Determining relative amounts of work done

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The discussion centers on understanding why the work done in moving a charge in an electric field is independent of the path taken. It is clarified that the electric field is conservative, meaning work depends only on the initial and final positions, not the path. The integral of E dot dl diverges near point charges, making certain paths invalid for calculation. Ultimately, the potential difference between points R and S is determined to be zero, leading to the conclusion that no work is done in moving the charge. The participants express gratitude for the insights shared, solidifying their understanding of the concept.
RoboNerd
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Homework Statement


workquestion.jpg

Homework Equations



Work = - (integral of) (E dot dl)

The Attempt at a Solution



Hi,

I know that the right answer is D zero, but I fail to understand why. I said that the answer was A as I have different charges, and I thought that depending on how I approach each individual charge by taking either the left-hand path or the right-hand path that my integral of E dot dl would be different resulting in different amounts of work done.

However, the solutions say that the amount of work done is the same regardless of path taken.
Could anyone please explain to me why this is the case (with the electric field being conservative)?

Thanks in advance for being willing to help.
 
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One of the definitions of a conservative force is that the work done by it is independent of the path taken. So the answer to your question is "by definition". Gravity is another conservative force. The work done by gravity when you move from one point to another depends on the height difference, not on how you go from one point to another. With electrostatics replace "height difference" with "potential difference". It's the same idea.

For this question, can you write expressions for the electric potential at points R and S? If these expressions are the same, then the potential difference is zero and the work done by the electric force will also be zero.
 
RoboNerd said:
the right answer is D zero
Do you mean C zero?
 
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RoboNerd said:
I thought that depending on how I approach each individual charge by taking either the left-hand path or the right-hand path that my integral of E dot dl would be different resulting in different amounts of work done.
The four lines labelled with length d are not intended to denote paths. They are intended to denote distances. You cannot integrate E dot dl over any of those lines because they all end up on point charges. The integrals all diverge (i.e. they are all infinite).

If you ignore the fact that the integrals are not well defined, it is clear by symmetry that the work done over a path that approaches the +Q charge from R is equal and opposite to that done on a path that recedes from the +Q charge toward S. They add to zero. Similarly for the work done on the paths approaching and receding from the +2Q charge.
 
The problem is asking for the work "required to move a charge -Q from point R to point S". One can chose a path that does not bring charge -Q near the given charges, in which case all integrals are well-defined. The work done by the electrical force along any path from R to S depends only on the end points because the electrical force is conservative. This work is the negative of the change in potential energy. The latter is equal to (-Q)ΔV so the question is what is ΔV between points R and S? If it's zero, there is no need to go further. If it's not zero, the work required to move charge -Q (presumably by an external agent) is the negative of the work done by the electrical force, i.e. -(-Q)ΔV = +QΔV.
 
haruspex said:
Do you mean C zero?
Yes, I actually did. :-) Thanks for catching that
 
jbriggs444 said:
The four lines labelled with length d are not intended to denote paths. They are intended to denote distances. You cannot integrate E dot dl over any of those lines because they all end up on point charges. The integrals all diverge (i.e. they are all infinite).
That is a very good point. I did not even think of that. Thanks for bringing that to my attention!

Thank you for bringing up very interesting points to my attention. I have determined that the potential differences are zero, and thus, no work is done.

Thank you again everyone for the help!
 

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