A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

U/U₀ = 2

(factor of 2 i.e U = 2U₀)

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

Explanation:

Energy stored in a capacitor can be expressed as;

U = 0.5CV^2 = Q^2/2C

And

C = ε₀ A/d

Where

C = capacitance

V = potential difference

Q = charge

A = Area of plates

d = distance between plates

So

U = Q^2/2C = dQ^2/2ε₀ A

The initial energy of the capacitor at d = d₀ is

U₀ = Q^2/2C = d₀Q^2/2ε₀ A 1

When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

The final energy stored in the capacitor at d = 2d₀ is

U = 2d₀Q^2/2ε₀ A ...2

The factor U/U₀ can be derived by substituting equation 1 and 2

U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )

Simplifying we have;

U/U₀ = 2

U = 2U₀

Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )

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