JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

A parallel plate air capacitor has a capacitance $C$ . When it is half filled as shown in figure with a dielectric constant $K = 5$ , the percentage increase in the capacitance is _________.

A
$33.34$
B
$66.67$
C
$200$
D
$400$

Solution

The original air capacitance is $C_{i} = \frac{ε _{0} A}{d} = C$ .

With the dielectric occupying half the gap, the arrangement is two capacitors in series, each of thickness $d /2$ :

$C_{1} = \frac{K ε _{0} A}{d /2} = 10 C, C_{2} = \frac{ε _{0} A}{d /2} = 2 C$

$C_{f} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{10 C \cdot 2 C}{12 C} = \frac{5 C}{3}$

Percentage increase $= \frac{C _{f} - C}{C} \times 100 = \frac{2}{3} \times 100 \approx 66.67%$ .

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