Question

A capacitor of capacitance ‘C’ is charged to a potential ‘V’ and disconnected from the battery. Now if the space between the plates is completely filled with a substance of dielectric constant ‘K’, the final charge and the final potential on the capacitor are respectively: 

• A. \( KCV \) and  \( \frac{V}{K} \)
• B. \( CV \) and  \( \frac{V}{K} \)
• C. \( \frac{CV}{K} \) and  \( KV \)
• D. \( \frac{CV}{K} \) and \( \frac{V}{K} \)

Hint:

When a capacitor is disconnected from a battery, the charge on it remains constant. Introducing a dielectric increases the capacitance and decreases the voltage accordingly.

Answer 1

The Correct Option is B

Step 1: Understanding the Effect of Dielectric on a Disconnected Capacitor Since the capacitor is disconnected from the battery, its charge remains constant. The charge on the capacitor is given by: \[ Q = C V \] Step 2: Effect of Introducing the Dielectric When a dielectric material with dielectric constant \( K \) is introduced, the capacitance of the capacitor increases as: \[ C' = K C \] However, since the charge remains constant, the new potential \( V' \) across the capacitor is given by: \[ V' = \frac{Q}{C'} = \frac{CV}{KC} = \frac{V}{K} \] Thus, the final charge remains \( CV \), and the new potential is \( \frac{V}{K} \).