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Electrostatics

JEE Main level, single correct, medium

4 of 311

Question 4 of 311

A parallel plate capacitor with plate separation $d$ has capacitance $C_0$ . A dielectric slab ( $K=4$ , thickness $t=d/2$ ) is placed inside. The new capacitance is:

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Answer: (B) $1.6 C_0$

C = \frac{\varepsilon_0 A}{d - t + t/K} = \frac{\varepsilon_0 A}{d - d/2 + d/8} = \frac{\varepsilon_0 A}{5d/8} = \frac{8}{5} C_0 = 1.6 C_0

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electrostatics, with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A parallel plate capacitor with plate separation ddd has capacitance C0C_0C0​. A dielectric slab (K=4K=4K=4, thickness t=d/2t=d/2t=d/2) is placed inside. The new capacitance is:

JEE Main question paper PDF

More Physics practice

A parallel plate capacitor with plate separation $d$ has capacitance $C_0$ . A dielectric slab ( $K=4$ , thickness $t=d/2$ ) is placed inside. The new capacitance is: