# ice_screenshot_20190305-165218.png A parallel plate capacitor is made of two square plates of side 'a', separated by a distance d The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

If K filled between the plates -

$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

- wherein

$\dpi{100} C\propto A$

$\dpi{100} C\propto\frac{1}{d}$

$\frac{y}{x} = \frac{d}{a}$

$dy = \frac{d}{a} dx$

$dC_{1} = \frac{\varepsilon_{o}adx}{d-y} dx$

$dC_{2} = \frac{K\varepsilon_{o}adx}{y}$

$dC_{eq} =\frac{dC_{1}dC_{2}}{dC_{1}+dC_{2}} = \frac{K\varepsilon_{o}adx}{Kd + (1-K)y}$

$C_{eq} = \int_{0}^{a} \frac{K\varepsilon_{o}adx}{Kd + (1-K)y}$

$= \frac{K\varepsilon_{o}a^{2}ln K}{ d(K - 1)}$

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