How much energy is stored in the capacitor?

In summary, Astronuc explained that the capacitor stores energy when the plates are disconnected from the battery. However, once the plates are reconnected and increased in separation from 2.75 mm to 11 mm, the same amount of energy is stored as before.
  • #1
FlipStyle1308
267
0
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.
 
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  • #3
Accidentally made new post instead of editting this post...whoops!
 
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  • #4
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?
 
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  • #5
Bump! Is anyone able to help me figure this problem out?
 
  • #6
You did use U = (1/2)C V2, right?

Then C = [tex]\frac{k\,\epsilon_o\,A}{d}[/tex].


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 [itex]k\,\epsilon_o\,A[/itex], then

C1d1 = C2d2. If one solves part a, then one can use this relationship for the second part.
 
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  • #7
Astronuc said:
You did use U = (1/2)C V2, right?

Then C = [tex]\frac{k\,\epsilon_o\,A}{d}[/tex].


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 [itex]k\,\epsilon_o\,A[/itex], 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?
 
  • #8
FlipStyle1308 said:
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?
 
  • #9
Doc Al said:
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?
 
  • #10
FlipStyle1308 said:
The voltage remains the same when the plates are separated, right?
No. Hint: After charging the capacitor, the battery was disconnected.
 
  • #11
FlipStyle1308 said:
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?
 
  • #12
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?
 
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  • #13
V, C, and Q are not all changed. Once the plates are disconnected from the battery, charge cannot enter or leave.
 
  • #14
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
FlipStyle1308 said:
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.
 
  • #16
Sorry, I quickly assumed charge referred to V, so it means that Q remains the same?
 
  • #17
That's right.
 
  • #18
Okay, I figured it out. Thank you for your help!
 
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FAQ: How much energy is stored in the capacitor?

How is energy stored in a capacitor?

Energy is stored in a capacitor through the buildup of electric charge on its plates. When a capacitor is connected to a power source, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores the energy in the form of potential energy.

What is the formula for calculating the energy stored in a capacitor?

The formula for calculating the energy stored in a capacitor is E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.

How does the capacitance of a capacitor affect the energy stored in it?

The capacitance of a capacitor is directly proportional to the amount of energy it can store. This means that a capacitor with a higher capacitance will be able to store more energy than one with a lower capacitance, given the same voltage.

Can the energy stored in a capacitor be increased?

Yes, the energy stored in a capacitor can be increased by either increasing the voltage or increasing the capacitance. However, it is important to note that increasing the voltage beyond the capacitor's maximum rating can cause it to fail.

How is the energy stored in a capacitor released?

The energy stored in a capacitor is released when the capacitor is connected to a circuit. As the capacitor discharges, the electric charge on its plates decreases, and the stored energy is converted into electrical energy to power the circuit.

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