$$V=k{q_s\over d}+V_{\infty}$$

where \(V_{\infty}\) is an arbitrary constant. That means that, if we place a target charge \(q_t\) at the star, then the system will have a potential energy of
$$\begin{eqnarray}
P&=&q_tV\\
&=&q_t\left(k{q_s\over d}+V_{\infty}\right)\\
&=&k{q_sq_t\over d}+q_tV_\infty\\
\end{eqnarray}$$
which is as we calculated earlier. (\(q_tV_\infty=P_\infty\))
Just as with potential energy, the formula for potential has an arbitrary constant \(V_\infty\) is the potential very far away from the source. What this means in practice is that we can choose the potential at *any one point* to be whatever we want it to be, and the other potentials will all be defined in comparison with that point. In this figure, four different people choose different values for the potential at the star, and find different potentials at the square. What they all agree on is that the potential at the square is 2 volts higher than the potential at the star, and that's the only thing that matters, physically speaking.

It's good to notice what *isn't* significant. We're used to thinking of "zero" as meaning the absence of something: if a car has zero velocity it's not moving, if a charge feels zero force then it's not being pushed, etc. But in the figure, Ms. Black thinks that the potential at the star is zero, while Ms. Blue thinks the square is at zero potential, and they're both right: clearly, it doesn't actually matter where the potential is zero. Similarly, it's meaningless to say that the potential at the star is "three times as big" as the potential at the square, as Ms. Purple thinks; no one else agrees with her. Some folks think the potential at the square is positive, some negative. In short, the *only* thing that matters, when it comes to potential, are the differences in potential, or how one potential compares to another.

Consider a negative source charge \(q_s=-7\u{nC}\). Find the electric potential 2 meters away, if...

(a) the potential at infinity is zero

According to the formula,
$$V=k{q_s\over d}+V_\infty=(9\ten9){-7\ten{-9}\over 2}+0=-31.5\u{V}$$

(b) the potential at infinity is 50V

When we raise the potential at infinity by 50 volts, every other potential is raised by 50 volts as well. Thus the potential is
$$-31.5\u{V}+50\u{V}=18.5\u{V}$$
Notice that the negative charge is creating a positive potential now! That's because "positive potential" is a meaningless concept.

(c) the potential *one meter* away from the charge is 50V.

Since we know the potential one meter away, we have
$$50=(9\ten9){-7\ten{-9}\over 1}+V_\infty$$
$$\implies V_\infty=50+63=113\u{V}$$
Knowing \(V_\infty\), we can then find the potential 2 meters away:
$$V=k{q_s\over d}+V_\infty=(9\ten9){-7\ten{-9}\over 2}+113\u{V}=81.5\u{V}$$
Notice that the potential twice as far away is *not* half the potential, or a quarter, or any even multiple. Multiples don't matter in potentials!

If you have more than one source, the potential at a location is additive like the electric field, but you only add \(V_{\infty}\) once at the end, not once for each source charge.

$$V=k{q_{s1}\over d_1}+k{q_{s2}\over d_2}+k{q_{s3}\over d_3}+\dots+V_{\infty}$$

Note that potentials are just numbers, not vectors: directions don't count! Also note that the distances \(d_1\), \(d_2\), etc are never negative. The only way a minus sign gets into a potential equation is through a charge.
Find the electric potential at the star, if \(V_\infty=-100\u{V}\).

$$\begin{eqnarray}
V&=&k{q_{s1}\over d_1}+k{q_{s2}\over d_2}+k{q_{s3}\over d_3}+\dots+V_{\infty}\\
&=&(9\ten9){5\ten{-9}\over 0.1}+(9\ten9){3\ten{-9}\over 0.1414}\\
&&+(9\ten9){-2\ten{-9}\over 0.1}-100\\
&=&450\u{V}+191\u{V}-180\u{V}-100\u{V}\\
&=&\boxed{361\u{V}}\\
\end{eqnarray}$$