I started off by getting the equation U = (1/2)QV = (1/2)CV^2 = (Q^2)/2C. Then I realized that I was given A and d, so I reworded the equation into U = (dQ^2)/(2EoA). At this point I was stuck because I do not have a Q to plug in.

I used U = (CV^2)/2 = (kEoAV^2)/2d, using k = 1.00059, Eo = 8.85 x 10^-12 C^2/Nm^2, A = 3.15 x 10^-4 n^2m d = 0.00275 m, and V = 575 V, obtaining an answer of 1.6768 x 10^-5 J as my answer for the first part. I tried the same method for part (b), but my answer of 4.192 x 10^-6 J is incorrect. Any ideas of where I went wrong?

Also, look at "A = 3.15 x 10^-4 n^2m" the units are strange. Area should be square of the basic length dimension, as in m^{2}. So, 315 cm^{2} = 315 x 10^{-4} m^{2} = 0.0315 m^{2}.

Also, since C = a/d where a is just [itex]k\,\epsilon_o\,A[/itex], then

C_{1}d_{1} = C_{2}d_{2}. If one solves part a, then one can use this relationship for the second part.

Sorry, I actually meant to put 315 x 10^-4 m^2 = A. I don't know why I put those units that I put lol. But using your method, I got the same answer I got before, 4.192 x 10^-6 J, which is incorrect. Using your equation, I solved for C2 = C1d1/d2 = (1.0143 x 10^-10 F)(0.00275 m)/(0.011 m) = 2.5358 x 10^-11, which I plugged into U = (1/2)CV^2, getting 4.192 x 10^-6 J, which again is incorrect. What am I doing wrong?

Looking at U = (1/2)QV = (1/2)CV^2 = (Q^2)/2C, and thinking about what Doc Al mentioned, what changes and what remains the same as the distance is changed and the capacitor is isolated?

Hmm, okay. That makes sense. So V remains the same. Does this also mean that after the separation increased to 11 mm, the same amount of energy is stored as before, which is 1.6768 x 10^-5 J?