecting in series
q q ¢ = = Charge on any capacitor
C
C
n
¢ =
\ V
nq
C
nV ¢ = =
5. In cor rect di a gram.
6. Charge on capacitor of capacitance
C C
V
2 2
30 = = C
V
q
C
MN
= = 30 V
7. For equilibrium,
qE mg =
q
V
d
r g =
4
3
3
p r
V
r
q
µ
3
71
A B
I
I
1
I
E
R
F
V
R
F D
C
I
2
R
C
3R/2
V/2
I
60V C C
2C
2C
A
B
M
N
60V C
A
B
C
—
2
E
I
V
C
R
B
G
C
R
A F
I
1
q
I–I
1
R
qE
mg
V
V
r
r
q
q
2
1
2
1
3
1
2
=
æ
è
ç
ç
ö
ø
÷
÷
´
Þ           V
2
4 = V
8. Electric field between the plates is uniform
but in all other regions it is zero.
9. Initially the capacitor offers zero resistance.
i =
+
=
12
1 2
4 A
10. q CV CE = =
11. In the steady state, capacitor behaves as
open circuit. the equivalent diagram is given
by
I
E
R R
E
R
=
+
=
3
4
4
7
But potential difference across capacitor,
V IR =
Þ 10
4
7
=
E
R
R
Þ E =17.5 V
12. As all the capacitors are connected in series
potential difference across each capacitor is
V
E
= = =
4
10
4
2.5 V
V V
A N
- = = 3V 7.5 V
V
A
= 7.5 V
V V
N B
- =2.5 V
V
B
= -2.5 V
13. Heat produced = Loss of energy
=
+
-
C C
C C
V V
1 2
1 2
1 2
2
2( )
( )
=
´ ´ ´
+ ´
-
- -
-
2 10 2 10
2 2 2 10
100 0
6 6
6
( )
( )
= ´ =
-
5 10 5
3
J mJ
14.  q qe
t
=
-
0
/h
I I e
t
=
-
0
/h
P I R I e R
t
= =
- 2
0
2 2 /h
=
-
Pe
t
0
2 /h
Þ     h
h
¢=
2
15. Common potential =
+
+
=
C V C V
C C
E
1 1 2 2
1 2
2
16. V V
A B
- = + ´ - + ´ = 6 3 2
9
1
3 3 12 V
72
I
R
3
—
4
E
R
R R
3R
E
R
R
E
R R R
1W
12V
6W
3W
12V
4A
1W
2W
17. In the steady state, current through battery
I =
+
=
12
6 2
3
2
A
Potential difference across the capacitor,
V
AB
= ´ = 6
3
2
9 N
\ q CV
AB
= = ´ = 2 9 18 mC
18. C
2
and C
3
are in parallel
Hence, V V
2 3
=
Again Kirchhoff’s junction rule
- + + = q q q
1 2 3
0
Þ q q q
1 2 3
= +
19. For the motion of electron
R
mu
eE
l = =
2
2 sin q
…(i)
and H
mu
eE
d = =
2 2
2
sin q
…(ii)
Dividing Eq. (ii) by Eq. (i),
tan q =
4d
l
20. V Ed = Þ d
V
E
V
= =
2
6
0
e
=
´ ´ ´
-
-
2 5 10
10
12
7
8.85
= ´ =
-
8.85 0.88 10
4
mm
21.
P and Q are at same potential, hence
capacitor connected between them have no
effect on equivalent capacitance.
22.
23. C
k A
d
k A
d
1
1 0 2 0
2 2
= +
e e

=
+ ( ) k k A
d
1 2 0
2
e
(Parallel grouping)
1
2 2
2 1 0 2 0
C
d
k A
d
k A
= +
e e
(Series grouping)
C
k k
k k
A
d
2
1 2
1 2
0
2
=
+
e
C
C
k k
k k
1
2
1 2
2
1 2
2
4
2 3
4 2 3
25
24
=
+
=
+
´ ´
=
( ) ( )
73
x
x
y
y
2mF 1mF
2mF
x y
8
—
3
mF
2mF
mF
2
—
3
A B
P Q R S
B
S
A
C C C C/2
R
C
C
B A
C C C 3C/2 C
3C/14
x y
1mF 1mF
1mF
2mF
I
A B
12V
2W
4W
2mF
6W
I
O
24.
