Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

Capacitor PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q61 with Solution

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Correct answer: C

Two capacitors $2C$ and $C$ are connected in parallel and charged to potential $V$ .

Since they are in parallel, both have the same potential difference $V$ .

So their initial charges are: $Q_1 = 2C \cdot V = 2CV$ $Q_2 = C \cdot V = CV$

Hence total initial charge on the isolated combination is $Q_{\text{total}} = 2CV + CV = 3CV$

After removing the battery, the capacitor combination is isolated. Therefore, the total charge remains conserved: $Q_{\text{total}} = 3CV$

The capacitor of capacitance $C$ is completely filled with dielectric constant $K$ . So its new capacitance becomes $C' = KC$

The other capacitor remains unchanged at $2C$ .

Thus new equivalent capacitance of the parallel combination is $C_{\text{eq,new}} = 2C + KC = C(K+2)$

Since the capacitors are still connected in parallel, they must have the same final potential difference $V'$ .

Using $Q_{\text{total}} = C_{\text{eq,new}} \cdot V'$ we get $3CV = C(K+2)V'$

Therefore, $V' = \frac{3CV}{C(K+2)} = \frac{3V}{K+2}$

$V' = \frac{3V}{K+2}$

So the correct option is C.