Question

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. \( V \). If one of the capacitors is inserted in liquid of dielectric constant \( K \), then potential difference of the other capacitor will become

• A. \( \frac{K}{V(K+1)} \)
• B. \( \frac{KV}{K+1} \)
• C. \( \frac{K+1}{KV} \)
• D. \( \frac{K}{V(1-K)} \)

Hint:

For capacitors in series, the voltage divides inversely proportional to their capacitance. A dielectric increases the capacitance of the capacitor in which it is inserted.

Answer 1

The Correct Option is B

Step 1: Understanding the capacitor's behavior.
For capacitors in series, the total capacitance is given by \( \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \). When a dielectric is inserted, the capacitance increases by a factor of \( K \), so the capacitance of the dielectric-inserted capacitor becomes \( K C_1 \).
Step 2: Applying the formula.
For the two capacitors, the total potential difference is \( V \), and we have the relation: \[ V = V_1 + V_2 \] Using the capacitance formula and the fact that \( C = \frac{Q}{V} \), we can solve for \( V_2 \) (the potential across the capacitor without the dielectric). After simplification, we get: \[ V_2 = \frac{KV}{K+1} \] Step 3: Conclusion.
The correct answer is (B), \( \frac{KV}{K+1} \).