Work required to bring charge from infinity

[henry3369]

Homework Statement

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.

Homework Equations

W = -ΔU

The Attempt at a Solution

W = -(U_final - U_initial)
U_initial = 0 because it is at infinity.
So W = -U_final = -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?

 

[H Smith 94]

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?

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 $$\Delta U = U_\text{final}-U_\text{initial}.$$

 

[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.

 

[ehild]

Homework Statement

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.
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?

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.