Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be \underline\hspace2cm V.

Capacitor PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q68 with Solution

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Correct answer: 6

The capacitors have capacitances $C$ and $3C$ , connected in parallel.

So, the initial equivalent capacitance is

C_{\text{eq, initial}} = C + 3C = 4C

They are charged to $18\,\text{V}$ , so total charge stored in the isolated combination is

Q_{\text{total}} = C_{\text{eq, initial}}\,V = 4C \times 18 = 72C

After disconnecting the battery, the capacitor combination is isolated, so the total free charge remains constant.

Thus,

Q_{\text{total}} = 72C

The capacitor of original capacitance $C$ is completely filled with dielectric constant $K=9$ .

Hence its new capacitance becomes

C' = K C = 9C

The other capacitor remains unchanged at

3C

So the new equivalent capacitance of the parallel combination is

C_{\text{eq, final}} = 9C + 3C = 12C

Since the total charge remains the same,

V_{\text{final}} = \frac{Q_{\text{total}}}{C_{\text{eq, final}}}

Substituting,

V_{\text{final}} = \frac{72C}{12C} = 6\,\text{V}

\boxed{6}