# Question Regarding Work Done By Capacitor

1. Jul 23, 2006

### verd

...This isn't a homework problem or anything, I'm just a bit confused about this and have a test coming up. So any clarification would be greatly appreciated...

http://www.synthdriven.com/images/deletable/24-4.jpg [Broken]

We were just recently introduced to the concept of electrostatic potential, and it was defined as U. However, when it comes to a capacitor, this quantity U pops up once again in the following equation:

$$U=\frac{1}{2}CV^2$$

Where U is explained as the total energy stored by the given capacitor...

Now, I know that U=-W. However, this phrasing has me confused. I know that the work is equivalent to the energy stored within a capacitor because the energy stored within the capacitor is the same as the amount of work done on the capacitor, right??

However, that formula doesn't seem to work in this situation because we're talking about a single test charge within the electric field of the capacitor... It's much more microscopic... So would I then use something like W=qV??

But even then, I need V, and in order to find V I'd need C...

I'm just a bit confused on how to go about this, I believe I'm confusing concepts, and any sort of explaination to sort this out would be greatly appreciated.

Thanks

Last edited by a moderator: May 2, 2017
2. Jul 23, 2006

### Päällikkö

Indeed you are mixing up concepts, the total energy of the capacitor (the formula you have) has little to do with moving one point charge in the capacitor.
How does an electric field and the force exerted on the point charge relate?

Last edited: Jul 23, 2006
3. Jul 23, 2006

### verd

Well... Coulomb's law says

$$E=\frac{kq}{r^2}$$
$$F=\frac{kq_{1}q_{2}}{r^2}$$

And I know that
$$V=Ed$$

...Am I going in the right direction??

4. Jul 23, 2006

### Päällikkö

Sort of.

I wouldn't use potential (V) in this problem. Again, how to relate E and F (you should see it from the equations you had in your last post, or think about the definition of a field)?

5. Jul 23, 2006

### Andrew Mason

You would not use Coulomb's law here. Coulomb's law just gives you the electric field, or force per unit charge, of a point charge. But the field here is not that of a point charge. It is uniform and it is given to you.

AM

6. Jul 23, 2006

### verd

Are you looking for E=Fq??

I mean I understand how to determine work given two point charges, but I've got something in between the plates of a capacitor... And I don't know the charge on the plates of the capacitor, it just know the field. So should I be looking for the charge on the plates of the capacitor?

7. Jul 23, 2006

### verd

Work in a non-electrical sense is W=Fd
In an electrical sense, it's W=qV

8. Jul 23, 2006

### Päällikkö

Well, almost: F = qE

You know the strength of the field generated by the two charged plates. Using the equation above, you can determine the force exerted on the point charge/test particle.

I should've asked the definition of work first, as Andrew Mason pointed out. Now's the time to put the equation into use, though.

9. Jul 23, 2006

### Andrew Mason

So, since electrical force = qE, you can express work as:

$$W = Fd = qEd$$

E = force per unit charge, q = charge, d = distance.

In this case, you are given E in terms of Energy per unit charge per unit of distance (which is just Force/unit charge).

So a coulomb of charge will gain 8 Joules if it moves one metre in that field. How much will 20e-6 Coulomb gain if it moves .05 m.?

AM

10. Jul 23, 2006

### verd

Wow. Thank you. This makes sense.

Thanks for straightening me out. I appreciate it guys. Thanks again.