Question

Two identical circular discs each of radius $14\text{cm}$ are kept $7\text{mm}$ apart. If there is air between the plates, then the capacitance of this arrangement is
$($Take ${\epsilon }_0=8.85\times {10}^{-12}{\text{Fm}}^{-1})$

• A. $44.88\text{pF}$
• B. $7.78\times {10}^{-10}\text{F}$
• C. $778.8\text{pF}$
• D. $77.88\text{pF}$

Answer 1

The correct option is D $77.88\text{pF}$

Given,

Radius of the disc, $R=14\text{cm}$,
Distance between the discs, $d=7\text{mm}$,

The capacitance of a parallel capacitor is given by,$$C=\frac{A{\epsilon }_0}{d}$$Where,
$A=\pi R^2$, is the area of the plates and $d$ is the distance between the plates.

$\Rightarrow C=\frac{\pi R^2{\epsilon }_0}{d}$

Substituting the values, we get

$C=\frac{22}{7}\times \frac{{(14\times {10}^{-2})}^2\times 8.85\times {10}^{-12}}{7\times {10}^{-3}}$

$\Rightarrow C=778.8\times {10}^{-13}\text{F}$

$\Rightarrow C=77.88\text{pF}$

Hence, option (d) is the correct answer.

Key concept: Capacitance of a paralllel plate capacitor with plates in the shape of circular discs.