FIGURE 26.1 shows the four key ideas of force, field, potential energy, and potential. The electric field and the electric potential were based on force and potential energy. We know, from Chapters 9 and 10, that force and potential energy are closely related. The focus of this chapter is to establish a similar relationship between the electric field and the electric potential. The electric potential and electric field are not two distinct entities, simply two different ways of describing how source charges alter the space around them.

FIGURE 26.1 The four key ideas.

If this is true, we should be able to find the electric potential from the electric field. Chapter 25 introduced all the pieces we need to do so. We used the potential energy of charge $q$ and the source charges to define the electric potential as

Potential energy is defined in terms of the work done by force $\vec{F}$ on charge q as it moves from position i to position f:

But the force exerted on charge q by the electric field is $\vec{F}=q\vec{E}$. Putting these three pieces together, you can see that the charge q cancels out and the potential difference between two points in space is

where s is the position along a line from point i to point f. That is, we can find the potential difference between two points if we know the electric field.

A graphical interpretation of Equation 26.3 is

Notice, because of the minus sign in Equation 26.3, that the area is subtracted from $V_{\text{i}}$.

EXAMPLE 26.1 Finding the potential

FIGURE 26.2 is a graph of $E_x$, the x-component of the electric field, versus position along the x-axis. Find and graph $V\left(x\right)$. Assume $V=0\text{V}$ at $x=0\text{m}$.

FIGURE 26.2 Graph of $E_x$ versus x.

FIGURE 26.2 Full Alternative Text

MODEL The potential difference is the negative of the area under the curve.

VISUALIZE $E_x$ is positive throughout this region of space, meaning that $\vec{E}$ points in the positive x-direction.

SOLVE We can see that $E_x=1000x\text{V}\text{/}\text{m}$, where x is in m. Thus

$$\begin{array}{lll}{V}_{\text{f}}\hfill & =\hfill & V(x)=0-(\mathrm{area}\mathrm{under}\mathrm{the}{E}_x\mathrm{curve})\hfill \\ \hfill & =\hfill & -\frac{1}{2}\times \text{base}\times \text{height}=-\frac{1}{2}(x)(1000x)=-500x^2\text{V}\hfill \end{array}$$

FIGURE 26.3 shows that the electric potential in this region of space is parabolic, decreasing from $0\text{V}$ at $x=0\text{m}$ to $-2000\text{V}$ at $x=2\text{m}$.

FIGURE 26.3 Graph of V versus x.

FIGURE 26.3 Full Alternative Text

REVIEW The electric field points in the direction in which V is decreasing. We’ll soon see that this is a general rule.

To see how this works, let’s use the electric field of a point charge to find its electric potential. FIGURE 26.4 identifies a point P at $s_{\text{f}}=r$ at which we want to know the potential and calls this position f. We’ve chosen position i to be at $s_{\text{i}}=\infty$ and identified that as the zero point of the potential. The integration of Equation 26.3 is straight inward along the radial line from i to f:

FIGURE 26.4 Finding the potential of a point charge.

The electric field is radially outward. Its s-component is

Thus the potential at distance r from a point charge q is

We’ve rediscovered the potential of a point charge that you learned in Chapter 25:

EXAMPLE 26.2 The potential of a parallel-plate capacitor

In Chapter 23, the electric field inside a capacitor was found to be

$$\vec{E}=\left({\displaystyle \frac{Q}{{ϵ}_{0}A}}\text{,}\text{from positive to negative}\right)$$

Find the electric potential inside the capacitor. Let $V=0\text{V}$ at the negative plate.

MODEL The electric field inside a capacitor is a uniform field.

VISUALIZE FIGURE 26.5 shows the capacitor and establishes a point P where we want to find the potential. We’ve chosen an s-axis measured from the negative plate, which is the zero point of the potential.

FIGURE 26.5 Finding the potential inside a capacitor.

FIGURE 26.5 Full Alternative Text

SOLVE We’ll integrate along the s-axis from $s_{\text{i}}=0$ (where $V_{\text{i}}=0\text{V})$ to $s_{\text{f}}=s$. Notice that $\vec{E}$ points in the negative s-direction, so $E_s=-Q\text{/}{ϵ}_{0}A\text{.}Q\text{/}{ϵ}_{0}A$ is a constant, so

$$V\left(s\right)=V_{\text{f}}=V_{\text{i}}-{\displaystyle {\int }_0^s{E}_{s}ds}=-\left(-{\displaystyle \frac{Q}{{ϵ}_{0}A}}\right){\displaystyle {\int }_0^{s}ds}={\displaystyle \frac{Q}{{ϵ}_{0}A}}s=Es$$

REVIEW $V=Es$ is the capacitor potential we deduced in Chapter 25 by working directly with the potential energy. The potential increases linearly from $V=0$ at the negative plate to $V=Ed$ at the positive plate. Here we found the potential by explicitly recognizing the connection between the potential and the field.