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**1. 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.

**2. Homework Equations**

W = Fd

c = Q/V

**3. 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!

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