Question

A parallel plate capacitor each with plate area A and separation 'd' is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in
Capacitance of the capacitor
Justify your answer in each case.

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Solution

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The capacitance increase due to the decrease in the potential difference and any dielectrics, K>1

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