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Question

I have a capacitor who’s potential difference between the plates is maintained at 5V using a battery. If the capacitance is 5 F and a dielectric of relative permittivity 3 is placed between the plates such that it fully fills the space, What is the change in energy?

A
Increases by 6.24×104J
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B
Decreases by 6.24×104J
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C
Increases by 9.36×104J
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D
Decreases by 9.36×104J
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Solution

The correct option is A Increases by 6.24×104J
Here the initial energy of the capacitor is
E=12CV2
E=12×5×106×52
E=3.12×104J
Now when calculating the final energy, we don’t have to change V since it is maintained at 5V by means of a battery. Only Capacitance becomes C’ = kC
E=12CV2E=123CV2E=3EE=9.36×104J
Energy change is, EE=6.24×104J

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