A parallel plate capacitor is charged to a potential difference V… - Vidyadip

Question

A parallel plate capacitor is charged to a potential difference V from a direct current source, while there is air between the plates. Without separating the capacitor from the battery, a dielectric medium of dielectric constant K has been filled in place of air. Explain with reasons which of the following changes will occur? Also give explanation of the answers.(i) Potential difference (ii) Electric field between the plates (iii) Capacitance (iv) charge (v) energy.

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Answer

(i) Since here the capacitor is connected to the battery, i.e. the electric potential difference V will remain unchanged.
(ii) Since both the potential difference and the distance between the plates are unchanged, the electric field $E=\frac{V}{D}$ will remain unchanged.
(iii) Capacitance will increase $\because C = KC _0$
(iv) Since V is constant and the capacitance becomes K times, the charge Q will also become K times.
(v) Initial energy $U_0=\frac{1}{2} C_0\left(V_0\right)^2$ and final energy $U =\frac{1}{2} K C _0\left(V_0\right)^2= K U _0$ i.e.,energy will also increase K times.

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