# Work is positive/negative potential energy

1. Oct 21, 2015

### Calpalned

I understand that $\Delta U = -W$ and that $\Delta V = \frac {\Delta U}{q}$
I don't understand why the example below has "set the change in potential energy equal to the (positive of the) work ...

2. Oct 21, 2015

### davidbenari

All this has to do with the notions of the work you have to do and the work which is done by the electric field. Maybe the second type is unfamiliar to you. The work you do is $\Delta U$ the work done by the field is $-\Delta U$

To take a more terrestrial example you can imagine a system composed of the sun and the earth . You know the work that the gravitational field does is calculated via $W=-\Delta U$. However the work that you have to do will be $\Delta U$ since you are opposing the gravitational field. If the field does x and amount of work you do -x. All this adds up to 0 and gives you motion with 0 kinetic energy all through out, therefore its the "minimum work".

Another way to look at this for this particular example is this way:

Your test charge has 0 potential energy at first. At the final stage it has $qV_{b}$. Where did the energy come from? You put it there. Therefore this way of choosing signs helps you identify you are the one depositing energy.