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Musya8 [376]
3 years ago
13

A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconne

cted, 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?
Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

Explanation:

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.

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a. Zin = 41.25 - j 16.35 Ω

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a.

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b.

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c.

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e.

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Pg = 478.4 w

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