Q122 / 150 · MHT CET 2026 (13 April Shift 1) · Physics
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
- Correct answerA$\dfrac{KV}{K+1}$
- B$\dfrac{KV}{K-1}$
- C$\dfrac{K+1}{KV}$
- D$\dfrac{K-1}{KV}$
Solution
Let the capacitance of each identical parallel plate air capacitor be $C$.
When one capacitor is inserted in a liquid of dielectric constant $K$, its capacitance becomes $C_1 = KC$. The capacitance of the other capacitor remains $C_2 = C$.
Since the two capacitors are connected in series across the battery of e.m.f. $V$, the equivalent capacitance of the circuit is $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{KC \cdot C}{KC + C} = \dfrac{KC}{K+1}$.
The total charge supplied by the battery is $Q = C_{eq}V = \dfrac{KCV}{K+1}$.
In a series combination, the charge on each capacitor is the same. The potential difference across the other capacitor (which remains in air) is $V' = \dfrac{Q}{C_2}$.
Substituting the value of $Q$, we get $V' = \dfrac{\dfrac{KCV}{K+1}}{C} = \dfrac{KV}{K+1}$.
Answer: $\dfrac{KV}{K+1}$
When one capacitor is inserted in a liquid of dielectric constant $K$, its capacitance becomes $C_1 = KC$. The capacitance of the other capacitor remains $C_2 = C$.
Since the two capacitors are connected in series across the battery of e.m.f. $V$, the equivalent capacitance of the circuit is $C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{KC \cdot C}{KC + C} = \dfrac{KC}{K+1}$.
The total charge supplied by the battery is $Q = C_{eq}V = \dfrac{KCV}{K+1}$.
In a series combination, the charge on each capacitor is the same. The potential difference across the other capacitor (which remains in air) is $V' = \dfrac{Q}{C_2}$.
Substituting the value of $Q$, we get $V' = \dfrac{\dfrac{KCV}{K+1}}{C} = \dfrac{KV}{K+1}$.
Answer: $\dfrac{KV}{K+1}$
Hence, the correct answer is option: 'A'.