A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^–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 50 pF?

Solution

Given,
Voltage rating for the design of parallel plate capacitor-V= 1 KV=1000 V
Dielectric constant of material-K= $ε_{r}$ = 3
dielectric strength of material = 10⁷ V/m

For safety, electric field at the most should be 10% of dielectric strength.
$\Rightarrow$ E=10% of 10⁷ = $\frac{10}{100}$ V/m

E= 10⁶ V/m

Area of the plates-A = ?
Capacitance of plates-C= $50 pF = 50 \times 10^{- 12} F$

Now, using the formula $E = \frac{V}{d}$
∴ $d = \frac{V}{E} = \frac{10^{3}}{10^{6}} = 10^{- 3} m$

Therefore, $C = \frac{ε_{0} ε_{r} A}{d}$

$\Rightarrow$ $A = \frac{Cd}{ε_{0} ε_{r}} = \frac{50 \times 10^{- 12} \times 10^{- 3}}{8.85 \times 10^{- 12} \times 3}$

$\Rightarrow$ $A = 1.9 \times 10^{- 3} m^{2}$

Hence, the minimum area of plates required is $1.9 \times 10^{- 3} m^{2}$ .

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