How much energy is stored in the capacitor?

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Question I need to solve:

A parallel-plate capacitor has plates with an area of 315 cm^2 and an air-filled gap between the plates that is 2.75 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.
(a) How much energy is stored in the capacitor?
(b) The separation between the plates is now increased to 11 mm. How much energy is stored in the capacitor now?
(c) How much work is required to increase the separation of the plates from 2.75 mm to 11 mm?

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.

 

[Astronuc]

Q = C * V

See response to Keyboard problem regarding calculating capacitance - https://www.physicsforums.com/showthread.php?t=131039

 

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Accidentally made new post instead of editting this post...whoops!

 

[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?

 

[FlipStyle1308]

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

 

[Astronuc]

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.

 

[FlipStyle1308]

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.

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?

 

[Doc Al]

But using your method, I got the same answer I got before, 4.192 x 10^-6 J, which is incorrect.

Check your arithmetic.

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.

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

 

[FlipStyle1308]

Check your arithmetic.

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

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

 

[Doc Al]

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

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

 

[Astronuc]

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

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?

 

[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?

 

[Doc Al]

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

 

[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?

 

[Doc Al]

So V remains the same.

V does not remain the same!

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

 

[FlipStyle1308]

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

 

[Doc Al]

That's right.

 

[FlipStyle1308]

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