What happens to the charge and amount of energy stored?

[Rijad Hadzic]

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
\Delta V = Qd/A\varepsilon_0
C = A\varepsilon_0 / d

The Attempt at a Solution

If Q = c\Delta V

and plugging in for c

Q = \Delta V A \varepsilon_0 / d

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

(1/2)Q = \Delta V A \varepsilon_0 / 2d, 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??

 

[Rijad Hadzic]

I think I can alter the equation

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

to

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

and then plugging in for C you get that the energy is halved.

But I just don't get how for one equation its halved, then for the other its doubled.

V can also be expressed by distance too, so the mindfuck adds on there.

Like for finding when D is doubled, finding charge,

Q = cV

well, C = Aepsilon /d , but V = Qd/ Aepsilon

So doubling the separation wouldn't do a thing here since the d's cancel out

I'm pretty confused guys. If anyone can point me in the right direction I would really appreciate it rn.

 

[Rijad Hadzic]

Is it maybe that I'm mistaken, and that \Delta V actually doesn't change, since it depends on the batteries voltage no matter what even if the distance is being changed between the plates?

 

[cnh1995]

Is it maybe that I'm mistaken, and that \Delta V actually doesn't change, since it depends on the batteries voltage no matter what even if the distance is being changed between the plates?

That's correct. The voltage across the capacitor remains unchanged, but since the distance between the plates is doubled, the capacitance is halved.