A parallel-plate air-filled capacitor having area 40 cm $^{2}$ and plate spacing 1.0 mm is charged to a potential difference of 600 V.Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, and (e) the energy density between the plates.

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Solution

(a) The capacitance is
$C = \frac{ε_{0} A}{d} = \frac{(8.85 \times 10^{- 12} C^{2} / N . m^{2}) (40 \times 10^{- 4} m^{2})}{1.0 \times 10^{- 3} m} = 3.5 \times 10^{- 11} = 35 p F$
(b) $q = C V = (35 p F) (600 V) = 2.1 \times 10^{- 8} C = 21 n C$
(c) $U = \frac{1}{2} C V^{2} = \frac{1}{2} (35 p F) (21 n C)^{2} = 6.3 \times 10^{- 6} J = 6.3 µ J$
(d) $E = V / d = 600 V / 1.0 \times 10^{- 3} m = 6.0 \times 10^{5} V / m .$
(e) The energy density (energy per unit volume) is :
$u = \frac{U}{A d} = \frac{6.3 \times 10^{- 6} J}{(40 \times 10^{- 4} m^{2}) (1.0 \times 10^{- 3} m)} = 1.6 J / m^{3}$

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