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Question

A 10 pF parallel plate capacitor is charged with a 4.0 V battery. While the capacitor is still connected to the battery, a dielectric slab (κ = 5.0) is inserted between the plates to completely fill the gap. How much electric potential energy is stored in the capacitor after inserting the dielectric?

A
5.3×1010J
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B
4.0×1010J
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C
4.6×1010J
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D
5.6×1010J
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Solution

The correct option is C 4.6×1010J
C=10pF=10×1012F
Voltage= Q=4.0V
Dielectric slab (K=5.0)
Electric potential energy= K.q12eV22
=12×4×10×1012
=20×1012J
Now, after inserting dielectric, energy available in the capacitor
=12×KC×(VK)2
=12×5×10×1012×4×45×5
=16×1012 .

Total electric potential= (20+16)×1012
=46×1012
=4.6×1010J .

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