Energy density applied to a capacitor

[Zareon]

Take a parallel plate capacitor, area A and distance d.

I know that E=sigma/e0, V=Ed, C=Ae0/d.

I have also shown that the energy stored is 1/2CV^2=(e0/2)(E^2)(Ad).

 

Now, if I increase the distance between the plates by an amount x, then E doesn't change and the increase in energy is (e0/2)(E^2)(Ax).

 

But the work done by increasing the distance is Fx=(QE)x=A(sigma)Ex=e0(E^2)(Ax). Not equal to (e0/2)(E^2)(Ax), but twice that!! What's going on here!?

[Externet]

Hi.

If you increase the distance between plates the capacity diminishes greatly.

Then, the energy does too.

Miguel

[Meir Achuz]

I can see you are thinking, but there is a resolution.

The force on one plate of a capacitor is not \sigma AE.

It is one half of that because E is due to both plates, but the force on plate 2 is only due to the E from plate 1.

[Zareon]

Thanks for the replies.

 

Externet:

I know the capacitance decreases (like 1/x), but the potential difference increases (like x^2) so the total energy is increased and gives the same (correct) value for the energy.

 

Meir Achuz:

That took me a while to grasp, but I think I see what you are saying. The plate can't exert a force on itself (just like I can't lift myself up by my hair). So I shouldn't consider the E-field created by that plate when calculating the work done, only the one from the other plate, which is 1/2E. Is that right?

Thanks a lot!

[Externet]

Sorry, cannot follow your thinking... I lost you.

 

Your first post says "...E doesn't change ..."

 

And on your post above, says "...the potential difference increases (like x^2)..."

 

No clue what you are thinking nor saying. Do not know what do you mean by E, by potential; by x.

 

If potential E does not change and capacitance decreases; energy will decrease.

Miguel

[Zareon]

E is not the energy. E is the electric field. Yeah, confusing... I'm gonna denote the stored energy by W.

 

So the electic field is constant: [math]E=\sigma/\epsilon_0[/math].

 

And the energy is: [math]W=1/2CV^2=\frac{\epsilon_0}{2}AdE^2[/math]

 

What I meant is that the capacitance decreases inversely to the distance between the plates. That's what I meant by goes like 1/x, or rather 1/d. But the potential V increases linearly with the distance (goes like x). So the energy 1/2CV^2 goes increases linearly with the distance between the plates (like x). And I now realize this is not exactly the same as what I said in my previous post where I said V goes like x^2...which is wrong

[Externet]

Still lost in space...

 

V^2 is VERY different from 1/V^2

 

And where do you get "potential V increases linearly with the distance" from ?

 

Miguel

[Zareon]

Still lost in space...

 

V^2 is VERY different from 1/V^2

Yes it is. I don't recall saying otherwise...

Still lost in space...

And where do you get "potential V increases linearly with the distance" from ?

 

V=Ed, where d is the distance between the plates. That's entirely trivial.