Question

The dielectric strength of air is $3.0\times {10}^6\text{V/m}$. A parallel-plate air-capacitor has area $20{\text{cm}}^2$ and plate separation $0.10\text{mm}$. Find the maximum $rms$ voltage of an $\text{AC}$ source which can be safely connected to this capacitor.

Answer 1

Given:-
Dielectric strength of air, $E=3\times {10}^6\text{V/m}$

Area of plate, $A=20{\text{cm}}^2$

Plate separation, $d=0.1\text{mm}={10}^{-4}\text{m}$

Let, the $rms$ value of voltage of $\text{AC}$ source is $V_{rms}$.

$\therefore$ Peak value of voltage on the plates of the capacitor is

$V_{max}=V_{rms}\sqrt{2}$

The dielectric strength is given by,

$E=\frac{V_{max}}{d}$

$\Rightarrow 3\times {10}^6=\frac{V_{rms}\sqrt{2}}{{10}^{-4}}$

$\Rightarrow V_{rms}=\frac{300}{\sqrt{2}}\text{V}\text{or}212.13\text{V}$