Electric Potential Energy in a Uniform Field

Let’s look at an electrical example of these basic concepts. In Fig. 1 a pair of charged parallel metal plates sets up a uniform, downward electric field with magnitude E. The field exerts a downward force with magnitude $F=q_0E$ on a positive test charge $q_0$. As the charge moves downward a distance d from point a to point b, the force on the test charge is constant and independent of its location. So the work done by the electric field is the product of the force magnitude and the component of displacement in the (downward) direction of the force:

$W_{a→b}=Fd=q_0Ed$

This work is positive, since the force is in the same direction as the net displacement of the test charge.

Fig.1: The work done on a point charge moving in a uniform electric field.

The y-component of the electric force, $F_y=-q_0E$ is constant, and there is no x- or z-component. This is exactly analogous to the gravitational force on a mass m near the earth’s surface; for this force, there is a constant y-component $F=-mg$ and the x- and z-components are zero. Because of this analogy, we can conclude that the force exerted on $q_0$ by the uniform electric field in Fig. 23.2 is conservative, just as is the gravitational force. This means that the work $W_{a→b}$ done by the field is independent of the path the particle takes from a to b. We can represent this work with a potential-energy function U, just as we did for gravitational potential energy in Section 7.1. The potential energy for the gravitational force $F_y=-mg$ was $U=mgy$ hence the potential energy for the electric force $F_y=-q_0E$ is

$U=q_0Ey$

When the test charge moves from height $y_a$ to height $y_b$ the work done on the charge by the field is given by

$W_{a→b}=-\Delta U=-(U_b-U_a)=-(q_0Ey_b-q_0Ey_a)=q_0E(y_a-y_b)d$

When $y_a$ is greater than $y_b$ (Fig. 23.3a), the positive test charge $q_0$ moves downward, in the same direction as $\mathbf{E}$; the displacement is in the same direction as the force $\mathbf{F}=q_0\mathbf{E}$ so the field does positive work and $U$ decreases. [In particular, if $y_a-y_b=d$ as in Fig. 23.2, Eq. (23.6) gives $W_{a→b}=q_0Ed$, in agreement with Eq. (23.4).] When $y_a$ is less than $y_b$ (Fig. 23.3b), the positive test charge $q_0$ moves upward, in the opposite direction to $\mathbf{E}$; the displacement is opposite the force, the field does negative work, and U increases.

Fig.2: A positive charge moving (a) in the direction of the electric field$\mathbf{E} and (b) in the direction opposite $\mathbf{E}$.

If the test charge $q_0$ is negative, the potential energy increases when it moves with the field and decreases when it moves against the field (Fig. 23.4).

Whether the test charge is positive or negative, the following general rules apply: U increases if the test charge $q_0$ moves in the direction opposite the electric force $\mathbf{F}=q_0\mathbf{E}$ (Figs. 23.3b and 23.4a); U decreases if moves in the $q_0$ same direction as $\mathbf{F}=q_0\mathbf{E}$ (Figs. 23.3a and 23.4b). 

Fig.3: A negative charge moving (a) in the direction of the electric field$\mathbf{E} and (b) in the direction opposite$\mathbf{E}. Compare with Fig. 2.

This is the same behavior as for gravitational potential energy, which increases if a mass m moves upward (opposite the direction of the gravitational force) and decreases if m moves downward (in the same direction as the gravitational force).

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