A parallel plate air filled capacitor of capacitance $C$ has plate area $A$ and the distance between the plates $d$. When a metal sheet of thickness $\left(\frac{d}{2}\right)$ and of the same area $A$ is introduced between the plates, its capacitance becomes $C_2$. The ratio $C_2: C_1$ is

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Aicharya · Accepted Answer

The capacitance of the air filled parallel plate capacitor is given by When a slab of dielectric constant $K$, and thickness $\mathrm{t}$ is introduced in between the plates of the capacitor, its new capacitance is given by $C^{\prime}$. This leads to parallel combination of two capacitors: $\frac{1}{C^{\prime}}=\frac{d-t}{\varepsilon_0 A}+\frac{t}{K \varepsilon_0 A}=\frac{d+t\left(\frac{1}{K}-1ight)}{\varepsilon_0 A}$ Since a metal sheet of thickness $d / 2$ is introduced, hence here, $t=d / 2$, $K=\propto$ (for metals) or $\frac{1}{K}=0$ Taking ratio of equation (2) and (1), $	herefore \frac{C^{\prime}}{C}=\frac{\frac{2 \varepsilon_0 A}{d}}{\frac{\varepsilon_0 A}{d}}=\frac{2}{1}=2: 1$

A parallel plate air filled capacitor of capacitance $C$ has plate area $A$ and the distance between the…

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