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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?

Respuesta :

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

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