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**1. Homework Statement**

A charged parallel plate capacitor is connected to a battery. The plates of the capacitor are pulled apart so that their separation doubles. What happens to the charge and the amount of energy stored by the capacitor?

**2. Homework Equations**

Q = cV

Ue = (1/2)(Q)^2 / c

[itex] \Delta V = Qd/A\varepsilon_0 [/itex]

[itex] C = A\varepsilon_0 / d [/itex]

**3. The Attempt at a Solution**

If [itex] Q = c\Delta V [/itex]

and plugging in for c

[itex] Q = \Delta V A \varepsilon_0 / d [/itex]

Is our original Q, now if D is seperated by 2, we get

[itex] (1/2)Q = \Delta V A \varepsilon_0 / 2d [/itex], thus our Q has been halved.

Now if potential energy = (1/2c) (Q)^2

so doubling our distance halves our capacitance. 2d = (1/2) c

which in turn, increases our potential energy by a factor of 2.

But my book is telling me Ue is halved. Why is this??