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Question

Two plates of a capacitor of the area 100 cm2 are separated by a distance of 8 cm, and charged by a battery of emf 56 V. If the separation is increased by 8.5 mm, then the new capacitance will be

A
1013 F
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B
1011 F
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C
1014 F
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D
1012 F
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Solution

The correct option is D 1012 F
Given:
A=100 cm2; d=8 cm; V=56 V

Let l be the increased distance between plates.

l=8.5 mm

It is pretty clear that the new capacitance of capacitor will be

Cnew=Aϵ0d+l

Change in voltage or charge will not influence the capacitance .

Let's substitute the values

A=100 cm2=102 m2

d=8 cm=8×102 m

l=8.5 mm=0.85×102 m

d+l=8.85×102 m

Cnew=102×8.85×10128.85×102

Cnew=1012 F
Key concept- Capacitance does not depend on its charge or voltage across it.


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