Question

Fill in the Blanks
Two parallel plate capacitors of capacitance $C$ and $2C$ are connected in parallel and charge to a potential difference $V$. The battery is then disconnected and the region between the plates of the capacitor $C$ is completely filled with a material of dielectric constant $K$. The potential difference across the capacitor now becomes.

Answer 1

The original capacitance of the parallel capacitance of the parallel combination of $C$ and $2C$ $=C+2C=3C$

Total charge, $Q=3CV$

When filled with dielectric, capacitance becomes $KC$

$\therefore C^{\prime }=KC+2C=C(K+2)$

But charge remains the same.

$\therefore$ Potential difference across the capacitors will be

$\frac{Q}{C^{\prime }}=\frac{3CV}{(K+2)C}=\frac{3V}{(K+2)}$