A parallel-plate capacitor with plates of area 540 cm2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.6 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?

Question

Rebel

Physics

9 October, 20:04

A parallel-plate capacitor with plates of area 540 cm2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.6 cm farther apart, the voltage between the plates increases by 100 V. What is the charge Q on the positive plate of the capacitor?

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Phillip Blair · Answer 1

Answer: 0.8 nC

Explanation: In order to explain this problem we have top use the expression for the capacitor of parallel plates, which is given by:

C=εo*A/d where A and d are the area and the separation of the plates.

Then we have

V=Q/C so ΔV = Q * Δd / (εo*A)

Q = 100 * 0.054*8.85*10^-12 / (0.06) = 0.8 nC