Point charges, such as electrons and protons, are among the fundamental building blocks of matter. Even when charge is distributed over a spherical conductor—such as a metal sphere—the electric field outside the sphere behaves exactly as if all the charge were concentrated at a single point at the center. This powerful result allows us to treat many real objects as point charges when calculating electric potential and electric field outside them.

Using calculus to compute the work required to bring a small test charge [latex]q[/latex] from very far away (where the potential is defined to be zero) to a distance [latex]r[/latex] from a point charge [latex]Q[/latex], and using the relationship

we find that the electric potential due to a point charge is

where

Several important conceptual points must be emphasized.

First, electric potential [latex]V[/latex] is a scalar quantity. It has magnitude but no direction. In contrast, the electric field [latex]\mathbf{E}[/latex] is a vector quantity. It has both magnitude and direction.

Second, notice how the distance dependence differs between potential and electric field. For a point charge:

while the electric field magnitude is

The electric field decreases with the square of distance, while the potential decreases only linearly with distance. This means that potential “spreads out” more gradually in space than the electric field does.

Third, because potential is a scalar, when multiple point charges are present, we simply add their individual potentials algebraically:

In contrast, electric fields must be added as vectors, taking direction into account. This difference reflects the deeper connection: potential relates to energy (a scalar), while electric field relates to force (a vector).

Example 13.2 What Is the Excess Charge on a Van de Graaff Generator?

A demonstration Van de Graaff generator has a metal sphere of diameter 25.0 cm that produces a voltage of 100 kV at its surface. What excess charge resides on the sphere?

Strategy

The radius of the sphere is

[latex]r = 12.5\ \text{cm} = 0.125\ \text{m}[/latex]

Using

[latex]V = \frac{kQ}{r}[/latex]

we solve for [latex]Q[/latex].

Solution

[latex]Q = \frac{rV}{k}[/latex]

[latex]Q = \frac{(0.125)(100\times10^{3})}{8.99\times10^{9}}[/latex]

[latex]Q = 1.39\times10^{-6}\ \text{C}[/latex]

[latex]Q = 1.39\ \mu\text{C}[/latex]

Discussion

This is only about one microcoulomb of excess charge, yet it produces a potential of 100,000 volts. This reinforces an important physical idea: it is relatively easy to create very large voltages with small isolated charges.

In medical and biological contexts, similar principles apply. For example, cell membranes maintain potential differences of about 70 mV across nanometer distances, producing extremely strong electric fields despite very small total charges.

In practice, voltage is always measured relative to a reference point. Often that reference is ground (Earth), which is assigned zero potential. Only potential differences have physical meaning. This is analogous to gravitational potential energy, where we may choose sea level to be zero. The choice of reference does not change physical predictions; only differences in potential matter.