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A metal plate of thickness $2 \mathrm{~mm}$ and area $36 \pi \mathrm{mm}^2$ - is slide in to a parallel plate capacitor of plate spacing $6 \mathrm{~mm}$ and area $36 \pi \mathrm{cm}^2$. The metal plate is at a distance $3 \mathrm{~mm}$ from one of the plates. What is the capacitance of this arrangement? (Let $\frac{\mathrm{I}}{4 \pi \mathrm{E}_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}$ )
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Verified Answer
The correct answer is:
$25 \mathrm{pF}$
When metal plate of thickness $2 \mathrm{~mm}$ is inserted into a parallel plate capacitor, then the arrangement is shown as,
Given, thickness of plate, $t=2 \mathrm{~mm}$ and area, $A=36 \pi \mathrm{cm}^2$
The above circuit can be redrawn as,
where, $C_1=\frac{\varepsilon_0 A}{d_1}$ and $C_2=\frac{\varepsilon_0 A}{d_2}$
Now, $C_{\text {eq }}=\frac{C_1 C_2}{C_1+C_2}$ (In the series combination)
$$
\begin{aligned}
& \Rightarrow C_{\text {eq }}=\varepsilon_0 A\left[\frac{\frac{1}{d_1} \times \frac{1}{d_2}}{\frac{1}{d_1}+\frac{1}{d_2}}\right]=\frac{\varepsilon_0 A}{d_1+d_2} \\
& \Rightarrow C_{\text {eq }}=\frac{1}{4 \pi \times 9 \times 10^9} \times \frac{36 \pi \times 10^{-4}}{3 \times 10^{-3}+1 \times 10^{-3}} \\
& \qquad\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\
& =\frac{1}{4} \times 10^{-10}=25 \times 10^{-12} \mathrm{~F}=25 \mathrm{pF}
\end{aligned}
$$
So, the correct option is (3).
Given, thickness of plate, $t=2 \mathrm{~mm}$ and area, $A=36 \pi \mathrm{cm}^2$
The above circuit can be redrawn as,
where, $C_1=\frac{\varepsilon_0 A}{d_1}$ and $C_2=\frac{\varepsilon_0 A}{d_2}$
Now, $C_{\text {eq }}=\frac{C_1 C_2}{C_1+C_2}$ (In the series combination)
$$
\begin{aligned}
& \Rightarrow C_{\text {eq }}=\varepsilon_0 A\left[\frac{\frac{1}{d_1} \times \frac{1}{d_2}}{\frac{1}{d_1}+\frac{1}{d_2}}\right]=\frac{\varepsilon_0 A}{d_1+d_2} \\
& \Rightarrow C_{\text {eq }}=\frac{1}{4 \pi \times 9 \times 10^9} \times \frac{36 \pi \times 10^{-4}}{3 \times 10^{-3}+1 \times 10^{-3}} \\
& \qquad\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\
& =\frac{1}{4} \times 10^{-10}=25 \times 10^{-12} \mathrm{~F}=25 \mathrm{pF}
\end{aligned}
$$
So, the correct option is (3).
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