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TS EAMCET 19-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 105
Total: 160
A parallel - plate capacitor of plate area
10
c
m
2
and plate separation
3
m
m
is charged to a potential difference
12
V
and then the battery is disconnected. A slab of dielectric constant 3 is then inserted between the plates. The work done on the system in the process of inserting the slab is
α
ε
0
. The value of
α
is
(Take
ε
0
as the permittivity of free space)
8
12
16
18
Validate
Solution:
Given, plate area,
A
=
10
cm
2
=
10
−
3
m
2
Plate separation,
d
=
3
mm
=
3
×
10
−
3
m
Potential difference,
V
=
12
V
Capacitance of parallel plate capacitor,
C
=
ε
0
A
d
=
ε
0
×
10
−
3
3
×
10
−
3
=
ε
0
3
⇒
C
=
ε
0
3
Charge stored in capacitor,
q
=
C
V
=
ε
0
3
×
12
=
4
ε
0
C
∵
Work done,
W
=
q
V
=
4
ε
0
⋅
12
=
48
ε
0
When dielectric material (with
K
=
3
) is introduced between the parallel plate capacitor, then new value of capacitance,
C
′
=
K
ε
0
A
d
=
K
C
=
3
×
ε
0
3
=
ε
0
Charge stored
q
′
=
C
′
V
′
⇒
q
′
=
ε
0
⋅
V
K
=
ε
0
12
3
=
4
ε
0
Work done
W
=
q
′
V
′
=
4
ε
0
⋅
V
K
=
4
ε
0
12
3
=
16
ε
0
=
α
ε
0
(Given)
α
=
16
© examsnet.com
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