A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconnected, then the two plates are pulled further apart to a new separation distance 2d. What happens to the potential difference across the capacitor, ΔV, and the charge on the capacitor, Q, as a result of this change?

Question

Physics

Justine Mckee

15 January, 10:34

A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconnected, then the two plates are pulled further apart to a new separation distance 2d. What happens to the potential difference across the capacitor, ΔV, and the charge on the capacitor, Q, as a result of this change?

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Answers (1)

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Reina Blankenship · Answer 1

Initial separation of plate = d

final separation = 2d

The capacitance of the capacitor will reduce from C to C/2 because

capacitance = ε A / d

d is distance between plates.

As the batteries are disconnected, charge on the capacitor becomes fixed.

Initial charge on the capacitor

= Capacitance x potential difference

Q = C ΔV

Final charge will remain unchanged

Final charge = C ΔV

Final capacitance = C/2

Final potential difference = charge / capacitance

= C ΔV / C/2

= 2 ΔV

Potential difference is doubled after the pates are further separated.