Two parallel plates, each having area A = 3557 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.59 cm. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm). What is the energy stored in this new capacitor?

Question

Jesse Durham

Physics

23 April, 19:47

Two parallel plates, each having area A = 3557 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.59 cm. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.18 cm). What is the energy stored in this new capacitor?

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Harmony Watts · Answer 1

Capacitance, C is equal to epsilon times area of the plate divided by distance. The permittivity of air is 8.84 x 10-12 F/m. Capacitance is equal to 2.66 x10^14 F. The energy is equal to 0.5 * C*V^2. Substituting, the energy is equal to 0.5 * 2.66 x10^14 F * 6^2 V^2 or equal to 4.80 x10^15 W.