Electric potentials and superposition

Question

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 $r$.

Can someone please help me out?

Answer 1

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.

Answer 2

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