# Work done assembling a system of charges?

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1. Feb 18, 2015

### x86

This is a concept that has long since plagued me. I will quote an article off of Google about the concept I am confused about:

"Consider a collection of static point charges located at position vectors (where runs from 1 to ). What is the electrostatic energy stored in such a collection? Another way of asking this is, how much work would we have to do in order to assemble the charges, starting from an initial state in which they are all at rest and very widely separated? " [http://farside.ph.utexas.edu/teaching/em/lectures/node56.html]

So, I understand the general concept. But I am confused in one of the cases. The case where the work that we do is negative (i.e. the work that we have to do to assemble the charge is < 0).

Here is an example that I will use to illustrate my confusion:

Say that we have two charges q in space, one positive, and one negative, and want to move them close to each other, to a distance of r.

The answerof course, is that the work that we need to do to bring the negative charge to the positive charge is then W = -kq^2/r < 0. I know this is the answer, and I know how to solve these problems. But I have no idea why this is true.

I am confused about the concept. I will try to explain what I am confused about below:

It is known that W = Fd. In this case , d = (infinity - r)

But I am confused about this, because essentially the charges attract one another. Therefore, we don't really have to apply any force to make them get close to each other, because an electric force will handle this for us. So our applied force F = 0. Then W = 0*d = 0. But of course, this is wrong. Because we know the answer to be W = -kq^2/r. But why?

Last edited by a moderator: Apr 15, 2017
2. Feb 18, 2015

It is known that W=F.d for neutral bodies but the case changes totally for charged bodies. We have equation of force F=qE for charged bodies so we can not put F=0 in case of charged bodies.

Last edited by a moderator: Apr 15, 2017
3. Feb 18, 2015

### Orodruin

Staff Emeritus
That the work required to assemble the charges is negative simply means that you would extract work from doing so. Unless you counter the electric force with a force in the other direction, the final configuration will involve moving charges, which simply means the work done by the electric force has instead gone into kinetic energy of the constituents.

4. Feb 18, 2015

### Orodruin

Staff Emeritus
There is absolutely nothing special about neutral bodies in the work-energy theorem. The only relevant issue is which force you consider and the path taken, which will determine if the force you considered did work or not.

5. Feb 18, 2015

### x86

Ah, so by doing negative work I'm essentially slowing the particles to a stop, at this snapshot in time?