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NEET 2020 (Phase 2) · Dielectrics

Dielectrics question from NEET 2020 (Phase 2)

Q89NEET 2020 · PHASE 2Medium

Dielectrics

A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K(=4)$ . The ratio of new capacitance to its original capacitance will be,

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Answer: (c)

Air $\frac{d}{2}$ and slab $\frac{d}{2}$ in series: $C=\frac{\epsilon_0A}{\frac{d}{2}+\frac{d}{2K}}$
With $K=4$ : $C=\frac{\epsilon_0A}{\frac{5d}{8}}=\frac{8}{5}C_0$
So the answer is (c) $8:5$ .

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