A capacitor has some dielectric between its plates, and the capacitor is connected to a $D C$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored, and the voltage will increase, decrease, or remain constant.

Question

A capacitor has some dielectric between its plates, and the capacitor is connected to a

D C

source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored, and the voltage will increase, decrease, or remain constant.

Toppr Admin · Accepted Answer

The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant $$ K $$ is given by $$ C=K\varepsilon _0 \frac{A}{d} $$
The capacitance of the parallel plate capacitor decreases with the removal of a dielectric medium as for air or vacuum $$ K=1 $$ and for dielectric $$ K>1. $$
If we disconnect the batter from the capacitor, then the charge stored will remain the same due to the conservation of charge.
The energy stored in an isolated charge capacitor $$ U=\dfrac{q^2}{2c} $$ as $$ q$$ is constant, energy stored $$ U \propto \dfrac{1}{C}.$$ As $$ C $$ decreases with the removal of the dielectric medium therefore energy stored increases.
The potential difference across the plates of the capacitor is given by $$ V=\dfrac{q}{C} $$ Since $$ q $$ is constant and $$ C $$ decreases which in tue=rn increases $$ V $$ and therefore $$ E $$ increases as $$ E=\dfrac{V}{d}. $$
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