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(A) Consider a small strip of width $dx$ at a distance $x$ from the left end. The height of the dielectric part is $y = (d/a)x$.This strip acts as two

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$\frac{K\varepsilon_0 a^2}{d(K-1)} \ln K$

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(A) Consider a small strip of width $dx$ at a distance $x$ from the left end. The height of the dielectric part is $y = (d/a)x$.This strip acts as two capacitors in series: one with air (thickness $d-y$) and one with dielectric (thickness $y$).The capacitance of the air part is $dC_1 = \frac{\varepsilon_0 a dx}{d-y}$ and the dielectric part is $dC_2 = \frac{K\varepsilon_0 a dx}{y}$.The equivalent capacitance $dC$ of the strip is given by $\frac{1}{dC} = \frac{1}{dC_1} + \frac{1}{dC_2} = \frac{d-y}{\varepsilon_0 a dx} + \frac{y}{K\varepsilon_0 a dx} = \frac{Kd - Ky + y}{K\varepsilon_0 a dx} = \frac{Kd - (K-1)y}{K\varepsilon_0 a dx}$.Thus,$dC = \frac{K\varepsilon_0 a dx}{Kd - (K-1)(d/a)x}$.Integrating from $x=0$ to $x=a$:$C = \int_0^a \frac{K\varepsilon_0 a dx}{Kd - \frac{(K-1)d}{a}x} = \frac{K\varepsilon_0 a}{d} \int_0^a \frac{dx}{K - \frac{(K-1)}{a}x}$.Let $u = K - \frac{(K-1)}{a}x$,then $du = -\frac{(K-1)}{a} dx$.$C = \frac{K\varepsilon_0 a}{d} \left( -\frac{a}{K-1} \right) [\ln(K - \frac{(K-1)}{a}x)]_0^a = \frac{-K\varepsilon_0 a^2}{d(K-1)} [\ln(1) - \ln(K)] = \frac{K\varepsilon_0 a^2}{d(K-1)} \ln K$.

$A$ parallel plate capacitor is made of two square plates of side $a$,separated by a distance $d$ $(d \ll a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$,as shown in the figure. The capacitance of this capacitor is

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