Potential energy of a capacitor

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Homework Help Overview

The discussion revolves around the potential energy of a capacitor, specifically examining the effects of doubling the length of the capacitor on its potential energy. The original poster presents a formula for potential energy and raises questions about the relationship between the capacitor's dimensions and its energy characteristics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of changing the length of the capacitor and its effect on capacitance and potential energy. Questions arise regarding the definition of "length of a capacitor" and how moving the capacitor affects its potential energy. Some participants seek clarification on the relevant formulas and concepts, such as the relationship between force, capacitance, and potential energy.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some have provided insights into the formulas related to capacitance and force, while others express uncertainty about the concepts involved. There is no explicit consensus yet, but guidance has been offered regarding the need to clarify the force involved in the scenario.

Contextual Notes

Participants note potential confusion regarding the definitions and relationships between the variables involved, particularly concerning the physical setup of the capacitor and the implications of moving its plates. The original poster's attempt to connect two approaches to the problem is acknowledged, but there is a lack of clarity on how to proceed with the second part of the question.

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 = (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?
 

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