Download Example 17-1 Electric Potential Energy Difference in a Uniform Field

Example 17-1 Electric Potential Energy Difference in a Uniform Field
An electron (charge q = 2e = 21.60 * 10219 C, mass m = 9.11 * 10231 kg) is released from rest in a uniform electric field.
The field points in the positive x direction and has magnitude 2.00 * 102 N>C. Find the speed of the electron after it has
moved 0.300 m.
Set Up
Because the electron has a negative charge,
s on the electron is directed
the force Fs = qE
opposite to the electric field. So the displacement of the electron will be at an angle u =
180° to the direction of the electric field.
If the conservative electric force is
the only force acting on the electron, then
mechanical energy is conserved. We’ll find
the change in electric potential energy using
Equation 17-2, and from this find the change
in kinetic energy of the electron. The electron’s initial kinetic energy is zero because it
starts at rest; once we know the final kinetic
energy, we can determine the electron’s final
speed.
Solve
Use Equation 17-2 to solve for the change in
electric potential energy.
Electric potential energy for a
charge in a uniform electric field:
Uelectric = -qEd cos u
E = 2.00 × 102 N/C
(17-2)
d = 0.300 m
–e
+x
Conservation of mechanical
energy:
Ki + Uelectric,i = Kf + Uelectric,f
(6-23)
Kinetic energy:
K =
1
mv 2
2
(6-8)
The change in electric potential energy equals the difference between the
final and initial electric potential energy:
Uelectric = Uelectric,f 2 Uelectric,i
From Equation 17-2,
Uelectric = Uelectric,f 2 Uelectric,i
= 2qEd cos u
= - 1 -1.60 * 10-19 C2 12.00 * 102 N>C2 10.300 m2 cos 180
= 2(29.60 * 10218 N # m)(21)
= 29.60 * 10218 N # m = 29.60 * 10218 J
Find the final kinetic energy and the final
speed of the electron.
The initial kinetic energy of the electron is Ki = 0. Solve the energy conservation equation for the final kinetic energy of the electron:
Kf = Ki + Uelectric,i 2 Uelectric,f = 0 2 (Uelectric,f 2 Uelectric,i)
= 0 2 (29.60 * 10218 J) = 9.60 * 10218 J
Finally, solve for the final speed of the electron:
1
mv 2f so
2
219.60 * 10-18 J2
2K f
=
= 4.59 * 106 m>s
vf =
B m
C 9.11 * 10-31 kg
Kf =
Reflect
The electric potential energy decreases when we release the electron in the electric field, in the same way that the gravitational potential energy decreases when you release a ball in the presence of Earth’s gravity. In both situations, the kinetic
energy increases by the same amount that the potential energy decreases.
You should verify that the force on the electron is very small, only 3.2 * 10217 N in magnitude. The electron has
only a tiny mass, however, and our results show that it acquires a ferocious speed (about 1.5% of the speed of light) after
moving just 0.300 m.