Question

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

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

Hint:

In a series combination the charge is the same on both, so the potential divides in inverse ratio of the capacitances.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Two capacitors in series connected to a battery carry the same charge. The voltage across each is \(V = Q/C\), so the capacitor with smaller capacitance takes the larger voltage.

Step 2: Capacitances.
Each capacitor has capacitance \(C\) in air. After one is placed in a liquid of dielectric constant \(K\), its capacitance becomes \(KC\).

Step 3: Find the charge.
\[ C_{eq} = \frac{C\cdot KC}{C + KC} = \frac{KC}{K + 1}, \qquad Q = C_{eq}V = \frac{KCV}{K + 1} \]

Step 4: Voltage across the air capacitor.
\[ V_{air} = \frac{Q}{C} = \frac{KV}{K + 1} \]

Step 5: Check.
The voltage across the liquid capacitor is \(\dfrac{V}{K + 1}\). The two add to \(V\), as they should.

Final Answer:
The potential difference of the other capacitor becomes \(\dfrac{KV}{K+1}\), option (A). \[ \boxed{\frac{KV}{K+1}} \]