A parallel plate capacitor is to be designed with a voltage ratin… - Vidyadip

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?

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Answer

Potential rating of a parallel plate capacitor, V = 1kV= 1000V
Dielectric constant of a matertal, $\in_{\text{r}}={3}$
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,
$\text{C}=\frac{\in_{0}\in_{\text{r}\text{A}}}{\text{d}}$
where, A = Area of each plate$\in_{0}$= Permittivity of free space $= 8.85 \times 10^{-12} N^{-1} C^2 m^{-2}$
$\therefore\ \text{A}=\frac{\text{Cd}}{\in_{0}\in_{\text{r}}}$ $=\frac{{50}\times{10}^{-12}\times{10}^{-3}}{{8.85}\times{10}^{-12}\times{3}}\approx{19}\text{cm}^{2}$
Hence, the area of each plate is about $19cm^2.$

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