- #1
Rijad Hadzic
- 321
- 20
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?
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]
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 separated 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??