Question

A parallel plate capacitor with no dielectric between the plates is connected to the constant voltage source. How would capacitance and charge change if the dielectric constant $K=2$ is inserted between the plates?
($C_0$ and $Q_0$ are the capacitance and charge of the capacitor before the introduction of the dielectric.)

• A. $C_0/2$ and $2Q_0$
• B. $2C_0$ and $Q_0/2$
• C. $C_0/2$ and $Q_0/2$
• D. $2C_0$ and $2Q_0$

Answer 1

The correct option is D $2C_0$ and $2Q_0$

The charge becomes $K$ times the previous charge on the capacitor.

So, $Q=K{Q}_0=2Q_0$

In the presence of the dielectric completely filling the space between the plates, the capacitance will increase by the factor of the dielectric constant of the material.

Thus, $C=\frac{Q}{V_o}=\frac{K{Q}_o}{V_o}=K{C}_0=2C_0$

If the battery remains connected across the capacitor, the voltage will remain the same i.e. equal to the voltage of the source.

Hence, option $\left(d\right)$ is correct.

Key concept- Capacitance & charge on a parallel plate capacitor with fully filled dielectric medium.