12.4.2 Electric Potential of a Point Charge

Consider two point charges: a source charge Q and a test charge q. We can obtain the EPE between these two point charges using the formula $\displaystyle k\frac{{Qq}}{r}$ . We can also obtain the EPE by using the formula $\displaystyle EPE=qV$ , where V is the electric potential (of the electric field) of Q at q’s position.

Since V is the EPE per unit charge, we can derive the electric potential of a source charge Q as

$\displaystyle \begin{aligned}V&=\frac{{EPE}}{q}\\&=k\frac{{Qq}}{r}\div q\\&=k\frac{Q}{r}\end{aligned}$

To include the sign or not?

When applying this formula, you should include the sign of the charge. So the formula will automatically compute a positive charge to produce positive potential around it, whereas a negative charge produces negative potential around it.

If we plot the variation of V with distance x from a point charge, we get these $\displaystyle y=\frac{k}{x}$ kind of graphs that approach zero at infinity.

If you think that the potential graphs resemble a mountain centred about a positive source charge, and a crater centred about a negative source charge, you’re quite right. Remember I said that you can think of electric potential as an indication of “electric height”? So it makes sense that a positive source charge raises an electric mountain around it, causing positive test charges to “roll down the slope” away from it. And a negative source charge produces a crater around it, causing positive test charges to “roll downslope” into it.

Besides field lines, we can also draw lines connecting points of the same V to represent an electric field. These lines are called the equipotential lines. For point charges, they are concentric circles which are spaced further and further apart (shown in green below). If you think that they look similar to contour lines on topography maps, you are spot on. As I said before, electric potential is kind of like “electric elevation”.

Example

Q_A and Q_B are point charges of +1.1 pC and -2.2 pC bound at vertices A and B of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate

a) the electric potential at point C

b) the work done (against the electric field) to move
i) a proton
ii) an electron
from C to another point where the electric potential is −1000 V.

Solution

Q_A and Q_B each individually produces electric potential at point C.

$\displaystyle {{V}_{A}}=k\frac{{{{Q}_{A}}}}{r}=(8.99\times {{10}^{9}})\frac{{1.1\times {{{10}}^{{-12}}}}}{{3.0\times {{{10}}^{{-6}}}}}=3296\text{ V}$

$\displaystyle {{V}_{B}}=k\frac{{{{Q}_{B}}}}{r}=(8.99\times {{10}^{9}})\frac{{-2.2\times {{{10}}^{{-12}}}}}{{3.0\times {{{10}}^{{-6}}}}}=-6593\text{ V}$