Question

A parallel-plate air capacitor is made from two plates 0.070 m square, spaced 9.2 mm apart. What must the potential difference between the plates be to produce an energy
density of 0.062 J/m3?

I am more concerned with the theory/work behind it than the actual answer - thank you.

Answer 1

Capacitance C of a parallel palte capacitor is given by C = KeoA/d

where A = area = pi r^2, e0 = constnat = 8.85*10^-12,

d   = distance between the plates, K = dieelctric constant (=1 for air)

Chareg Q = CV where V = Volatge

Energy U = 0.5QV = 0.5 CV^2 = Q^2/2C

eletric field E = V/d

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so here we apply energy density u = 0.5 eo E^2

so E^2 = 2 * 0.062/8.85 *10^-12

E^2   = 1.4011*10^10

Electric field E = 1.183 *10^5 N/C

so Potential diff V = Ed

V = 1.183 *10^5 * 9.2*10^-3

V = 1088.92 volts