The electric potential is an abstract idea, and it will take some practice to see just what it means and how it is useful. We’ll use multiple representations—words, pictures, graphs, and analogies—to explain and describe the electric potential.

To begin, consider charged particles that move in a vacuum without collisions. If we know the electric potential in a region of space, we can determine whether a charged particle speeds up or slows down as it moves through that region. FIGURE 25.16 illustrates this idea. Here a group of source charges, which remains hidden offstage, has created an electric potential V that increases from left to right. A charged particle q, which for now we’ll assume to be positive, has electric potential energy $U_{\text{elec}}=qV$. If the particle moves to the right, its potential energy increases and so, by energy conservation, its kinetic energy must decrease. A positive charge slows down as it moves into a region of higher electric potential.

FIGURE 25.16 In the absence of other applied forces, a charged particle speeds up or slows down as it moves through a potential difference.

It is customary to say that the particle moves through a potential difference $\text{Δ}V=V_{\text{f}}-V_{\text{i}}$. The potential difference between two points is often called the voltage. The particle that travels to the right moves through a positive potential difference $(\text{Δ}V>0$ because $V_{\text{f}}>V_{\text{i}})$, so we can say that a positively charged particle slows down as it moves through a positive potential difference.

The particle that moves to the left in Figure 25.16 travels in the direction of decreasing electric potential—through a negative potential difference—and loses potential energy. It speeds up as it transforms potential energy into kinetic energy. A negatively charged particle would slow down because its potential energy $qV$ would increase as V decreases. TABLE 25.2 summarizes these ideas.

Because $U_{\text{elec}}=qV$, we can write the conservation of mechanical energy statement $K_{\text{f}}+U_{\text{f}}=K_{\text{i}}+U_{\text{i}}$ for a charged particle in a vacuum as

Conservation of energy is the basis of a powerful problem-solving strategy.

EXAMPLE 25.6 Moving through a potential difference

A proton with a speed of $2.0\times {10}^5\text{m}\text{/}\text{s}$ enters a region of space in which there is an electric potential. What is the proton’s speed after it moves through a potential difference of 100 V? What will be the final speed if the proton is replaced by an electron?

MODEL The system is the charge plus the unseen source charges that create the potential. Assume that the proton moves in a vacuum. This is an isolated system, so mechanical energy is conserved.

VISUALIZE FIGURE 25.17 is a before-and-after pictorial representation of a charged particle moving through a potential difference. A positive charge slows down as it moves into a region of higher potential $(K\to U)$. A negative charge speeds up $\left(U\to K\right)$.

FIGURE 25.17 A charged particle moving through a potential difference.

FIGURE 25.17 Full Alternative Text

SOLVE The potential energy of charge q is $U=qV$. Conservation of energy, now expressed in terms of the electric potential V, is $K_{\text{f}}+q{V}_{\text{f}}=K_{\text{i}}+q{V}_{\text{i}}$, or

$$K_{\text{f}}=K_{\text{i}}-q\text{Δ}V$$

where $\text{Δ}V=V_{\text{f}}-V_{\text{i}}$ is the potential difference through which the particle moves. In terms of the speeds, energy conservation is

$$\frac{1}{2}m{v}_{\text{f}}{}^2=\frac{1}{2}m{v}_{\text{i}}{}^2-q\text{Δ}V$$

We can solve this for the final speed:

$$v_{\text{f}}=\sqrt{v_{\text{i}}{}^2-\frac{2q}{m}\text{Δ}V}$$

For a proton, with $q=e$, the final speed is

$$\begin{array}{ll}{\left(v_{\text{f}}\right)}_{\text{p}}\hfill & =\sqrt{{\left(2.0\times {10}^5\text{m/s}\right)}^2-{\displaystyle \frac{2\left(1.60\times {10}^{-19}\text{C}\right)\left(100\text{V}\right)}{1.67\times {10}^{-27}\text{kg}}}}\hfill \\ \hfill & =1.4\times {10}^5\text{m/s}\hfill \end{array}$$

An electron, though, with $q=-e$ and a different mass, reaches speed $(v_{\text{f}}){}_e=5.9\times {10}^6\text{m}\text{/}\text{s}$.

REVIEW The proton slowed down and the electron sped up, as we expected. Note that the electric potential already existed in space due to other charges that are not explicitly seen in the problem. The electron and proton have nothing to do with creating the potential. Instead, they respond to the potential by having potential energy $U=qV$.