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: Use the voltage divider rule for capacitors.
In series, $V_1 : V_2 = \dfrac{1}{C_1} : \dfrac{1}{C_2}$.

Step 2: Apply.
$V_{air} : V_{liquid} = \dfrac{1}{C} : \dfrac{1}{KC} = K : 1$.

Step 3: Share V.
$V_{air} = V\cdot\dfrac{K}{K + 1}$.

Final Answer:
Option (A). \[ \boxed{\frac{KV}{K+1}} \]