Question

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 :

• A. $\frac{V}{K}$
• B. $\frac{3V}{K}$
• C. $\frac{V}{K+2}$
• D. $\frac{3V}{K+2}$

Hint:

In problems where a charged capacitor system is isolated from the battery, the key principle is the conservation of total charge. The potential difference and stored energy will change if the capacitance is altered.

Answer 1

The Correct Option is D

Initially, the two capacitors (2C and C) are in parallel and charged to a potential V.
The initial equivalent capacitance is $C_{initial} = 2C + C = 3C$.
The total charge stored in the system is $Q = C_{initial} \times V = (3C)V = 3CV$.
When the battery is removed, this total charge Q is conserved.
Next, the capacitor with capacitance C is filled with a dielectric of constant K. Its new capacitance becomes $C' = KC$.
The two capacitors (2C and C') are still connected in parallel.
The final equivalent capacitance is $C_{final} = 2C + C' = 2C + KC = C(2+K)$.
The new potential difference ($V'$) across the parallel combination is given by the total charge divided by the final equivalent capacitance.
$V' = \frac{Q}{C_{final}}$
$V' = \frac{3CV}{C(2+K)}$
$V' = \frac{3V}{K+2}$