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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 $U_a = 0$?

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3 Answers 3

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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

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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 $-U_{b}$.

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  • $\begingroup$ I think you are mixing formula for work of field force and work of external force. There is a difference in sign. $\endgroup$
    – tobix10
    Commented Oct 12, 2015 at 20:41
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Simple Answer:

Since x=2a is closer to the +ve charge, it must be of higher potential. so, you (external force) must do work to bring a +ve charge from infinity overcoming the repulsion of field. Hence, it is a +ve work.

To summarize, In this case:

  • External Force does +ve work
  • Electric Field does -ve work
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