Equations 26.3 for V in terms of $E_s$ and 26.9 for $E_s$ in terms of V have profound implications for the geometry of the potential and the field. FIGURE 26.9 shows two equipotential surfaces, with $V_{+}$ positive relative to $V_{-}$. To learn about the electric field $\vec{E}$ at point P, allow a charge to move through the two displacements $\text{Δ}\vec{s}{}_1$ and $\text{Δ}\vec{s}{}_2$. Displacement $\text{Δ}\vec{s}{}_1$ is tangent to the equipotential surface, hence a charge moving in this direction experiences no potential difference. According to Equation 26.9, the electric field component along a direction of constant potential is $E_s=-dV\text{/}ds=0$. In other words, the electric field component tangent to the equipotential is $E_{\parallel }=0$.

FIGURE 26.9 The electric field at P is related to the shape of the equipotential surfaces.

Displacement $\text{Δ}\vec{s}{}_2$ is perpendicular to the equipotential surface. There is a potential difference along $\text{Δ}\vec{s}{}_2$, hence the electric field component is

You can see that the electric field is inversely proportional to $\text{Δ}{s}_2$, the spacing between the equipotential surfaces. Furthermore, because $(V_{+}-V_{-})>0$, the minus sign tells us that the electric field is opposite in direction to $\text{Δ}\vec{s}{}_2$. In other words, $\vec{E}$ is perpendicular to the equipotential surfaces and points “downhill” in the direction of decreasing potential.

These important ideas are summarized in FIGURE 26.10.

FIGURE 26.10 The geometry of the potential and the field.

Mathematically, we can calculate the individual components of $\vec{E}$ at any point by extending Equation 26.9 to three dimensions:

where $\partial V\text{/}\partial x$ is the partial derivative of V with respect to x while y and z are held constant. You may recognize from calculus that the expression in parentheses is the gradient of V, written $\nabla V$. Thus, $\vec{E}=-\nabla V$. More advanced treatments of the electric field make extensive use of this mathematical relationship, but for the most part we’ll limit our investigations to those we can analyze graphically.

EXAMPLE 26.5 Finding the electric field from the equipotential surfaces

In FIGURE 26.11 a $1\text{cm}\times 1\mathrm{cm}$ grid is superimposed on a contour map of the potential. Estimate the strength and direction of the electric field at points 1, 2, and 3. Show your results graphically by drawing the electric field vectors on the contour map.

FIGURE 26.11 Equipotential lines.

FIGURE 26.11 Full Alternative Text

MODEL The electric field is perpendicular to the equipotential lines, points “downhill,” and depends on the slope of the potential hill.

VISUALIZE The potential is highest on the bottom and the right. An elevation graph of the potential would look like the lower-right quarter of a bowl or a football stadium.

SOLVE Some distant but unseen source charges have created an electric field and potential. We do not need to see the source charges to relate the field to the potential. Because $E\approx -\text{Δ}V\text{/}\text{Δ}s$, the electric field is stronger where the equipotential lines are closer together and weaker where they are farther apart. If Figure 26.11 were a topographic map, you would interpret the closely spaced contour lines at the bottom of the figure as a steep slope.

FIGURE 26.12 shows how measurements of $\text{Δ}s$ from the grid are combined with values of $\text{Δ}V$ to determine $\vec{E}$. Point 3 requires an estimate of the spacing between the 0 V and the 100 V surfaces. Notice that we’re using the 0 V and 100 V equipotential surfaces to determine $\vec{E}$ at a point on the 50 V equipotential.

FIGURE 26.12 The electric field at points 1 to 3.

FIGURE 26.12 Full Alternative Text

REVIEW The directions of $\vec{E}$ are found by drawing downhill vectors perpendicular to the equipotentials. The distances between the equipotential surfaces are needed to determine the field strengths.