Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity…
Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will [A = area of plate. $t_1, t_2$ and $t_3$ are thickness of dielectric slabs, $k_1$, $\mathrm{k}_2$ and $\mathrm{k}_3$ are dielectric constant.]
These are three capacitors connected in series.
$\begin{aligned}
& \frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{\mathrm{t}_1}{\varepsilon_0 \mathrm{Ak}_1}+\frac{\mathrm{t}_2}{\varepsilon_0 \mathrm{Ak}_2}+\frac{\mathrm{t}_3}{\varepsilon_0 \mathrm{Ak}_3} \\
& \mathrm{C}=\frac{\mathrm{A} \varepsilon_0}{\left[\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}+\frac{\mathrm{t}_3}{\mathrm{k}_3}\right]}
\end{aligned}$
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