In a parallel plate capacitor,two dielectric | vedclass

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Question

(A) parallel plate capacitor with dielectric slabs of thicknesses $t_1$ and $t_2$ can be treated as two capacitors in series.The capacitance of a para

Vedclass · Accepted Answer

$\frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2}}$

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(A) parallel plate capacitor with dielectric slabs of thicknesses $t_1$ and $t_2$ can be treated as two capacitors in series.The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\varepsilon_0 A}{t/K}$.For two slabs in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.Substituting the values,we get $\frac{1}{C_{eq}} = \frac{t_1}{\varepsilon_0 A K_1} + \frac{t_2}{\varepsilon_0 A K_2}$.$\frac{1}{C_{eq}} = \frac{1}{\varepsilon_0 A} (\frac{t_1}{K_1} + \frac{t_2}{K_2})$.Therefore,$C_{eq} = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2}}$.

In a parallel plate capacitor,two dielectric slabs of thicknesses $t_1$ and $t_2$ and dielectric constants $K_1$ and $K_2$ respectively are inserted. What is the capacitance of this capacitor?

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