Question

A metal plate of thickness $2\, mm$ and area $36 \,\pi \,cm ^{2}-$ is slide into a parallel plate capacitor of plate spacing $6 \,mm$ and area $36 \,\pi \,cm ^{2}$. The metal plate is at a distance $3\, mm$ from one of the plates. What is the capacitance of this arrangement? $\left(\right.$ Let $\left.\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right)$

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

Answer 1

Answer: (C)

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

Given, thickness of plate, $t=2 \,mm$ and area, $A=36\, \pi \,cm ^{2}$
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, $C_{ eq }=\frac{C_{1} C_{2}}{C_{1}+C_{2}}$ ( In the series combination)
$\Rightarrow \,C_{ eq }=\varepsilon_{0} A\left[\frac{\frac{ l }{d_{1}} \times \frac{ l }{d_{2}}}{\frac{ l }{d_{1}}+\frac{ l }{d_{2}}}\right]=\frac{\varepsilon_{0} A}{d_{1}+d_{2}}$
$\Rightarrow C_{ eq }=\frac{1}{4 \pi \times 9 \times 10^{9}} \times \frac{36 \pi \times 10^{-4}}{3 \times 10^{-3}+1 \times 10^{-3}}$
$\left(\because \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} / C ^{2}\right)$
$=\frac{1}{4} \times 10^{-10}=25 \times 10^{-12} F =25\, pF$