# Electric potentials and superposition

I had a question regarding the addition of electric potentials. Consider two positively charged particles $q_1$ and $q_2$ at distance $R$ apart. Let the charges have magnitudes $q_1$ and $q_2$. For a moment, let me remove $q_2$ and calculate the potential a small distance $r$ from $q_1$. Now, let me put $q_2$ back and calculate the potential at the same point ($q_1$, $q_2$, $r$ lie on the same straight line and $r$ is between $q_1$ and $q_2$). When is the potential greater and why? I did get the explanation that because the potential is scalar you just add individual potentials and hence it's greater when both the charges are present, but I didn't understand this thoroughly enough. When I looked at it, I saw that when both charges are present, the force on a test charge at $r$ is smaller in magnitude (because the test charge experiences two forces in opposite directions) and so the potential (which is integral $F$.$dr$) would be smaller than if only $q_1$ were present. But then again, the question was from where to where do I integrate, from infinity to $r$ or from the point between $q_1$ and $q_2$ where the net force is zero (and hence potential is 0) to $r$.

• Don't forget the constant of integration! – Anurag B. Apr 27 '18 at 12:03

What you must remember is that you are evaluating $\int E\,dr$ all the way from infinity to the point midway between the two charges.