Parallel plate capacitor A parallel plate capacitor has plates of area 500 cm2 a

Question

Parallel plate capacitor A parallel plate capacitor has plates of area 500 cm2 and is connected across the terminals of a battery. After some time has passed, the capacitor is disconnected from the battery. Then the plates are moved 0.40 cm further apart but the charge on each plate remains constant and the potential difference between the plates increases by 100 V (a) What is the magnitude of the charge on each plate? (b) Determine the change in stored energy in the capacitor due to the movement of the plates. d0.004 m

Explanation / Answer

here by using the formula

Q = e0 * E * A = e0 * A * delta V / delta d

e0 = 8.85 * 10^-12 C^2 / N*m^2

A = 500 cm^2 = 500 * 10^-4 m^2

delta V = 100 V

delta d = 0.40 cm = 0.004 m

then by using these values in the fomula

Q = 8.85 * 10^-12 * 500 * 10^-4 * 100 / 0.004 = 11 *10^-9 C

so the charge on each plate is 11 nC

b)

delta U = 0.5 * Q * delta V

delta U = 0.5 * 11 * 10^-9 * 100 = 0.55 * 10^-6 J

so the energy in the capacitor due to the movement of the plates is 0.55 *10^-6 J

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