When a test charge $q_0$ is placed in an electric field $\mathbf{E}$ created by some source charge distribution, the electric force acting on the test charge is $q_0\mathbf{E}$. The force $q_0\mathbf{E}$ is conservative because the force between charges described by Coulomb’s law is conservative. When the test charge is moved in the field by some external agent, the work done by the field on the charge is equal to the negative of the work done by the external agent causing the displacement. This is analogous to the situation of lifting an object with mass in a gravitational field—the work done by the external agent is $mgh$ and the work done by the gravitational force is $mgh$.
When analyzing electric and magnetic fields, it is common practice to use the notation $d\mathbf{s}$ to represent an infinitesimal displacement vector that is oriented tangent to a path through space. This path may be straight or curved, and an integral performed along this path is called either a path integral or a line integral (the two terms are synonymous).
For an infinitesimal displacement $d\mathbf{s}$ of a charge, the work done by the electric field on the charge is $\mathbf{F}d\mathbf{s}=q_0\mathbf{E}.d\mathbf{s}$. As this amount of work is done by the field, the potential energy of the charge–field system is changed by an amount $U=-q_0\mathbf{E}.d\mathbf{s}$. For a finite displacement of the charge from point A to point B, the change in potential energy of the system $\Delta U=U_B-U_A$ is
$\Delta U=-q_0 \int_{A}^{B}\mathbf{E}.d\mathbf{s}$
The integration is performed along the path that $q_0$ follows as it moves from A to B. Because the force $q_0\mathbf{E}$ is conservative, this line integral does not depend on the path taken from A to B.
For a given position of the test charge in the field, the charge–field system has a potential energy U relative to the configuration of the system that is defined as U = 0. Dividing the potential energy by the test charge gives a physical quantity that depends only on the source charge distribution. The potential energy per unit charge $U/q_0$ is independent of the value of $q_0$ and has a value at every point in an electric field. This quantity $U/q_0$ is called the electric potential (or simply the potential) V. Thus, the electric potential at any point in an electric field is
$V=\frac{U}{q_0}$
The fact that potential energy is a scalar quantity means that electric potential also is a scalar quantity.
As described by Equation 25.1, if the test charge is moved between two positions A and B in an electric field, the charge–field system experiences a change in potential energy. The potential difference $\Delta V=V_B-V_A$ between two points A and B in an electric field is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge $q_0$:
$\Delta V \equiv \frac{U}{q_0}=-\int_{A}^{B}\mathbf{E}.d\mathbf{s}$
Just as with potential energy, only differences in electric potential are meaningful. To avoid having to work with potential differences, however, we often take the value of the electric potential to be zero at some convenient point in an electric field.
Potential difference should not be confused with difference in potential energy. The potential difference between A and B depends only on the source charge distribution (consider points A and B without the presence of the test charge), while the difference in potential energy exists only if a test charge is moved between the points. Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field.
Equations 25.1 and 25.3 hold in all electric fields, whether uniform or varying, but they can be simplified for a uniform field. First, consider a uniform electric field directed along the negative y axis, as shown in Figure 1. Let us calculate the potential difference between two points A and B separated by a distance $|\mathbf{s}|=d$, where $\mathbf{s}$ is parallel to the field lines. Equation 25.3 gives
$V_B-V_A=\Delta V =-\int_{A}^{B}\mathbf{E}.d\mathbf{s}$
$V_B-V_A=\Delta V =-\int_{A}^{B}(E \ cos \ 0^0).ds=-\int_{A}^{B}Eds$
Because E is constant, we can remove it from the integral sign; this gives
$\Delta V =-E \int_{A}^{B}ds=-Ed$
The negative sign indicates that the electric potential at point B is lower than at point A; that is, $V_B > V_A$. Electric field lines always point in the direction of decreasing electric potential, as shown in Figure 1a.
