Question

Two condensers of capacities, '2C' and 'C' are joined in parallel and charged upto potential 'V'. The battery is then disconnected and condenser of capacity 'C' is filled completely with a medium of dielectric constant 'K'. The potential difference across the capacitors in the second case is

• A. \(\frac{3\text{V}}{(\text{K}+2)}\)
• B. \(\frac{\text{V}}{(\text{K}+2)}\)
• C. \(\frac{5\text{V}}{(\text{K}+2)}\)
• D. \(\frac{2\text{V}}{(\text{K}+2)}\)

Hint:

The battery is disconnected, so total charge stays constant.

Answer 1

The Correct Option is A

Step 1: Check limiting case:
If $K = 1$ nothing changes and the voltage must stay $V$.

Step 2: Charge conservation:
$3CV = (2C + KC)V'$.

Step 3: Solve:
$V' = \dfrac{3V}{K+2}$. At $K=1$ it gives $V$, so option (A) fits and the others do not.

Final Answer:
The potential becomes 3V / (K + 2). \[ \boxed{\text{(A) }\dfrac{3V}{K+2}} \]