13.9 Electrical potential due to a point charge  (Page 3/2)

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.

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 [link] .) What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

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

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

Solution

Solving for $Q$ and entering known values gives

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

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.