Parallel-plate capacitor problem

1. Sep 7, 2006

FlipStyle1308

Question I need to solve:

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.

2. Sep 7, 2006

3. Sep 7, 2006

FlipStyle1308

Last edited: Sep 8, 2006
4. Sep 8, 2006

FlipStyle1308

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?

Last edited: Sep 8, 2006
5. Sep 8, 2006

FlipStyle1308

Bump! Is anyone able to help me figure this problem out?

6. Sep 9, 2006

Staff: Mentor

You did use U = (1/2)C V2, right?

Then C = $$\frac{k\,\epsilon_o\,A}{d}$$.

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 m2. So, 315 cm2 = 315 x 10-4 m2 = 0.0315 m2.

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

C1d1 = C2d2. If one solves part a, then one can use this relationship for the second part.

Last edited: Sep 9, 2006
7. Sep 9, 2006

FlipStyle1308

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?

8. Sep 9, 2006

Staff: Mentor

The capacitance changes as Astronuc explained. But what remains the same when the plates are separated?

9. Sep 9, 2006

FlipStyle1308

The voltage remains the same when the plates are separated, right?

10. Sep 10, 2006

Staff: Mentor

No. Hint: After charging the capacitor, the battery was disconnected.

11. Sep 10, 2006

Staff: Mentor

I mislead you with the U = (1/2)C V2.

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?

12. Sep 10, 2006

FlipStyle1308

As the distance is changed and the capacitor is isolated, U stays the same, and V, C, and Q are changed, right? What do I do next?

Last edited: Sep 10, 2006
13. Sep 11, 2006

Staff: Mentor

V, C, and Q are not all changed. Once the plates are disconnected from the battery, charge cannot enter or leave.

14. Sep 11, 2006

FlipStyle1308

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?

15. Sep 11, 2006

Staff: Mentor

V does not remain the same!

Reread my last post for a hint about what does remain the same as the plates are separated.

16. Sep 11, 2006

FlipStyle1308

Sorry, I quickly assumed charge referred to V, so it means that Q remains the same?

17. Sep 11, 2006

Staff: Mentor

That's right.

18. Sep 11, 2006

FlipStyle1308

Okay, I figured it out. Thank you for your help!

Last edited: Sep 11, 2006