Potential energy of a capacitor

In summary: 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.
  • #1
silenzer
54
0

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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  • #2
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.
 
  • #3
Sorry, d is the length of the space between the capacitor.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 

1. What is potential energy of a capacitor?

The potential energy of a capacitor is the amount of energy stored in the electric field between its two plates when it is charged. It is also known as the energy density of the capacitor.

2. How is the potential energy of a capacitor calculated?

The potential energy of a capacitor can be calculated using the formula U = 1/2 CV^2, where U is the potential energy, C is the capacitance, and V is the voltage across the capacitor.

3. What factors affect the potential energy of a capacitor?

The potential energy of a capacitor is affected by its capacitance, voltage, and the distance between the plates. A larger capacitance or voltage will result in a higher potential energy, while a larger distance between plates will decrease the potential energy.

4. Can the potential energy of a capacitor be negative?

No, the potential energy of a capacitor cannot be negative. It is always a positive value, as it represents the stored energy in the electric field.

5. How is potential energy of a capacitor used in practical applications?

The potential energy of a capacitor is used in a variety of practical applications, such as in electronic circuits, power systems, and energy storage devices. It is also utilized in technologies like cameras, flashlights, and defibrillators.

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