.A capacitor is charged to potential V and is disconnected . A dielectric of dielectric
constant 4 is inserted filling the whole space between the plates. How do the following
change? –Capacitance, potential difference, Field between the plates, energy stored by
the capacitors.

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Solution

(i) The capacitance increases as the dielectric constant K > 1. K=4

(ii) Potential difference V = Q/C. As C increases and Q remains the same since the battery is disconnected, the p.d. between the plates decreases.

(iii) Electric field E = V/d where V is the p.d. and d the separation between the plates. As V decreases and d remains the same, electric field also decreases.

(iv) Energy stored in a capacitor U = ½ Q2/C. As Q is constant and C increases, U decreases.

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.A capacitor is charged to potential V and is disconnected . A dielectric of dielectric constant 4 is inserted filling the whole space between the plates. How do the following change? –Capacitance, potential difference, Field between the plates, energy stored by the capacitors.

.A capacitor is charged to potential V and is disconnected . A dielectric of dielectric
constant 4 is inserted filling the whole space between the plates. How do the following
change? –Capacitance, potential difference, Field between the plates, energy stored by
the capacitors.