Work equation for electrostatics and electricity

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SUMMARY

The discussion centers on the equations for calculating work in electrostatics, specifically in the context of moving a charge within a parallel plate capacitor. Two equations are presented: W=qV and W=-PE. The example provided involves a +3 charge being moved from the negative to the positive plate of a capacitor connected to a 10 V battery, yielding results of 30 J and -30 J, respectively. The ambiguity in the problem statement regarding who is performing the work—either the system or the individual—is highlighted as crucial for determining the correct interpretation of the work done.

PREREQUISITES
  • Understanding of electrostatics principles
  • Familiarity with the concept of electric potential (V)
  • Knowledge of potential energy (PE) in electric fields
  • Basic grasp of vector forces and work calculations
NEXT STEPS
  • Study the concept of electric potential energy in capacitors
  • Learn how to apply the work-energy principle in electrostatics
  • Explore the differences between work done by external forces versus work done by the system
  • Investigate the implications of force direction on work calculations in electrostatics
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Students and educators in physics, electrical engineers, and anyone interested in understanding the principles of work in electrostatics and electricity.

Benny851
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I have a question about the equation for work as it pertains to electrostatics and electricity. One book I read says equation for work done by electric field is W=qV, but another book says W=-PE. However, when I try both equations for the following problem I get different answer. Could someone please explain to me which equation I should use and why? Thanks.

Parallel plate capacitor with 10 V battery. +3 charge is on negative plate and you need to move it to positive plate. calculate work.

w=-pe -> -(3*10) = -30

w=qv -> 3*10= 30
 
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Benny851 said:
I have a question about the equation for work as it pertains to electrostatics and electricity. One book I read says equation for work done by electric field is W=qV, but another book says W=-PE. However, when I try both equations for the following problem I get different answer. Could someone please explain to me which equation I should use and why? Thanks.

Parallel plate capacitor with 10 V battery. +3 charge is on negative plate and you need to move it to positive plate. calculate work.

w=-pe -> -(3*10) = -30

w=qv -> 3*10= 30

Something I like to do is to put myself into the problem.

Imagine for a moment that you are able to shrink yourself down to the size of the charge and physically push it. If the force that you apply on the charge is in the same direction as the charge's displacement, then it can be said that "you have done positive work on the system."

On the other hand, if you are pushing the charge forward, but the charge ends up going backward such that the force that you exert is in the opposite direction as the charge's displacement, then it can be said that "you have done negative work on the system." This is equivalent to saying "the system has done positive work on you."

So the correct answer involves who is doing work on what.

In this problem there are two forces involved: the force that you exert on the system and the equal and opposite force that the system exerts on you (or exerts on the charge if you were not there to keep it from accelerating).

Assuming a constant force, one can define the work as
[tex]W = \vec F \cdot \vec{\Delta x}[/tex]
If [itex]\vec F[/itex] and [itex]\vec{\Delta x}[/itex] are in the same direction the work is positive, if they are in opposite directions, the work is negative. Of course the answer depends on which force you are talking about!

So let's revisit this problem as it was worded:

Parallel plate capacitor with 10 V battery. +3 charge is on negative plate and you need to move it to positive plate. calculate work.

The question, as it was worded, is sort of ambiguous. It's not clear if its talking about the work done by the system or the work done by you. But if I had to guess, I would say it wants you give the work done by you, on the system, when you move the charge from the negative plate to the positive plate.
 
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