Earlier in the chapter we found an expression for the electric potential energy of a charged particle between the plates of a parallel-plate capacitor. Now let’s investigate the electric potential. FIGURE 25.18 shows two parallel electrodes, separated by distance d, with surface charge density $\pm \eta$. As a specific example, we’ll let $d=3.00\text{mm}$ and $\eta =4.42\times {10}^{-9}\text{C}\text{/}{\text{m}}^2$. The electric field inside the capacitor, as you learned in Chapter 23, is

This electric field is due to the source charges on the capacitor plates.

FIGURE 25.18 A parallel-plate capacitor.

In Section 25.1, we found that the electric potential energy of a charge q in the uniform electric field of a parallel-plate capacitor is

$U_{\text{elec}}$ is the energy of q interacting with the source charges on the capacitor plates.

Our new view of the interaction is to separate the role of charge q from the role of the source charges by defining the electric potential $V=U_{q+\text{sources}}\text{/}q$. Thus the electric potential inside a parallel-plate capacitor is

where s is the distance from the negative electrode. The electric potential, like the electric field, exists at all points inside the capacitor. The electric potential is created by the source charges on the capacitor plates and exists whether or not charge q is inside the capacitor.

FIGURE 25.19 illustrates the important point that the electric potential increases linearly from the negative plate, where $V_{-}=0$, to the positive plate, where $V_{+}=Ed$. Let’s define the potential difference $\text{Δ}{V}_{\text{C}}$ between the two capacitor plates to be

In our specific example, $\text{Δ}{V}_{\text{C}}=(500\text{ N/C})(0.0030\text{ m})=1.5\text{ V}$. The units work out because $1.5(\text{N m})\text{/C}=1.5\text{ J/C}=1.5\text{ V}$.

FIGURE 25.19 The electric potential of a parallel-plate capacitor increases linearly from the negative to the positive plate.

Equation 25.26 has an interesting implication. Thus far, we’ve determined the electric field inside a capacitor by specifying the surface charge density $\eta$ on the plates. Alternatively, we could specify the capacitor voltage $\text{Δ}{V}_{\text{C}}$ (i.e., the potential difference between the capacitor plates) and then determine the electric field strength as

In fact, this is how E is determined in practical applications because it’s easy to measure $\text{Δ}{V}_{\text{C}}$ with a voltmeter but difficult, in practice, to know the value of $\eta$.

Equation 25.27 implies that the units of electric field are volts per meter, or $\text{V}\text{/}\text{m}$. We have been using electric field units of newtons per coulomb. In fact, as you can show as a homework problem, these units are equivalent to each other. That is,

Returning to the electric potential, we can substitute $E=\text{Δ}{V}_{\text{C}}\text{/}d$ into Equation 25.25 for $V_{\text{cap}}$. Thus the electric potential inside a capacitor is

The potential increases linearly from $V_{-}=0\text{V}$ at the negative plate $(s=0)$ to $V_{+}=$ $\text{Δ}{V}_{\text{C}}$ at the positive plate $(s=d)$.