Question

A parallel plate air capacitor with no dielectric between the plates is connected to the constant voltage source. What will be the new capacitance and charge on the capacitor if a dielectric with dielectric constant $K=5$ is inserted between the plates? Intially capacitance and charge on the capacitor are $50\mu \text{F}$ and $200\mu \text{C}$ respectively before the introduction of the dielectric.

• A. $50\mu \text{F}$ and $200\mu \text{C}$
• B. $50\mu \text{F}$ and $1000\mu \text{C}$
• C. $250\mu \text{F}$ and $1000\mu \text{C}$
• D. $250\mu \text{F}$ and $200\mu \text{C}$

Answer 1

The correct option is C $250\mu \text{F}$ and $1000\mu \text{C}$

Given:
$C_0=50\mu \text{F};Q_0=200\mu \text{C};K=5$

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, $C=K{C}_0=5\times 50=250\mu \text{F}$

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

From, $Q=CV$, the charge will also become $K$ times the previous charge on the capacitor.

So, $Q=K{Q}_0=5\times 200\mu \text{C}=1000\mu \text{C}$

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

Key concept - Capacitance & charge on parallel plate capacitor increases due to fully filled dielectric, if battery remains connected.