Question

How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00 cm2?

Answer 1

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is $9 \times 10^{-4} m^{2}$. As the capacitance of capacitor is given as follows.

            C = $\frac{\epsilon_{o}A}{d}$

It is known that the dielectric strength of air is as follows.

               E = $3 \times 10^{6} V/m$

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = $\frac{\epsilon_{o} A}{d} \times E \times d$

                   = $\epsilon_{o}AE$

                   = $8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}$

                   = $1.062 \times 10^{-8} C$

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.