A capacitor has capacitance $C_0$ when there is no dielectric between it's plates. 2 slabs of dielectric constant $K_1, K_2$ respectively with area equal to area of plates but thickness half of the distance between the plates are placed in between the plates. Then the new capacitance is

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Aicharya · Accepted Answer

Initial capacitance of capacitor, $C_0=\frac{\varepsilon_0 A}{d}$

Now, capacitor is filled with two dielectrics each slab has area $A$ and distance $d / 2$ as shown

Above arrangement is a series combination of two capacitors as shown

so, capacity of above arrangement is $\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}$ $\begin{aligned} & =\frac{1}{\frac{\varepsilon_0 K_1 A}{(d / 2)}}+\frac{1}{\varepsilon_0 K_2 A /(d / 2)} \\ & =\frac{d}{2 \varepsilon_0 A}\left[\frac{1}{K_1}+\frac{1}{K_2}\right]\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{1}{C_{\mathrm{eq}}}=\frac{d}{2 \varepsilon_0 A}\left[\frac{K_1+K_2}{K_1 K_2}\right] \\ & \Rightarrow C_{\mathrm{eq}}=\frac{2 \varepsilon_0 A}{d}\left[\frac{K_1 K_2}{K_1+K_2}\right] \\ & \Rightarrow C_{\mathrm{eq}}=2 C_0\left(\frac{K_1 K_2}{K_1+K_2}\right)\end{aligned}$

A capacitor has capacitance $C_0$ when there is no dielectric between it's plates. 2 slabs of dielectric…

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