19.3 Electrical Potential Due to a Point Charge

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge $q$ from a large distance away to a distance of $r$ from a point charge $Q$, and noting the connection between work and potential $\left(W=–q\mathrm{\Delta }V\right)$, it can be shown that the electric potential $V$ of a point charge is

where k is a constant equal to $9.0\times {\text{10}}^{\text{9}}\text{N}\text{·}{\text{m}}^{\text{2}}\text{/}{\text{C}}^{\text{2}}$.

The potential at infinity is chosen to be zero. Thus $V$ for a point charge decreases with distance, whereas $\mathbf{\text{E}}$ for a point charge decreases with distance squared:

Recall that the electric potential $V$ is a scalar and has no direction, whereas the electric field $\mathbf{\text{E}}$ is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that $V$ is closely associated with energy, a scalar, whereas $\mathbf{\text{E}}$ is closely associated with force, a vector.

Example 19.7

What Is the Excess Charge on a Van de Graaff Generator

A demonstration Van de Graaff generator has a 25.0 cm diameter metal sphere that produces a voltage of 100 kV near its surface. (See Figure 19.11.) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

Figure 19.11 The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.

Strategy

The potential on the surface will be the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using the equation

$$V=\frac{\text{kQ}}{r}.$$

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Solution

Solving for $Q$ and entering known values gives

$$\begin{array}{lll}Q& =& \frac{\text{rV}}{k}\\ & =& \frac{\left(0\text{.}\text{125}\text{m}\right)\left(\text{100}\times {\text{10}}^3\text{V}\right)}{8.99\times {\text{10}}^9\text{N}·{\text{m}}^2/{\text{C}}^2}\\ & =& \text{1.39}\times {\text{10}}^{\mathrm{–6}}\text{C}=\text{1.39 µC.}\end{array}$$

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Discussion

This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted in Electric Potential Energy: Potential Difference, this is analogous to taking sea level as $h=0$ when considering gravitational potential energy, ${\text{PE}}_{\mathrm{g}}=\text{mgh}$.