Question

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

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

Hint:

In series combination, charge remains same on both capacitors. Use \(Q = C_1V_1 = C_2V_2\) and \(V_1+V_2 = V\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Two identical capacitors in series connected to battery \(V\). One capacitor is filled with dielectric constant \(K\). We need the new potential difference across the other capacitor.

Step 2: Key Formula or Approach:
Initially, each capacitance = \(C\). In series, equivalent capacitance = \(C/2\), charge on each = \(Q = (C/2)V\).
After inserting dielectric, one capacitor becomes \(KC\), the other remains \(C\).

Step 3: Detailed Explanation:
Initial: \(C_1 = C\), \(C_2 = C\). Series equivalent = \(\frac{C}{2}\). Charge on each = \(Q = \frac{C}{2}V\).
After dielectric: \(C_1' = KC\), \(C_2' = C\). Same battery still connected, so total voltage \(V\) remains. Let \(V_1\) and \(V_2\) be new potentials. In series, charge is same on both: \(Q' = C_1' V_1 = C_2' V_2\). Also \(V_1 + V_2 = V\). From \(KC \cdot V_1 = C \cdot V_2 \implies V_2 = K V_1\). Then \(V_1 + K V_1 = V \implies V_1 = \frac{V}{K+1}\). Then \(V_2 = \frac{KV}{K+1}\). Thus potential difference across the other capacitor (the one without dielectric) is \(V_2 = \frac{KV}{K+1}\).

Step 4: Final Answer:
Option (C) is correct.