A parallel plate capacitor with plate area A | vedclass

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(B) Consider an element of width $dx$ at a distance $x$ $(x < d/2)$ from the left plate. The differential capacitance $dC$ of this element is given by

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$\frac{kA}{2 \ln \left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$

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(B) Consider an element of width $dx$ at a distance $x$ $(x For the first half of the capacitor ($0$ to $d/2$),the capacitors are in series. The equivalent capacitance $C_1$ is given by:$\frac{1}{C_1} = \int_{0}^{d/2} \frac{1}{dC} = \frac{1}{A} \int_{0}^{d/2} \frac{dx}{\varepsilon_{0} + kx}$Integrating this,we get:$\frac{1}{C_1} = \frac{1}{kA} [\ln(\varepsilon_{0} + kx)]_{0}^{d/2} = \frac{1}{kA} \ln\left(\frac{\varepsilon_{0} + kd/2}{\varepsilon_{0}}\right) = \frac{1}{kA} \ln\left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)$So,$C_1 = \frac{kA}{\ln\left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$.Due to symmetry,the capacitance of the second half $(C_2)$ is the same as $C_1$. Since the two halves are in series,the total equivalent capacitance $C_{eq}$ is:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C_1}{2} = \frac{kA}{2 \ln \left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$.

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