A parallel plate capacitor is charged by a battery. After sometim… - Vidyadip

Question

A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected?
Justify your answer in each case.

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Answer

Capacltance $\text{C} = \frac{\text{K}\varepsilon_{0}\text{A}}{\text{d}},$ Hence capacitance Increases K times.
Potential difference $\text{V} = \frac{\text{V}_{o}}{\text{K}},$Hence potential difference decreases by a factor K.
Energy stored E $ =\frac{1}{2}\text{CV}^{2},$ As capacitance becomes K times & potential difference becomes 1/K times therefore energy stored becomes 1/K times.

Alternate Answer

Energy stored = Q²/2C. As capacitance increases by a factor K, the energy stored will decrease by the same factor.

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