Suppose we have two point charges in the Cartesian coordinate system. $q_1= e, q_2 = 2e$, where $q_1$ is positioned at $(0,0,0)$ and $q_2$ at $(a,0,0)$ for $a > 0 $. Further there is a point charge $q_3 = -e $ starting at infinty and I wanna now how much work does it need to come to the point $A = (0,a,0)$. I have two different ideas how to approach this:

  1. I know oppositely charged objects attract each other so it will just "fly" there without any help, more I can say it has a higher potential at infinty than at the point A

  2. I know this equation for the potential of a charge in a system of charges $Q_1,\dots,Q_N$ $$\phi(\vec{x}) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^N \frac{Q_i}{\|\vec{x}-\vec{x}_i\|} $$ Considering $$\phi(\vec{x}) = \int_{-\infty}^{\vec{x}} \vec{E}(\vec{x}) d\vec{x}$$ these two equations I should get the desired result. (I wonder why this equation doesn't depend on my test charge.)

My question is which approach is the correct one? Where are my mistakes?

EDIT: As pointed out in the comments i calculate the potential: $$\phi(\vec{x}) = \frac{e}{4\pi \epsilon_0 a} (1+\sqrt{2})$$ I still wonder why approach one isn't true and why the potential doesnt depend on my test charge, though i think i can read the second in a book and will try to do so.

This is my first physics related question on English so please tell me if there are obscurities.

  • $\begingroup$ finding the potential is correct. i believe your question basically wanted to know the work that needed to be done to bring the test charge to hte required point quasistatically. Otherwise , the charge would accelerate to an infinite velocity on its way to the designated point. $\endgroup$
    – Lelouch
    Jul 19, 2016 at 8:53

1 Answer 1


Those are correct. You don't need the second equation, the first equation is constructed from the second, with $\phi = 0$ at infinity.
You can use the work done by the electric force: $$W = q_3(V_a - V_b)$$ Where $V_a = 0$ is the potential at the starting point, and $V_b$ at the end point.
Note it has higher potential at $A$. the potential at infinity is $0$.
Note that this is the work done by the electric force (Not by you), so approach 1 should be true, but you use approach 2 to solve it anyway.
The potential will not depend on the test charge because the test charge cannot have any force due it's own electric field.


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