Work and Electric Potential

[MedPR]

---

Members do not see this ad.

In regard to electric potentials.

Two charges Q1=Q2=1.1*10^-8C are 2*10^-2 meters apart. Point A is exactly between them, and point B is 10^-1 meters from both of them. How much work is required to move a charge q=10^-9C from A to B?

Va=kQ/da + kQ/da
Va=2*10^4 Nm/C = 2*10^4 V

Vb = kQ/db +kQ/db
Vb= 2*10^3 V

W=q(Vb-Va)
W= -1.8*10^-5 J

It makes sense to me why work is negative, considering the charges Q are both repelling A, but in opposite directions so those forces cancel out. Since point B is far away, the potential is less at point B, and no work is needed to move a charge from higher to lower potential (right?).

However, what does the negative work indicate? Is it just the amount of work necessary to move charge A back to "exactly between" Q1 and Q2 from point B?

The explanation (from NOVA) is this:

Since the test charge is positive, we imagine the two charges Q are mountains next to each other, and A is a mountain pass. Point B is further down, so the energy change is negative. No energy is required, but instead energy can be derived. Therefore the negative answer is justified.

I know the answer to my question is the second to last sentence, but I don't understand what it is saying.

 

[milski]

Work is the difference between the energy in the two states. Negative work in that case means that you gained energy. You lost some potential energy by moving the charge further away and that energy has to be accounted somehow. Negative work is the way to do that.

I just read what I wrote and sounds awfully confusing. So let's try again. When the charge was moved further away it lost some potential energy. That's the same as saying that negative work was done on it. Of course, that energy has to go somewhere - it will be transferred to whatever you used to stop the charge in that place. Could be heat, if you used friction, some other form of potential energy, like a spring, that's not so interesting. The important part is that the energy of charge decreased and that negative difference is the work done on the charge in this case.