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?

Respuesta :

Explanation:

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

            C = [tex]\frac{\epsilon_{o}A}{d}[/tex]

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

               E = [tex]3 \times 10^{6} V/m[/tex]

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

                   = [tex]\frac{\epsilon_{o} A}{d} \times E \times d[/tex]

                   = [tex]\epsilon_{o}AE[/tex]

                   = [tex]8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}[/tex]

                   = [tex]1.062 \times 10^{-8} C[/tex]

or,                = 10.62 nC

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

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