Question

A metal plate of thickness $2mm$ and area $36\pi c{m}^2-$ is slide into a parallel plate capacitor of plate spacing $6mm$ and area $36\pi c{m}^2$. The metal plate is at a distance $3mm$ from one of the plates. What is the capacitance of this arrangement? $($ Let $\frac{1}{4\pi {ϵ}_0}=9\times 10^{9}N{m}^2{C}^{-2})$

• A. 8 pF
• B. 15 pF
• C. 25 pF
• D. 20 pF

Answer 1

Answer: (C)

When metal plate of thickness $2mm$ is inserted into a parallel plate capacitor, then the arrangement is shown as,

Given, thickness of plate, $t=2mm$ and area, $A=36\pi c{m}^2$
The above circuit can be redrawn as,

where, $C_1=\frac{{\epsilon }_{0}A}{d_1}$ and $C_2=\frac{{\epsilon }_{0}A}{d_2}$
Now, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$ ( In the series combination)
$\Rightarrow C_{eq}={\epsilon }_{0}A\left[\frac{\frac{l}{d_1}\times \frac{l}{d_2}}{\frac{l}{d_1}+\frac{l}{d_2}}\right]=\frac{{\epsilon }_{0}A}{d_1+d_2}$
$\Rightarrow C_{eq}=\frac{1}{4\pi \times 9\times 10^9}\times \frac{36\pi \times 10^{-4}}{3\times 10^{-3}+1\times 10^{-3}}$
$\left(\because \frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2/C^2\right)$
$=\frac{1}{4}\times 10^{-10}=25\times 10^{-12}F=25pF$