The derivative in Equation 26.9 gives $E_s$, the component of the electric field parallel to the displacement $\text{Δ}\vec{s}$. It doesn’t tell us about the electric field component perpendicular to $\text{Δ}\vec{s}$. Thus Equation 26.9 is most useful if we can use symmetry to select a coordinate axis that is parallel to $\vec{E}$ and along which the perpendicular component of $\vec{E}$ is known to be zero.

For example, suppose we knew the potential of a point charge to be $V=q\text{/}4\pi {ϵ}_{0}r$ but didn’t remember the electric field. Symmetry requires that the field point straight outward from the charge, with only a radial component $E_r$. If we choose the s-axis to be in the radial direction, parallel to $\vec{E}$, we can use Equation 26.9 to find

This is, indeed, the well-known electric field of a point charge.

Equation 26.9 is especially useful for a continuous distribution of charge because calculating V, which is a scalar, is usually much easier than calculating the vector $\vec{E}$ directly from the charge. Once V is known, $\vec{E}$ is found simply by taking a derivative.

A geometric interpretation of Equation 26.9 is that the electric field is the negative of the slope of the V-versus-s graph. This interpretation should be familiar. You learned in Chapter 10 that the force on a particle is the negative of the slope of the potential-energy graph: $F=-dU\text{/}ds$. In fact, Equation 26.9 is simply $F=-dU\text{/}ds$ with both sides divided by q to yield E and V. This geometric interpretation is an important step in developing an understanding of potential.

EXAMPLE 26.4 Finding E from the slope of V

FIGURE 26.7 is a graph of the electric potential in a region of space where $\vec{E}$ is parallel to the x-axis. Draw a graph of $E_x$ versus x.

FIGURE 26.7 Graph of V versus position x.

FIGURE 26.7 Full Alternative Text

MODEL The electric field is the negative of the slope of the potential graph.

SOLVE There are three regions of different slope:

$$\begin{array}{ll}0<x<2\text{cm}\hfill & \{\begin{array}{c}\text{Δ}V\text{/}\text{Δ}x=\left(20\text{V}\right)\text{/}\left(0.020\text{m}\right)=100\text{V/m}\hfill \\ E_x=-1000\text{V/m}\hfill \end{array}\hfill \\ 2<x<4\text{cm}\hfill & \{\begin{array}{c}\text{Δ}V\text{/}\text{Δ}x=0\text{V/m}\hfill \\ E_x=0\text{V/m}\hfill \end{array}\hfill \\ 4<x<8\text{cm}\hfill & \{\begin{array}{c}\text{Δ}V\text{/}\text{Δ}x=\left(-20\text{V}\right)\text{/}\left(0.040\text{m}\right)=-500\text{V/m}\hfill \\ E_x=500\text{V/m}\hfill \end{array}\hfill \end{array}$$

The results are shown in FIGURE 26.8.

FIGURE 26.8 Graph of $E_x$ versus position x.

FIGURE 26.8 Full Alternative Text

REVIEW The electric field $\vec{E}$ points to the left ($E_x$ is negative) for $0<x<2\text{cm}$ and to the right ($E_x$ is positive) for $4<x<8\text{cm}$. Notice that the electric field is zero in a region of space where the potential is not changing.