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?
2×10−2J
2.5×10−2
6.5×10−2J
4×10−2J
A
2×10−2J
B
6.5×10−2J
C
2.5×10−2
D
4×10−2J
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Solution
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Given:
Voltage=V=100V
C1=2μF (Initial capacitance)
C2=10μF (Final capacitance)
As E=12CV2
Now
E1=12C1V2=12×2×10−6×V2
E2=12C2V2=12×10×10−6×V2
ΔE=E2−E1=12×10×10−6×V2−12×2×10−6×V2
ΔE=12×8×10−6×V2
ΔE=12×8×10−6×1002
ΔE=4×10−2J
Hence the correct option is (D).
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