# Why isn't the work minus the potential energy when bringing a charge in from infinity?

This is an example in my physics textbook, and there is just one step that I don't understand.

Two point charges are located on the x-axis, $q_1 = -e$ at $x = 0$ and $q_2 = +e$ at $x=a$. Find the work that must be done by an external force to bring a third point charge $q_3 = +e$ from infinity to $x=2a$.

So I understand that $W_{a\rightarrow b} = U_a-U_b$ is the equation we will use to solve the problem, where $U$ at point $a$ is $$U_a = \frac{q_0}{4\pi \epsilon_0}\cdot (\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+\cdots)$$ So all we need to compute is $U_a$, or the potential energy when the third point is infinitely far away, and $U_b$, when it is at position $x=2a$. I understand that $U_a = 0$, but the textbook says that, from this, we conclude that $W=U$, and they determine the answer $W$ by effectively computing $U_b$. My question is: why is it not $-U_b$, if the equation is $W=U_a-U_b$ and $U_a = 0$?

So for this question, you have to keep in mind that at a distance of infinity, the potential energy between two charges is 0 (no matter if it is between 2 of the same charges or two opposite charges)

Also, when two opposite charges get closer, potential energy decreases (-) and when two like charges get closer, potential energy increases (+)

How you would approach this problem would be to add the potential energies of the third point charge in respect to each of the other 2 charges.

So it would be:

Work = (potential energy between q1 and q3) + (potential energy between q2 and q3)

Work = (k*q1*q3)/2a + (k*q2*q3)/a

Note that a negative sign is not needed in the first part of the equation because q1 is negative already. Also k is Coulomb's constant.

By plugging in the known values, the answer can be obtained.

Hope that helped! Jack

You was asked to find external force so the answer in a book is correct. If the question was about the work that electrostatic field is doing then it is $-U_{b}$.

• I think you are mixing formula for work of field force and work of external force. There is a difference in sign. Commented Oct 12, 2015 at 20:41