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 Active Figure 25.2a. Let’s calculate the potential difference between two points 훽 and 훾 separated by a distance d, where the displacement Ss points from 훽 toward 훾 and is parallel to the field lines.
Equation 25.3 gives
V훾2V훽5 DV5 23
훾 훽 E
S
?dSs 5 23
훾
훽 E ds 1cos 0°2 5 23
훾 훽 E ds Because E is constant, it can be removed from the integral sign, which gives
DV5 2E 3
훾 훽
ds5 2Ed (25.6) Potential difference X
between two points in a uniform electric field Pitfall Prevention 25.2 Voltage
A variety of phrases are used to describe the potential difference between two points, the most com- mon being voltage, arising from the unit for potential. A voltage applied to a device, such as a televi- sion, or across a device is the same as the potential difference across the device. Despite popular language, voltage is not something that moves through a device.
Pitfall Prevention 25.3 The Electron Volt
The electron volt is a unit of energy, NOT of potential. The energy of any system may be expressed in eV, but this unit is most convenient for describing the emission and absorp- tion of visible light from atoms.
Energies of nuclear processes are often expressed in MeV.
훽 훾
E
S
Figure 25.1 (Quick Quiz 25.1) Two points in an electric field.
25.2 | Potential Difference in a Uniform Electric Field 713
The negative sign indicates that the electric potential at point 훾 is lower than at point 훽; that is, V훾 , V훽. Electric field lines always point in the direction of decreasing electric potential as shown in Active Figure 25.2a.
Now suppose a test charge q0 moves from 훽 to 훾. We can calculate the change in the potential energy of the charge–field system from Equations 25.3 and 25.6:
DU 5 q0 DV 5 2q0Ed (25.7) This result shows that if q0 is positive, then DU is negative. Therefore, in a system con- sisting of a positive charge and an electric field, the electric potential energy of the system decreases when the charge moves in the direction of the field. Equivalently, an electric field does work on a positive charge when the charge moves in the direc- tion of the electric field. That is analogous to the work done by the gravitational field on a falling object as shown in Active Figure 25.2b. If a positive test charge is released from rest in this electric field, it experiences an electric force q0E
S
in the direction of E
S
(downward in Active Fig. 25.2a). Therefore, it accelerates downward, gaining kinetic energy. As the charged particle gains kinetic energy, the potential energy of the charge–field system decreases by an equal amount. This equivalence should not be surprising; it is simply conservation of mechanical energy in an iso- lated system as introduced in Chapter 8.
The comparison between a system of an electric field with a positive test charge and a gravitational field with a test mass in Active Figure 25.2 is useful for con- ceptualizing electrical behavior. The electrical situation, however, has one feature that the gravitational situation does not: the test charge can be negative. If q0 is negative, then DU in Equation 25.7 is positive and the situation is reversed. A system consisting of a negative charge and an electric field gains electric potential 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. 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 훽 and 훾 in a uniform electric field such that the vector Ss is not parallel to the field lines as shown in Figure 25.3. In this case, Equation 25.3 gives
DV5 23
훾 훽
E
S
?dsS5 2E
S
?3
훾 훽
dsS5 2E
S
?Ss (25.8)
where again E
S
was removed from the integral because it is constant. The change in potential energy of the charge–field system is
DU5q0DV5 2q0E
S
?Ss (25.9)
Change in potential between W
two points in a uniform electric field
When a positive test charge moves from point 훽 to point 훾, the
electric potential energy of the charge–field system decreases.
When an object with mass moves from point 훽 to point 훾, the
gravitational potential energy of the object–field system decreases.
E
S
⫹ d q0
훾 훽
a
Sg
d
훾m
훽
b
(a) When the electric field E
S
is directed downward, point 훾 is at a lower electric potential than point 훽. (b) An object of mass m moving downward in a gravitational field gS.
ACTIVE FIGURE 25.2
훽 d
훾
u 훿
E
S
Ss
Point 훾 is at a lower electric potential than point 훽.
Points 훾 and 훿 are at the same electric potential.
Figure 25.3 A uniform electric field directed along the positive x axis.
E x a m p l e 25.1 The Electric Field Between Two Parallel Plates of Opposite Charge
A battery has a specified potential difference DV between its terminals and estab- lishes that potential difference between conductors attached to the terminals.
A 12-V battery is connected between two parallel plates as shown in Figure 25.5.
The separation between the plates is d 5 0.30 cm, and we assume the electric field between the plates to be uniform. (This assumption is reasonable if the plate sepa- ration is small relative to the plate dimensions and we do not consider locations near the plate edges.) Find the magnitude of the electric field between the plates.
SOLUTION
Conceptualize In Chapters 23 and 24, we investigated the uniform electric field between parallel plates. The new feature to this problem is that the electric field is related to the new concept of electric potential.
Categorize The electric field is evaluated from a relationship between field and potential given in this section, so we categorize this example as a substitution problem.
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 that in Figure 25.3, where the potential difference V훾 2 V훽 is equal to the potential differ- ence V훿 2 V훽. (Prove this fact to yourself by working out two dot products for E
S
?Ss: one for Ss훽S훾, where the angle u between E
S
and Ss is arbitrary as shown in Figure 25.3, and one for Ss훽S훿, where u 5 0.) Therefore, V훾 5 V훿. The name equipoten- tial surface is given to any surface consisting of a continuous distribution of points having the same electric potential.
The equipotential surfaces associated with a uniform electric field consist of a family of parallel planes that are all perpendicular to the field. Equipotential sur- faces associated with fields having other symmetries are described in later sections.
Quick Quiz 25.2 The labeled points in Figure 25.4 are on a series of equipo- tential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves from 훽 to 훾, from 훾 to 훿, from 훿 to , and from to .
9 V 8 V 7 V 6 V
훾 훽
훿
Figure 25.4 (Quick Quiz 25.2) Four equipotential surfaces.
V = 12 V A
B
d
Figure 25.5 (Example 25.1) A 12-V battery connected to two parallel plates. The electric field between the plates has a magnitude given by the potential difference DV divided by the plate separation d.
Use Equation 25.6 to evaluate the magnitude of the elec- tric field between the plates:
E5 0VB2VA0
d 5 12 V
0.3031022 m5 4.0 3 103 V/m The configuration of plates in Figure 25.5 is called a parallel-plate capacitor and is examined in greater detail in Chapter 26.
E x a m p l e 25.2 Motion of a Proton in a Uniform Electric Field
A proton is released from rest at point 훽 in a uniform electric field that has a magnitude of 8.0 3 104 V/m (Fig. 25.6). The proton under- goes a displacement of magnitude d 5 0.50 m to point 훾 in the direction of E
S
. Find the speed of the proton after completing the displacement.
SOLUTION
Conceptualize Visualize the proton in Figure 25.6 moving downward through the potential difference. The situation is analogous to an object falling through a gravitational field.
d
훽
훾
E
S
v훾
S
v훽 0
S
Figure 25.6 (Example 25.2) A
proton accelerates from 훽 to 훾 in the direction of the electric field.