Consider a capacitor constructed from two square metal plates, each...

Consider an elemental strip of width $dx$ at a distance $x$ from the left end of the capacitor.

The area of the plates for this strip is $dA = a \, dx$.

This strip can be considered as two capacitors in series: one with dielectric $K_1 and the other with dielectric$K_2$.<br><br>From the geometry of the figure, the thickness of the dielectric$K_2$at distance$x$is$y = \dfrac{d}{a}x$.<br><br>The thickness of the dielectric$K_1$at distance$x$is$d - y = d - \dfrac{d}{a}x$.<br><br>The capacitance of the upper part (with$K_1$) is:<br><br>$dC_1 = \dfrac{K_1 \varepsilon_0 a \, dx}{d - \dfrac{d}{a}x}$<br><br>The capacitance of the lower part (with$K_2$) is:<br><br>$dC_2 = \dfrac{K_2 \varepsilon_0 a \, dx}{\dfrac{d}{a}x}$<br><br>Since these two capacitors are in series, the equivalent capacitance$dC$of the elemental strip is given by:<br><br>$\dfrac{1}{dC} = \dfrac{1}{dC_1} + \dfrac{1}{dC_2}$<br><br>$\dfrac{1}{dC} = \dfrac{d - \dfrac{d}{a}x}{K_1 \varepsilon_0 a \, dx} + \dfrac{\dfrac{d}{a}x}{K_2 \varepsilon_0 a \, dx}$<br><br>$\dfrac{1}{dC} = \dfrac{d}{\varepsilon_0 a \, dx} \left[ \dfrac{1}{K_1} - \dfrac{x}{a K_1} + \dfrac{x}{a K_2} \right]<br><br>\dfrac{1}{dC} = \dfrac{d}{\varepsilon_0 a \, dx} \left[ \dfrac{1}{K_1} + \dfrac{x}{a} \left( \dfrac{K_1 - K_2}{K_1 K_2} \right) \right]

$dC = \dfrac{\varepsilon_0 a \, dx}{d \left[ \dfrac{1}{K_1} + \dfrac{x}{a} \left( \dfrac{K_1 - K_2}{K_1 K_2} \right) \right]}$

The total capacitance $C$ is obtained by integrating $dC$ from $x = 0$ to $x = a$:

$C = \int_{0}^{a} \dfrac{\varepsilon_0 a \, dx}{d \left[ \dfrac{1}{K_1} + \dfrac{x}{a} \left( \dfrac{K_1 - K_2}{K_1 K_2} \right) \right]}$

Let v = \dfrac{1}{K_1} + \dfrac{x}{a} \left(\dfrac{K_1 - K_2}{K_1 K_2} \right).

Differentiating with respect to $x$, we get $dv =$\dfrac{K_1 - K_2}{a K_1 K_2}$dx$, which gives $dx =$\dfrac{a K_1 K_2}{K_1 - K_2}$dv$.

The limits of integration change as follows:
When $x = 0$, $v =$\dfrac{1}{K_1}.
When $x = a$, $v =$\dfrac{1}{K_1} + \dfrac{K_1 - K_2}{K_1 K_2} = \dfrac{K_2 + K_1 - K_2}{K_1 K_2} = \dfrac{1}{K_2}.

Substituting these into the integral:

$C = \int_{1/K_1}^{1/K_2}$\dfrac{\varepsilon_0 a}{d \cdot v} $\left( \dfrac{a K_1 K_2}{K_1 - K_2} \right)$dv$<br><br>$C = $\dfrac{\varepsilon_0 a^2 K_1 K_2}{d(K_1 - K_2)} \int_{1/K_1}^{1/K_2} \dfrac{dv}{v}<br><br>$C = $\dfrac{\varepsilon_0 a^2 K_1 K_2}{d(K_1 - K_2)} [\ln$v]_{1/K_1}^{1/K_2}$<br><br>$C = $\dfrac{\varepsilon_0 a^2 K_1 K_2}{d(K_1 - K_2)} \left( \ln \dfrac{1}{K_2} - \ln \dfrac{1}{K_1} \right)$

$C = \dfrac{\varepsilon_0 a^2 K_1 K_2}{d(K_1 - K_2)} \ln \left( \dfrac{K_1}{K_2} \right)<br><br>Answer: \dfrac{\varepsilon_0 K_1 K_2 a^2 \ln(K_1/K_2)}{(K_1 - K_2)d}$