FIGURE 25.3 shows a charged particle inside a parallel-plate capacitor with electrode spacing d. This is a uniform electric field, and the situation looks very much like Figure 25.2 for a mass in a uniform gravitational field. The one difference is that $\vec{g}$ always points down whereas the positive-to-negative electric field can point in any direction. To deal with this, let’s define a coordinate axis s that points from the negative plate, which we define to be $s=0$, toward the positive plate. The electric field $\vec{E}$ then points in the negative s-direction, just as the gravitational field $\vec{g}$ points in the negative y-direction. This s-axis, which is valid no matter how the capacitor is oriented, is analogous to the y-axis used for gravitational potential energy.

FIGURE 25.3 The electric field does work on the charged particle.

A positive charge q inside the capacitor speeds up and gains kinetic energy as it “falls” toward the negative plate. Is the charge losing potential energy as it gains kinetic energy? Indeed it is, and the calculation of the potential energy is just like the calculation of gravitational potential energy. The electric field exerts a constant force $F=qE$ on the charge in the direction of motion; thus the work done on the charge by the electric field is

where we again have to be careful with the signs because $s_{\text{f}}<s_{\text{i}}$.

The work done by the electric field causes the electric potential energy to change by

Comparing the initial and final terms on the two sides of the equation, we see that the electric potential energy of charge q in a uniform electric field is

where s is measured from the negative plate. We could add a constant to $U_{\text{elec}}$, but we’ve made the choice that $U_{\text{elec}}=0$ at $s=0$. Equation 25.10 was derived with the assumption that q is positive, but it is valid for either sign of q.

FIGURE 25.4 shows positive and negative charged particles moving in the uniform electric field between the plates of a parallel-plate capacitor. For a positive charge, $U_{\text{elec}}$ decreases and K increases as the charge moves toward the negative plate (decreasing s). Thus a positive charge is going “downhill” if it moves in the direction of the electric field. A positive charge moving opposite the field direction is going “uphill,” slowing as it transforms kinetic energy into electric potential energy.

FIGURE 25.4 A charged particle exchanges kinetic and potential energy as it moves in an electric field.

According to Equation 25.10, a negative charged particle has negative potential energy. You learned in Chapter 10 that there’s nothing wrong with negative potential energy—it’s simply less than the potential energy at some arbitrarily chosen reference location. The more important point, from Equation 25.10, is that the potential energy increases (becomes less negative) as a negative charge moves toward the negative plate. A negative charge moving in the field direction is going “uphill,” transforming kinetic energy into electric potential energy as it slows.

FIGURE 25.5 is an energy diagram for a positively charged particle in an electric field. Recall that an energy diagram is a graphical representation of how kinetic and potential energies are transformed as a particle moves. For positive q, the electric potential energy given by Equation 25.10 increases linearly from 0 at the negative plate (with $U_0=0$) to $qEd$ at the positive plate. The total mechanical energy—which is under your control—is constant. If $E_{\text{mech}}<qEd$, as shown here, a positively charged particle projected from the negative plate will gradually slow (transforming kinetic energy into potential energy) until it reaches a turning point where $U_{\text{elec}}=E_{\text{mech}}$. But if you project the particle with greater speed, such that $E_{\text{mech}}>qEd$, it will be able to cross the gap to collide with the positive plate.

FIGURE 25.5 The energy diagram for a positively charged particle in a uniform electric field.

EXAMPLE 25.1 Conservation of energy

A $2.0\text{cm}\times 2.0\text{cm}$ parallel-plate capacitor with a 2.0 mm spacing is charged to $\pm 1.0\text{nC}$. First a proton and then an electron are released from rest midway between the capacitor plates. Assume the motion takes place in a vacuum.

1. What is each particle’s energy?

2. What is each particle’s speed as it reaches the plate?

MODEL We’ll define the system to be the charged particle and the charged electrodes. This is an isolated system with no dissipation, so the mechanical energy is conserved.

VISUALIZE FIGURE 25.6 is a before-and-after pictorial representation, as you learned to draw in Part II. Each particle is released from rest $(K=0)$ and moves “downhill” toward lower potential energy. Thus the proton moves toward the negative plate, the electron toward the positive plate.

FIGURE 25.6 A proton and an electron in a capacitor.

FIGURE 25.6 Full Alternative Text

SOLVE

1. The s-axis was defined to point from the negative toward the positive plate of the capacitor. Both charged particles have $s_{\text{i}}=\frac{1}{2}d$, where $d=2.0\text{mm}$ is the plate separation. The proton $(q=e)$ has energy

   $$E_{\text{mech p}}=K_{\text{i}}+U_{\text{i}}=0+eE\left(\frac{1}{2}d\right)=\frac{1}{2}eEd$$

   while the electron $(q=-e)$ has

   $$E_{\text{mech e}}=K_{\text{i}}+U_{\text{i}}=0-eE\left(\frac{1}{2}d\right)=-\frac{1}{2}eEd$$

   The electric field inside the parallel-plate capacitor, from Chapter 23, is

   $$E={\displaystyle \frac{Q}{{ϵ}_{0}A}}=2.82\times {10}^5\text{N}\text{/}\text{C}$$

   Thus the particles’ energies can be calculated to be

   $$E_{\text{mech p}}=4.5\times {10}^{-17}\text{J and }E{}_{\text{mech e}}=-4.5\times {10}^{-17}\text{J}$$

   Notice that the electron’s mechanical energy is negative.

2. Conservation of mechanical energy requires $K_{\text{f}}+U_{\text{f}}=$ $K_{\text{i}}+U_{\text{i}}=E_{\text{mech}}$. The proton collides with the negative plate, so $U_{\text{f}}=0$, and the final kinetic energy is $K_{\text{f}}=\frac{1}{2}{m}_{\text{p}}{v}_{\text{f}}{}^2=E_{\text{mech p}}$. Thus the proton’s impact speed is

   $$(v_{\text{f}}){}_p=\sqrt{{\displaystyle \frac{2E_{\text{mech p}}}{m_{\text{p}}}}}=2.3\times {10}^5\text{m}\text{/}\text{s}$$

   Similarly, the electron collides with the positive plate, where $U_{\text{f}}=qEd=-eEd=2E_{\text{mech e}}$. Thus energy conservation for the electron is

   $$K_{\text{f}}=\frac{1}{2}{m}_{\text{e}}{v}_{\text{f}}{}^2=E_{\text{mech e}}-U_{\text{f}}=E_{\text{mech e}}-2E_{\text{mech e}}=-E_{\text{mech e}}$$

   We found the electron’s mechanical energy to be negative, so $K_{\text{f}}$ is positive. The electron reaches the positive plate with speed

   $$(v_{\text{f}}){}_e=\sqrt{{\displaystyle \frac{-2E_{\text{mech e}}}{m_{\text{e}}}}}=1.0\times {10}^7\text{m}\text{/}\text{s}$$

REVIEW Even though both particles have mechanical energy with the same magnitude, the electron has a much greater final speed due to its much smaller mass.