Work Done on a Parallel Plate Capacitor

[vertabatt]

Homework Statement

An isolated 5.60 microfarad parallel-plate capacitor has 4.90 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.00 microfarads.

Homework Equations

W = Fd
c = Q/V

The Attempt at a Solution

My first instinct was to solve to the potential of the system in it's initial conditions:

C = Q/V

5.60*10^-6 F = .0049C/V
V = 875 V

Now is where I become a little confused. I'm not sure how I can solve for the distance between the plates in the final condition. Also, I'm not sure what values stay constant as the plates are separated.

Any help is much appreciated!

 

[Doc Al]

Use C = Q/V. Note that V = Ed, where E is the field between the plates. Does the field change as the plates are separated?