# Homework Help: Work to move two plates

1. Oct 3, 2007

### indigojoker

two parallel plates of area A is seperated by a distance d. One plate has positive charge distribution with area chrage density $\sigma$ and the other plate has an area charge density $-\sigma$

how much work does it take to move the plates 2d?

W=Fd=QEd'

i'm not sure what to use for the Q or the E. Should it be for both plates or for just one plate?

Assuming that it's just for one plate, we get: $W = \sigma A \frac{\sigma}{2 \epsilon_o} d'$

and that would be it right?

then we just let d'=2d

2. Oct 3, 2007

### learningphysics

Just use the energy stored in a capacitor... What is the initial energy stored in the capacitor? What is the final energy?

Final energy - initial energy = work done.

3. Oct 3, 2007

### indigojoker

right, so this gets at my question of whether to use E for one plate or two

we have U=1/2QV=(0.5)QEd

now i can use what i did on the previous post and use Q for one plate $\sigma A$

or I could use the Q on both plates $2\sigma A$

same goes for E

4. Oct 3, 2007

### learningphysics

You need to use Q on one plate... ie: Q = $$\sigma A$$

The field E is $$\frac{\sigma}{\epsilon}$$

5. Oct 3, 2007

### indigojoker

seems like you are using the charge of one plate and the E field due to both the plates.

you would get what I got: $W = \sigma A \frac{\sigma}{2 \epsilon_o} d'$

so why did I have to use the E field of just one plate for what I did to work?

6. Oct 3, 2007

### learningphysics

you won't get that answer if you use the E field for just one plate.

7. Oct 4, 2007

### indigojoker

do the plates are a distance d apart.

to move them 2d apart, we move a distance d.

Using Q = $$\sigma A$$

E = $$\frac{\sigma}{\epsilon}$$

we get the work needed is:
$$W = \sigma A \frac{\sigma}{ \epsilon_o} d'$$

so the reason why we are using Q of one plate is because we are saying that we want to move one plate a distance of d (instead of two plates a distance of d/2), and this plate is being moving though the E-field that is contributing from both plates, so we use $\frac{\sigma}{\epsilon}$

So we actually move it d apart, which is why i used d instead of 2d

do i make sense?

8. Oct 4, 2007

### learningphysics

Ah... I think I'm understanding now... yes, when you use Work = Force * d, to get the force on one plate... you need to use the field due to just the other plate (because the field due to a plate does not create a force on itself...) in other words $$\frac{\sigma}{2\epsilon}$$ is the field needed to get the force on one plate. And the charge is Q (the force on one plate is the field due to the other plate times the charge of the first plate). so the force on one plate is $$Q*\frac{\sigma}{2\epsilon} = \sigma A\frac{\sigma}{2\epsilon}$$

So this way you get that factor of (1/2) in there... just like when you use final energy - initial energy.

So work is force*d = $$\sigma A\frac{\sigma}{2\epsilon}d$$

9. Oct 4, 2007

### indigojoker

okay, now i'm confused, why is $$\frac{\sigma}{2\epsilon}$$ used

I convinced myself that it was $$\frac{\sigma}{\epsilon}$$ because that is the field in which the plates are being moved

10. Oct 4, 2007

### learningphysics

There are two ways to do the problem...

1) Get the final energy of the capacitor and subtract the initial energy of the capacitor.

2) calculate the force need to move the plate. multiply that force by d.

First try the problem using method 1). show your steps.

11. Oct 4, 2007

### indigojoker

why for total energy of both plates are you using the charge for one plate?

12. Oct 4, 2007

### learningphysics

That's just what the energy of a parallel plate capacitor is... The energy of the capacitor is (1/2)CV^2 = Q^2/(2C) where Q is the charge on the positive plate...

the net charge on the capacitor (both plates) is Q + -Q = 0.

13. Oct 4, 2007

### indigojoker

then the energy of the negative charge is -Q^2/(2C)

in such case, the total energy is zero?

14. Oct 5, 2007

### learningphysics

no. that doesn't make sense... then the energy of the capacitor is always 0. I just pointed out that the net charge is always 0 for a parallel plate capacitor... so it doesn't make sense to use that for the enrgy...

All the equations for a parallel plate capacitor... Q = CV, E = (1/2)CV^2, E = Q^2/(2C)

In all these equations Q is the charge on the positive plate...

The energy of the capacitor (the entire capacitor, both plates...) is Q^2/(2C) where Q is the charge on the positive plate...