Q57 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is $\_\_\_\_$.
- A1.0
- B2.5
- Correct answerC2.0
- D1.5
Solution
Initial capacitor has plate separation 5 mm with capacitance $C_1 = \epsilon_0 A / 5$.
After inserting 2 mm mica sheet, the gap remaining is 3 mm of air. The effective separation becomes: $d_{eff} = 3 + 2/k$ where $k$ is the dielectric constant of mica.
New capacitance: $C_2 = \frac{\epsilon_0 A}{3 + 2/k}$
Since battery remains connected, voltage is constant. The charge is: $Q = CV$
Ratio of charges: $\frac{Q_2}{Q_1} = \frac{C_2}{C_1} = \frac{\epsilon_0 A/(3 + 2/k)}{\epsilon_0 A/5} = \frac{5}{3 + 2/k}$
Given that charge increases by 25%: $Q_2 = 1.25 Q_1$
$\frac{5}{3 + 2/k} = 1.25$
$5 = 1.25(3 + 2/k)$
$5 = 3.75 + 2.5/k$
$1.25 = 2.5/k$
$k = 2$
After inserting 2 mm mica sheet, the gap remaining is 3 mm of air. The effective separation becomes: $d_{eff} = 3 + 2/k$ where $k$ is the dielectric constant of mica.
New capacitance: $C_2 = \frac{\epsilon_0 A}{3 + 2/k}$
Since battery remains connected, voltage is constant. The charge is: $Q = CV$
Ratio of charges: $\frac{Q_2}{Q_1} = \frac{C_2}{C_1} = \frac{\epsilon_0 A/(3 + 2/k)}{\epsilon_0 A/5} = \frac{5}{3 + 2/k}$
Given that charge increases by 25%: $Q_2 = 1.25 Q_1$
$\frac{5}{3 + 2/k} = 1.25$
$5 = 1.25(3 + 2/k)$
$5 = 3.75 + 2.5/k$
$1.25 = 2.5/k$
$k = 2$
Hence, the correct answer is option: 'C'.