Calculating Work on Equipotential Surfaces

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Homework Help Overview

The discussion revolves around calculating the work done in moving a charge between two equipotential surfaces at different voltages. The subject area is electrostatics, specifically focusing on the concept of work in electric fields.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind the work being zero when moving a charge between equipotential surfaces, questioning the implications of voltage differences and the nature of work in electric fields.

Discussion Status

Several participants express agreement with the idea that the work done is zero, discussing the reasoning involving equal magnitudes of work with opposite signs during the charge's movement. There is an ongoing exploration of the underlying concepts without a definitive conclusion.

Contextual Notes

Participants note the potential confusion arising from multiple-choice questions, indicating a concern about the clarity of the problem setup and the assumptions being made regarding work and voltage.

threewingedfury
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The work in joules required to carry a 6.0 C charge from a 5.0 V equipotential surface to a 6.0V equipotential surface and back again to the 5.0V surface is:

A) 0
B) 1.2 X 10^-5
C) 3.0 X 10^-5
D) 6.0 X 10^-5
E) 6.0X10^-6

I was thinkin the work is 0, but then again that seems too easy
 
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Do you have a reason for thinking it's 0 ?
 
Is it 0 because of W=F*d=qEd?? The work done would be equal to 6C(1V)=6J when you bring it from 5V to 6V (6-5=1) and the work done would be equal to 6C(-1V)=-6J when it is going down the voltage from 6V to 5V (5-6=-1). Therefore when you add 6J and -6J you get 0? Is my solution correct?
 
I can't argue with that.
 
thats what I was thinking, I just didn't know because mult choice questions are so tricky!

thanks!
 
Zero..Work is of equal manitude but of opposite signs..they cancel to make net work 00000
 

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