Potential energy of a capacitor

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SUMMARY

The potential energy of a capacitor, defined by the formula U = 1/(2C) * Q, is directly affected by the distance d between its plates. When the length d is doubled, the capacitance C is halved, resulting in a doubling of the potential energy U. To analyze the work done in moving the capacitor plates, one must apply Coulomb's law to determine the force F acting on the charges. Understanding the relationship between capacitance, electric field, and force is crucial for solving the problem.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and potential energy.
  • Familiarity with Coulomb's law and electric fields in electrostatics.
  • Knowledge of the relationship between force, work, and distance in physics.
  • Basic grasp of parallel-plate capacitor equations, including C = (εr*ε0*A/d).
NEXT STEPS
  • Study the derivation of the capacitance formula for parallel-plate capacitors.
  • Learn how to calculate electric fields within capacitors and their implications on potential energy.
  • Explore the relationship between force and potential energy in electrostatics.
  • Investigate the implications of changing plate distance on capacitance and energy storage.
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone seeking to deepen their understanding of capacitor behavior and electrostatics.

silenzer
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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 = (εr*ε0*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?
 

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