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! Commented Apr 27, 2018 at 12:03

Just for illustration assume that both positive charges are equal the potential is zero at infinity, and you are looking at the electric field and potential midway between the two charges. The electric field at that point (zero - neutral point)) when both charges are present is certainly smaller than the electric field when only one of the charges is present.
It is also true that with both charges present the potential at that point is larger than the potential is only one charge is present.

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.
With two charges there may only be a small contribution to the integral when close to the two charges but the electric field further from those two charges is larger than that due to one charge and approximates to the field due a charge of twice that of one charge when sufficiently far away and so the contribution to the integral is larger at larger distances.
Overall the integral for both charges is larger than for just one charge ie the potential is larger when two charges are present.

• 1) why is E larger at midpoint M when both charges are present. when both charges are present E=0 at M, but with only one charge present E is non zero at M. similarly isnt the potential V=0 at M when both charges are present, because a test charge at M will remain there. if only one charge is present then V is non zero at M. so what do you mean? Commented Apr 28, 2018 at 12:44
• 2) also i am a bit confused about the limits of ∫E.dr . why is it from infinity to some point P between the charges. Or is it from M to P , because the potential at M is 0 and hence it can be the reference? i know the potential at P is by definition work done to bring a test charge from infinity to P, but if thats only because potential is 0 at infinity why cant we just measure work to bring it from M to P? Commented Apr 28, 2018 at 12:48
• @sanjay 1) is a horrible typo which I have corrected. I read your question as asking why it was that when the electric field between the two charges decreased the potential increased. Commented Apr 28, 2018 at 13:11

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 q1 were present.

You are confusing two different quantities, force and potential. They are not proportional to each other. Force is related to the electric field (force on a test charge). The electric field is related to the gradient (think slope) of the potential. For any given value of electric field, the slope could be huge or zero, so the force on the test charge tells you nothing about the potential at the point.

The integral to find the potential would be $$\int \vec{E}\cdot\mathrm d \vec{r}$$, not $$\vec{F}$$, and you need to do two integrals, one for each charge, but that's already been done if you use $$\phi=\frac{q}{4\pi\epsilon_0 r}$$ for each charge