A parallel plate capacitor of capacitance C h | vedclass

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(A) The capacitor is equivalent to a series combination of $N$ capacitors,each of thickness $\delta = dx = \frac{d}{N}$.For a large $N$,we can treat t

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$1$

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(A) The capacitor is equivalent to a series combination of $N$ capacitors,each of thickness $\delta = dx = \frac{d}{N}$.For a large $N$,we can treat this as a continuous variation where $\frac{m}{N} = \frac{x}{d}$.The dielectric constant varies as $K(x) = K(1 + \frac{x}{d})$.The equivalent capacitance $C_{eq}$ for capacitors in series is given by $\frac{1}{C_{eq}} = \int \frac{dx}{C(x)}$,where $C(x) = \frac{K(x) \varepsilon_0 A}{dx}$.Substituting the values: $\frac{1}{C_{eq}} = \int_0^d \frac{dx}{\frac{K(1 + x/d) \varepsilon_0 A}{dx}} = \frac{1}{K \varepsilon_0 A} \int_0^d \frac{dx}{1 + x/d}$.Let $u = 1 + \frac{x}{d}$,then $du = \frac{dx}{d}$,so $dx = d \cdot du$.$\frac{1}{C_{eq}} = \frac{d}{K \varepsilon_0 A} \int_1^2 \frac{du}{u} = \frac{d}{K \varepsilon_0 A} [\ln u]_1^2 = \frac{d}{K \varepsilon_0 A} \ln 2$.Thus,$C_{eq} = \frac{K \varepsilon_0 A}{d \ln 2}$.Comparing this with the given expression $C = \alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$,we get $\alpha = 1$.

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