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A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10⁷Vm^–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?

Vidyadip · Accepted 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 10⁷ = 10⁶ $\frac{\text{V}}{\text{m}}$

Capacitance of the parallel plate capacitor, C = 50pf = 50 × 10^-12F 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 × 10^-12 N^-1 C² 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 19cm².

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