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Q.

A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$ , where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given C₀ = capacitance of the capacitor with air as medium between plates.)

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a

$\frac{3 K C_{0}}{3 + K}$

b

$\frac{4 K C_{0}}{3 + K}$

c

$\frac{K}{4 + K}$

d

$\frac{3 + K}{4 K C_{0}}$

answer is A.

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Detailed Solution

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$x + y + \frac{3 d}{4} = d x + y = \frac{d}{4}$

$\frac{A \in_{0}}{d} = C_{0}$

$∆ V = Ex + \frac{E}{k} \times \frac{3 d}{4} + Ey = \frac{3 Ed}{4 k} + E (x + y)$

$∆ V = E [\frac{3 d}{4 k} + \frac{d}{4}] ∆ V = \frac{σ}{\in_{0}} [\frac{3 d + dk}{4 k}] = \frac{Qd}{A \in_{0}} [\frac{3 + k}{4 k}]$

$\frac{Q}{∆ V} = C = \frac{A \in_{0}}{d} [\frac{4 k}{k + 3}]$

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A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given C0 = capacitance of the capacitor with air as medium between plates. )