# Homework Help: Work required to bring charge from infinity

1. Apr 16, 2015

### henry3369

1. The problem statement, all variables and given/known data
Two point charges are located on the x-axis, q1 = -e at x=0 and q2 = +e at x=a. Find the work that must be done by an external force to bring a third point charge q3 = +e from infinity to x = 2a.

2. Relevant equations
W = -ΔU

3. The attempt at a solution
W = -(Ufinal - Uinitial)
Uinitial = 0 because it is at infinity.
So W = -Ufinal = -kq3(q1/r1 + q2/r2). After doing this I get:
-(e^2)/(8πεa) which has the correct magnitude but wrong sign. My book uses W = U instead of W = -U. Is there a reason for this? Work is equal to change in NEGATIVE potential energy, so why would it be positive?

2. Apr 16, 2015

### H Smith 94

This is a issue with conventions stemming from the fact you've used a negative Coulomb potential. By convention the Coulomb force $\vec{F}(\vec{r})$ is negative to imply attraction: $$\vec{F}(\vec{r}) = -\frac{q_1q_2}{4\pi\epsilon_0}\frac{1}{r}\cdot \hat{r}$$ which gives rise to the Coulomb (or electrostatic) potential $U(\vec{r})$ found by, $$U(r) = \int F(r) \, dr$$ and so $$U(r) = -\frac{q_1q_2}{4\pi\epsilon_0} \int \frac{1}{r^2} dr = -\frac{q_1q_2}{4\pi\epsilon_0} \left( - \frac{1}{r} \right) = \frac{q_1q_2}{4\pi\epsilon_0} \frac{1}{r}.$$
Either that, or $$\Delta U = U_\text{final}-U_\text{initial}.$$

3. Apr 17, 2015

### rude man

You know from the basic charge layout that it will take positive work to bring the positive charge to x=2a. Try to think that way in general instead of relying on the math to give you the correct signs.

4. Apr 17, 2015

### ehild

The work of the electric field is equal to change in negative potential energy, but the work that must be done by an external force is opposite.