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| Fig.1: (a) When the electric field E is directed downward, point B is at a lower electric potential than point A. When a positive test charge moves from point A to point B, the charge–field system loses electric potential energy. (b) When an object of mass m moves downward in the direction of the gravitational field g, the object–field system loses gravitational potential energy. |
Now suppose that a test charge $q-0$ moves from A to B. We can calculate the change in the potential energy of the charge–field system from Equations 25.3 and 25.6:
$\Delta U=q_0\Delta V=-q_0Ed$
From this result, we see that if $q_0$ is positive, then $\Delta U$ is negative. We conclude that a system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field. This means that an electric field does work on a positive charge when the charge moves in the direction of the electric field. (This is analogous to the work done by the gravitational field on a falling object, as shown in Figure 25.2b.) If a positive test charge is released from rest in this electric field, it experiences an electric force $q_0\mathbf{E}$ in the direction of E (downward in Fig. 25.2a). Therefore, it accelerates downward, gaining kinetic energy. As the charged particle gains kinetic energy, the charge–field system loses an equal amount of potential energy.
If $q_0$ is negative, then $\Delta U$ in Equation 25.7 is positive and the situation is reversed: A system consisting of a negative charge and an electric field gains electric po tential energy when the charge moves in the direction of the field. If a negative charge is released from rest in an electric field, it accelerates in a direction opposite the direction of the field. In order for the negative charge to move in the direction of the field, an external agent must apply a force and do positive work on the charge.
Now consider the more general case of a charged particle that moves between A and B in a uniform electric field such that the vector s is not parallel to the field lines, as shown in Figure 25.3. In this case, Equation 25.3 gives
$\Delta V =-\int_{A}^{B}\mathbf{E}.d\mathbf{s}=-\mathbf{E}\int_{A}^{B}.d\mathbf{s}=\mathbf{E}.\mathbf{s}$
where again we are able to remove E from the integral because it is constant. The change in potential energy of the charge–field system is
$\Delta U=q_0\Delta V=-q_0\mathbf{E}.\mathbf{s}$
Finally, we conclude from Equation 25.8 that all points in a plane perpendicular to a uniform electric field are at the same electric potential. We can see this in Figure 25.3, where the potential difference $V_B-V_A$ is equal to the potential difference $V_C - V_A$. (Prove this to yourself by working out the dot product $\mathbf{E}.\mathbf{s}$ for $\mathbf{s}_{A→B}$, where the angle $\theta$ between $\mathbf{E}$ and $\mathbf{s}$ is arbitrary as shown in Figure 25.3, and the dot product for $\mathbf{s}_{A→C}$, where $\theta =0$) Therefore, $V_B = V_C$. The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.
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| Fig.2: A uniform electric field directed along the positive x axis. Point B is at a lower electric potential than point A. Points B and C are at the same electric potential. |
The equipotential surfaces of a uniform electric field consist of a family of parallel planes that are all perpendicular to the field. Equipotential surfaces for fields with other symmetries are described in later sections.
Example 1
A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A $+12 \mu C$ charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).
(a) What is the change in the potential energy of the charge field system?
(b) Through what potential difference does the charge move?
Solution:
Given:
Electric Field E = 250 V/m directed along the positive
Charge (q): $12 \mu C=12 \times 10^{-6} \ C$
Initial Position: (0,0)
Final Position: (x,y) = (20.0 cm, 50.0 cm) = (0.20 m, 0.50 m)
Electric Potential Energy Change $\Delta U$
$\Delta U =q\Delta V$
where $\Delta V$ is the potential difference.
Potential Difference ($\Delta V$):
In a uniform electric field, $\Delta V=-E.\Delta x$ where is the displacement along the electric field's direction (-axis).
The displacement in the
$\Delta x=x_f-x_i=0.20 \ m$
(a) Change in Potential Energy ($\Delta U$)
$\Delta U =q\Delta V=q(-E\Delta x)$
$\Delta U =-(12 \times 10^{-6} \ C)(250 \ V/m)(0.20 \ m)$
$\Delta U = -6.00 \times 10^{-4} \ J$
So, the change in potential energy is: $-0.60 \ mJ$
(b) Potential Difference:
From the calculation above, the potential difference is:
$\Delta V=-E.\Delta x = -(250 \ V/m)(0.20 \ m)$
$\Delta V=-50.0 \ V$
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