Is Work Required to Move Capacitor Plates Apart?

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Discussion Overview

The discussion revolves around whether work is required to move the plates of a capacitor apart while maintaining a constant potential difference. Participants explore the implications of mechanical work, electrical forces, and energy changes in the context of capacitors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that moving the plates apart while maintaining the potential difference may not require additional work, as they believe no work is done against the potential field.
  • Others argue that the attractive force between the plates due to the electric field must be considered, which implies that work is needed to separate the plates.
  • It is noted that as the plates are pulled apart, the capacitance decreases, and if a battery is connected to maintain the voltage, energy must be supplied to the system.
  • Some participants express uncertainty about the relationship between capacitance and distance, with one noting that doubling the distance would halve the capacitance.
  • There are discussions about the equations governing energy stored in capacitors and how to calculate changes in capacitance and work done.
  • Participants explore different formulations for calculating electrical work and mechanical work, with various equations being proposed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether work is required to move the plates apart, as multiple competing views remain regarding the definitions of work and the implications of maintaining a constant potential difference.

Contextual Notes

There are unresolved assumptions regarding the definitions of mechanical work and electrical work, as well as the dependence on specific conditions such as the connection to a battery and the nature of the dielectric material.

Physgeek64
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If I have a capacitor and I maintain the potential difference between the two plates as I move them apart would I have to do any work (other than mechanical work) to move them apart? my logic tells be I shouldn't have to as I'm not doing any work again the potential field? Just wondered what you guys thought?
 
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That depends on what you include in "mechanical work".

Besides gravitational force, the plates attract each other, due to the electric field between the plates.

The latter force can be calculated from: Eelec = ½*V2*C , V is constant here.

When you pull the plates apart, C will be decreased. Now say that V is maintained by some connected battery, the battery will be charged ( supplied by energy ).

Now calculate the total change in energy in the capacitor+battery. You must add this energy to the system, when you take apart the plates.
 
Last edited:
Hesch said:
That depends on what you include in "mechanical work".

Besides gravitational force, the plates attract each other, due to the electric field between the plates.

The latter force can be calculated from: Eelec = ½*V2*C , V is constant here.

When you pull the plates apart, C will be decreased. Now say that V is maintained by some connected battery, the battery will be charged ( supplied by energy ).

Now calculate the total change in energy in the capacitor+battery. You must add this energy to the system, when you take apart the plates.

Ahh yes, I forgot that charge must be added to keep the potential constant, and I don't think the capacitance would be varied? But adding this charge requires work to be done. Silly me. Thank you :)
 
BvU said:
Oops: there is a formula for this capacitance
Ahh okay! But would this not best be explained through the addition of charge? I accept that thee capacitance will change, but for calculation purposes?
 
REad on in the link: there is even an equation for the energy stored in a capacitor !
 
BvU said:
REad on in the link: there is even an equation for the energy stored in a capacitor !
I saw- I get that this is the energy stored on the capacitor, but suppose we don't know the dielectric strength, and V is kept constant. E= 1/2 dC V^2, but how would you go about finding dC (i.e. the change in capacitance). Sorry if I'm being really dumb and missing something extremely obvious
 
Correction: ##E = {1\over 2} CV^2##
Suppose you double the distance ##d##. What happens to C ?
You maintain V, and you have an expression for E before and after changing ##d##. The difference is the work done -- mechanical plus electrical.
How would you go about to calculate one of the two -- and thereby determine the other as well ?
 
BvU said:
Correction: ##E = {1\over 2} CV^2##
Suppose you double the distance ##d##. What happens to C ?
You maintain V, and you have an expression for E before and after changing ##d##. The difference is the work done -- mechanical plus electrical.
How would you go about to calculate one of the two -- and thereby determine the other as well ?
If you double the distance then c would half. Okay I think I've got it. Thank you.

From this would it be possible to calculate the force per unit area. i.e. would the use of F=-dW/dx be valid ?
 
  • #10
That's the mechanical one. Before embarking on that: what is the electrical work ?
 
  • #11
BvU said:
That's the mechanical one. Before embarking on that: what is the electrical work ?

W=Q integral E.dr

E= sigma/ epsilon naught?

W= Q sigma / epsilon-naught dx ?
 
  • #12
I'd say ##W = \int VI\;dt = V \int I\;dt = V\Delta Q ## (since V is constant). Much easier ...
 

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