First slide

Capacitance

Question

A parallel-plate capacitor is to be designed, using a dielectric of dielectric constant 5, so as to have a dielectric strength of $10^{9} {Vm}^{- 1}$ . If the voltage rating of the capacitor is 12 kV, the minimum area of each plate required to have a capacitance of 80 pF is :

Moderate

Solution

$\begin{array}{l} d = \frac{V}{E} = \frac{12 \times 10^{3}}{10^{9}} = 12 \times 10^{- 6} m \\ C = \frac{k ε_{0} A}{d} \Rightarrow A = \frac{C d}{k ε_{0}} = \frac{80 \times 10^{- 12} \times 12 \times 10^{- 6}}{5 \times 8.8 \times 10^{- 12}} \\ = 21.8 \times 10^{- 6} m^{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App