The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is:

First we note the basic relation for any capacitor, namely the charge-voltage relation $$Q = C\,V$$ where $$C$$ is the capacitance, $$V$$ the potential difference across its plates and $$Q$$ the magnitude of charge stored on either plate.

The two given capacitors are air-filled parallel-plate capacitors having capacitances

$$C_1 = C \qquad\text{and}\qquad C_2 = nC.$$

Because they are connected in parallel to the same battery, the potential difference across each capacitor is initially the battery voltage $$V$$. Using $$Q = C\,V$$ we obtain their individual charges just after connection to the battery:

$$Q_1^{\text{(initial)}} = C_1 V = C\,V,$$

$$Q_2^{\text{(initial)}} = C_2 V = nC\,V.$$

The total charge residing on the combination is therefore

$$Q_{\text{total}}^{\text{(initial)}} \;=\; Q_1^{\text{(initial)}} + Q_2^{\text{(initial)}} \;=\; C\,V + nC\,V \;=\; (n + 1)\,C\,V.$$

Now the battery is carefully removed. When the external circuit is opened in this way, no further charge can flow into or out of the parallel combination, so the total charge remains fixed at the value just calculated:

$$Q_{\text{total}} = (n + 1)\,C\,V \quad\text{(conserved)}.$$

After the battery has been detached, a dielectric slab of dielectric constant $$K$$ is inserted fully between the plates of the first capacitor only. Inserting a dielectric multiplies the capacitance by $$K$$, so the new capacitances are

$$C_1' = K\,C \qquad\text{and}\qquad C_2' = nC \;(\text{unchanged}).$$

Because the connecting wires joining the two capacitors are still intact, the two capacitors remain in parallel and must therefore share a common new potential difference. Let us denote this unknown final potential difference by $$V'$$.

Using $$Q = C\,V$$ once more, the individual charges after insertion of the dielectric become

$$Q_1' = C_1' V' = K\,C\,V',$$

$$Q_2' = C_2' V' = nC\,V'.$$

The total charge after the insertion is thus

$$Q_{\text{total}}^{\text{(final)}} = Q_1' + Q_2' = K\,C\,V' + nC\,V' = (K + n)\,C\,V'.$$

But charge is conserved, so this final total charge must equal the initial total charge:

$$(K + n)\,C\,V' = (n + 1)\,C\,V.$$

We divide both sides by $$C$$ to eliminate the common factor and then solve for $$V'$$:

$$(K + n)\,V' = (n + 1)\,V,$$

$$V' = \frac{(n + 1)\,V}{K + n}.$$

This is the new potential difference across each capacitor in the parallel combination after the dielectric has been introduced.

Hence, the correct answer is Option A.