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Question

A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10^7Vm^ –1 . (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10 % of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50pF ?

Accepted Answer

Potential rating of a parallel plate capacitor $, V = 1kV= 1000V$
Dielectric constant of a matertal, $\epsilon_{\text{r}}={3}$
Dielectric strength $={10}^{7}\frac{\text{V}}{\text{m}}$
For safety, the field intensity never exceeds $10\%$ of the dielectric strength.
Hence, electric field intensity $, E = 10\% of 10^7 = 10^6$ $\frac{\text{V}}{\text{m}}$
Capacitance of the parallel plate capacitor $, C = 50pf = 50 \times 10^{-12}F$
Distance between the plates is given by,$\text{d}=\frac{\text{V}}{\text{E}}$
$=\frac{1000}{{10}^{6}}={10}^{-3}\text{m}$
Capacitance i given by the relation, $\text{C}=\frac{\epsilon_{0}\epsilon_{\text{r}\text{A}}}{\text{d}}$ where $, A =$ Area of each plate $\epsilon_{0} =$ Permittivity of free space $= 8.85 \times 10^{-12} N^{-1} C^2 m^{-2}$
$\therefore\ \text{A}=\frac{\text{Cd}}{\epsilon_{0}\epsilon_{\text{r}}}$
$=\frac{{50}\times{10}^{-12}\times{10}^{-3}}{{8.85}\times{10}^{-12}\times{3}}\approx{19}\text{cm}^{2}$
Hence, the area of each plate is about $19\ cm^2$.

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