Question

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of
(i) capacitance.
(ii) potential difference between the plates.
(iii) electric field between the plates.
(iv) the energy stored in the capacitor.

Answer 1

$\left(i\right)$ The capacitance increases as the dielectric constant $K>1$.

$\left(ii\right)$ Potential difference $V=\frac{Q}{C}$. As $C$ increases and $Q$ remains the same since the battery is disconnected, the p.d. between the plates decreases.

$\left(iii\right)$ Electric field $E=\frac{V}{d}$ 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.

$\left(iv\right)$ Energy stored in a capacitor $U=\frac{1}{2}\frac{Q^2}{C}$. As $Q$ is constant and $C$ increases, $U$ decreases.