What happens to the capacitance after increasing distance?

[eprparadox]

Homework Statement

A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the:

A. capacitance
B. surface charge density on each plate
C. stored energy
D. electric field between the two places
E. charge on each plate

Homework Equations

$C = \frac{\epsilon_0 A}{d}$

$C = \frac{Q}{V}$

$U = \frac{1}{2}CV^2$

The Attempt at a Solution

I know the answer is C but am confused on how to get there.

After we charge the capacitor and then disconnect it, the charge is fixed. I set these two expressions equal:

$\frac{\epsilon_0 A}{d} = \frac{Q}{V}$

And I thought that if the separation doubles that the potential difference across the plates must also double. And then plugging into $\frac{1}{2}CV^2$, the energy would increase by 4 times.

Without the equations, it does make sense that if we do work to double the separation that that energy we did will go into the stored energy of the field. But I'm not sure where I'm going wrong with the equations.

Any thoughts?

 

[Dadface]

It's true that V doubles but C, ( Q/V ) halves.

 

[Dr Dr news]

Since C = ∈A/d = Q/V, increasing d to 2d decreases C to C/2 and increases V to 2V since by the conservation of charge Q = con., So what happens to the energy, CV^2/2?