25. Cases (a), (b) and (c) are balanced
Wheatstone bridge.
26. The given arrangement can be considered as
the combination of three capacitors as shown
in figure.
Hence, C
k A
d
1
1 0
2
=
e
C
k
A
d
k A
d
2
2 0
2 0 2
2
= =
e
e
/
C
k
A
d
k A
d
3
3 0
3 0 2
2
= =
e
e
/
Effective capacitance,
C C
C C
C C
A
d
k k k
k k
= +
+
= +
+
é
ë
ê
ù
û
ú 1
2 3
1 2
0 1 2 3
2 3
2
e
27. Here, plate 1 is connected to plate 5 and
plate 3 is connected to plate 6.
Capacitance of all other capacitance is same,
i.e., C
A
d
=
e
0
but that of formed by plates 4
and 5 is
C
2
as distance between these two
plates is 2d.
The equivalent circuit is shown in figure.
\ C C
A
d
eq
F F = = = ´ =
11
7
11
7
11
7
7 11
0
e
m m
74
A B
4C
—
7
A B
4C
—
3 C
C C
11C
—
7
A
2 1 5 4 4 3
C
—
2
C
5 6
C 2
3
C
—
3
C
C
C
B
A B
C
C
1
2
3
4
5
6
d
d
d
2d
d
A
B
C
1
C
2
C
3
B
A
C C C
C C
C
P
A
B
Q
A
B
C C
Q
P
C
C C
A
B
C C
C
C C
A
B
C
C
—
2
C
—
2
A
B
2C
C
JEE Corner
Assertion and Reason
1. Capacitance =
q
V
is constant for a given
capacitor.
2. Reason correctly explains the assertion.
3. U qV =
1
2
, W qV =
4. For discharging of capacitor
q q e
t
=
-
0
/ t
dq
dt
q
e
t
= -
- 0
t
t /
= -
-
q
RC
e
t 0 /t
Hence, more is the resistance, less will be
the slope.
5. Charge on two capacitors will be same only if
both the capacitors are initially uncharged.
6. As potential difference across both the
capacitors is same, charge will not flow
through the switch.
7. C and R
2
are shorted.
8. Time constant for the circuit,
t = RC
9. In series, charge remains same
and U
q
C
=
2
2
Þ U
C
µ
1
10. In series charge remains same
\ V
q
C
1
1
= , V
q
C
2
2
=
On inserting dielectric slab between the
plates of the capacitor, C
2
increases and
hence, V
2
decreases. So more charge flows to
C
2
.
Objective Questions (Level 2)
1. E
Q
A
x d
Q
A
d x d
Q
A
d x
=
- <
- < <
< <
4
2
2
4
2 3
0
0
0
e
e
e
i
i
i
^
^
^
for
for
for

d
ì
í
ï
ï
ï
î
ï
ï
ï
2. Let E
0
= external electric field
and E = electric field due to sheet
\ E E E
1 0
8 = - =
E E E
2 0
12 = + =
Þ     E =2 V/m  Þ
s
e 2
2
0
=
s e = 4
0
3. When the switch is just closed, capacitors
behave like short circuit, no current pass
through either 6 W or 5 W resistor.
4. For charging of capacitor
I I e
t
=
-
0
/ t
ln log I I
t
= -
0
t
ln ln I
V
R
t
RC
= -
But, I I
01 02
=
Þ
V
R
V
R
1
1
2
2
=
Also,
1 1
1 1 2 2
R C R C
>
Þ R C R C
2 2 1 1
>
As only two parameters can be different,
C C
1 2
=
R R
2 1
>
and V V
2 1
>
5. Charge on capacitor at the given instant.
q
q CE
= =
0
2 2
Heat produced = Energy stored in capacitor
= =
q
C
CE
2 2
2 8
Heat liberated inside the battery,
=
+
´
r
r r 2
Total heat produced
=
CE
2
24
75
```

122 docs

## DC Pandey Solutions for NEET Physics

122 docs

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