JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

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 ______.

A
2.5
B
2.0
C
1.5
D
1.0

Solution

Original capacitance: $C = \frac{ε _{0} A}{d} = \frac{ε _{0} A}{5 \times 1 0 ^{- 3}}$ .

After insertion, the capacitor is two capacitors in series: a 3 mm air gap and a 2 mm mica slab.

$C_{1} = \frac{ε _{0} A}{3 \times 1 0 ^{- 3}}$ , $C_{2} = \frac{K ε _{0} A}{2 \times 1 0 ^{- 3}}$ .

$C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{K ε _{0} A}{2 + 3 K} \cdot \frac{1}{1 0 ^{- 3} \cdot \frac{1}{6} \cdot 6} = \frac{6 K ε _{0} A}{( 2 + 3 K ) \cdot 1 0 ^{- 3}}$ .

Wait, recompute: $\frac{1}{C _{e q}} = \frac{3 \times 1 0 ^{- 3}}{ε _{0} A} + \frac{2 \times 1 0 ^{- 3}}{K ε _{0} A} = \frac{1 0 ^{- 3} ( 3 K + 2 )}{K ε _{0} A}$ .

$C_{e q} = \frac{K ε _{0} A}{( 3 K + 2 ) \cdot 1 0 ^{- 3}}$ .

Since the battery remains connected, $V$ is constant and charge scales with capacitance:

$\frac{Q _{2}}{Q _{1}} = \frac{C _{e q}}{C} = \frac{K ε _{0} A / [( 3 K + 2 ) \cdot 1 0 ^{- 3} ]}{ε _{0} A / [ 5 \times 1 0 ^{- 3} ]} = \frac{5 K}{3 K + 2}$ .

Setting $\frac{Q _{2}}{Q _{1}} = 1.25$ : $\frac{5 K}{3 K + 2} = \frac{5}{4} \Rightarrow 20 K = 15 K + 10 \Rightarrow K = 2$ .

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