Calculating Work Done in Moving Multiple Charges in a Cube Configuration

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Homework Help Overview

The problem involves calculating the work done in moving multiple charges, specifically eight charges of varying magnitudes, from the corners of a cube configuration to a point far away from each other. The context is rooted in electrostatics, focusing on potential energy and work associated with charge interactions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the potential energy calculations for individual charges and the implications of removing charges sequentially versus all at once. Questions arise regarding the validity of calculating total work done as a simple multiple of individual work.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how to approach the calculation of work done. Some guidance has been offered regarding the nature of work done when assembling versus disassembling the charge configuration, but no consensus has been reached.

Contextual Notes

Participants are grappling with the assumptions about potential energy and the effects of charge interactions as they consider different methods for calculating work. There is an acknowledgment of the complexity involved in integrating potential changes over distance.

cupid.callin
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The question is like this:

8 charges of magnitude q and different sighs are placed at corners of a cube of side a.
Find the work done in taking them far away from each other.

Homework Equations



U = kq1q2/r
W = ΔU

The Attempt at a Solution



First i found out the potential energy of a +q charge

U = -3kq2/a + 3kq2/√2a - Kq2/√3a

and the same comes out to be for -q charge (of course)

So for 1 charge,
W = Uf - Ui
W = 0 - ( -3kq2/a + 3kq2/√2a - Kq2/√3a )

W = 3kq2/a - 3kq2/√2a + Kq2/√3a

So work done for 8 charges = 8W ... Right?

But its wrong.

Some help please.
 

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Each time you remove a charge from the group you get less work than from the last one because there are fewer and fewer charges contributing to the field.

The same thing happens in reverse, too, when you assemble charges. The first one is free -- no competing charge to work against or be attracted to. The next one sees the first charge, so there's work to be done. The next one after that sees both of the first two, and so on.
 
well what if i take all of them together?

initial U for all is same.

and obviously we won't consider value of U in b/w the process

...

So then shouldn't work done be 8W?
 
cupid.callin said:
well what if i take all of them together?

initial U for all is same.

and obviously we won't consider value of U in b/w the process

...

So then shouldn't work done be 8W?

The work done (or gained back) to take the arrangement apart will the the negative of the work done to assemble it in the first place. Either way, you need to calculate the sum of all the individual works.

I suppose you could work out how the potential changes when the cube is scaled by some factor, and then integrate from 1 to infinity. But that's going to require even more brain sweat than just calculating the work done assembling the cube one charge at a time.
 
i'll try your way and get back on this asap
 

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