# Potential energy of a capacitor

• silenzer
It is written as E = -(μ*σ*ε), where μ is the charge per unit area, σ is the surface charge density, and ε is the permittivity of dielectric.I think you're right, I do need to remember that equation. So, once you have F, how do you find Q?Once you have F, you can find Q by multiplying it by the area of the capacitor.

## Homework Statement

What happens to the potential energy of a capacitor with length d when said length d is doubled? Use the formula for the potential energy of a capacitor [U = 1/(2C) * Q]. After that, refer to the work done by moving the capacitor the distance d. These two approaches should yield the same result!

## Homework Equations

U = 1/(2C) * Q, and, I guess, the traditional formula for work, W = F * d.

## The Attempt at a Solution

I know how to solve the first part, because C is halved when d is doubled, so the potential energy is doubled. But I have absolutely no clue as to how to even begin working on the second part. Any tips?

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What does the "length of a capacitor" mean? How does moving a capacitor change its potential energy which apparently depends on Q and C only? I'm not a EE, but I don't understand this problem.

Sorry, d is the length of the space between the capacitor.

So are you really asking about moving one plate of the capacitor a distance d, thus changing its potential energy? If so, what formulas do you have relating C, d, Q, and F? I'm asking because I don't know them and I think you need them to work the problem.

silenzer said:
and, I guess, the traditional formula for work, W = F * d.
...
But I have absolutely no clue as to how to even begin working on the second part. Any tips?
yeah, that's the right place to start for the second part. What is the force in this case? p.s. sorry for jumping in.

LCKurtz said:
So are you really asking about moving one plate of the capacitor a distance d, thus changing its potential energy? If so, what formulas do you have relating C, d, Q, and F? I'm asking because I don't know them and I think you need them to work the problem.

Well, we have F according to Coulomb's law, and U = 1/(2C) * Q, and then there is C = (εr0*A/d, where A is the area of the capacitor and d is the aforementioned length.

The trick, I think, I think, is to figure out exactly what F is. If we find out F, the rest should be easy.

yep. And true, it is electrostatics, so it is technically Coulomb's law for each charge. But you have many charges. Do you remember the equation for the electric field inside a parallel-plate capacitor?