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Question

The capacitance of a variable capacitor joined with the battery of 100V is changed from 2μF to 10μF. What is the change in the energy stored in it?

A
2×102J
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B
2.5×102
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C
6.5×102J
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D
4×102J
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Solution

The correct option is D 4×102J
Given:
Voltage=V=100V
C1=2μF (Initial capacitance)
C2=10μF (Final capacitance)
As E=12CV2
Now
E1=12C1V2=12×2×106×V2
E2=12C2V2=12×10×106×V2
ΔE=E2E1=12×10×106×V212×2×106×V2
ΔE=12×8×106×V2
ΔE=12×8×106×1002
ΔE=4×102J
Hence the correct option is (D).



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