# Theory Question about Electric Potential Energy

1. Feb 20, 2010

I'm having a little bit of trouble understanding the concept that the potential electrical energy between two opposite charges is negative while between like charges it is positive.
Can someone please explain in detail why this is so? Thanks in advance.

2. Feb 20, 2010

### Stonebridge

I think you need to reformulate your question as it doesn't tie in with the facts.
The electrical potential energy (of a unit positive charge, which is how PE is defined here) between two like charges is positive if the charges are positive, and negative if the charges are negative.
Between two opposite charges it can be either positive or negative depending on where you are in the field in relation to the two charges.
The definition of electrical potential energy at a point is in terms of the work done bringing a unit positive charge from infinity to that point.

3. Feb 20, 2010

### saunderson

In generally the work expended against a force field is positive.
The work expended towards a force field is negative.

With this arbitrary definition we obtain the result you have stated.

So if you have a charge $$q_1>0$$ at the origin and a charge $$q_2<0$$ at $$r_2$$ then the work you must expend on $$q_2$$ to pull the charge from infinity to $$r_2$$ is

$$W = V(r_2) = - q_2 \, \int \limits_{\infty}^{r_2} \mathrm{d} \vec r ~ \vec E_1(r) = q_1 \,\int \limits_{\infty}^{r_2} \mathrm{d} \vec r ~ \vec \nabla \phi_1(r) = q_1 \Bigl[\phi_1(r_2) - \phi_1(\infty) \Bigr] = q_2 q_1 \frac{1}{4\pi \varepsilon_0 r_2}$$​

The fact that $$W<0$$ (with the above definition of $$q_2, q_1$$) shows, that you have to expend the work towards the force field to bring the charge $$q_2$$ from infinty to $$r_2$$ (the force between opposite charges is attractive).

4. Feb 20, 2010

### saunderson

I don't agree with you! Compare electric potential with electric potential energy. Because two like charges are always repulsing each other the electric potential energy between them is always positive (cause you have to expend work against the force field to get one charge from infinity to any position).

5. Feb 20, 2010

### Stonebridge

Not if the charges are negative.
Electrical potential refers to the potential energy of a unit POSITIVE charge.

6. Feb 20, 2010

### saunderson

$$V(r) = E - T ~ \ne ~ q_{+} \, \phi(r) \qquad \mbox{with} ~ q_{+} ~ \mbox{as unit POSITIVE charge}$$​
where $$T$$ is the kinetic energy of the particle and $$E$$ the total energy.