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Test Index
Electrostatic Potential and Capacitance Part 1
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Section:
Physics
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© examsnet.com
Question : 1
Total: 100
A source of potential difference
V
is connected to the combination of two identical capacitors as shown in the figure. When key '
K
' is closed, the total energy stored across the combination is
E
1
. Now key '
K
' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now
E
2
. The ratio
E
1
∕
E
2
will be :
[26-Jul-2022-Shift-2]
‌
1
10
‌
2
5
‌
5
13
‌
5
26
Validate
Solution:
(1) Switch is closed
C
eq
−
2
C
Energy
E
1
=
‌
1
2
C
eq
V
2
=
‌
1
2
2
C
×
V
2
E
1
=
CV
2
(ii) When switch is opened charge on right capacitor remain CV while potential on left capacitor remain same
Dielectric
K
=
5
C
′
=
KC
C
′
=
5
C
E
2
=
‌
1
2
(
5
C
)
V
2
+
‌
(
CV
)
2
2
(
5
C
)
E
2
=
‌
5
CV
2
2
+
‌
CV
2
10
E
2
=
‌
13
CV
2
5
‌
E
1
E
2
=
‌
C
V
2
13
C
V
2
5
=
‌
5
13
‌
E
1
E
2
=
‌
5
13
© examsnet.com
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