Download 648 CHAPTER 17. ELECTRIC POTENTIAL ENERGY AND THE

648
CHAPTER 17.
ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL
1
Let v be the final speed of the electron. Since the electron began at rest, ∆KE = mv2 , so
2
r
2∆KE
(1)
.
v=
m
Convert the value of ∆KE found in part a) from electron-volts to joules, and use it in equation (1),
s
r
2∆KE
2(1.00 × 103 eV ) 1.602 × 10−19 J
=
= 1.88 × 107 m/s .
v=
m
9.11 × 10−31 kg
1 eV
b)
17.42
From Equation 14.13 on page 645 of the text, we know the average kinetic energy per particle is
KEaverage =
3
3
6.21 × 10−21 J
kT = (1.381 × 10−23 J/K )(300 K ) =
= 3.88 × 10−2 eV
2
2
1.602 × 10−19 J / eV
The expression for the average kinetic energy per particle is independent of the mass of the particle and depends only on the temperature and Boltzmann’s constant. Hence, oxygen molecules at the same temperature
have the same average kinetic energy per particle.
Note that since
1
= 4.00 × 10−2 ,
25
the above result for the average kinetic energy at (a warm) room temperature can be expressed by saying
1
eV . This is the way the average
that the average kinetic energy per particle when T = 300 K is ≈
25
chemist or physicist remembers the kinetic energy of a particle at room temperature.
17.43
a) The direction of the electric field is from regions of higher to regions of lower electric potential. Hence,
in Figure P.43 on page 800 of the text, the electric field is directed from the bottom sheet to the top sheet.
b) The potential energy of the electron near the top sheet is
PE = qV = (−1.602 × 10−19 C )(−100 V ) = 1.60 × 10−17 J .
If you prefer to work in units of eV , the answer would be
PE = qV = (−e)(−100 V ) = 100 eV .
c)
The potential energy of the electron near the bottom sheet is
PE = qV = (−1.602 × 10−19 C )(100 V ) = −1.60 × 10−17 J .
If you prefer to work in units of eV , the answer would be
PE = qV = (−e)(100 V ) = −100 eV .
d) Under the influence of the electric force alone, the electron will move to lower its potential energy.
Therefore, you should release the electron near the upper sheet so it will accelerate towards the lower sheet.
e)
The electric force is the only force acting on the electron. Using the CWE theorem,
Wnonconservative = ∆KE + ∆PE
=⇒ 0 J = [KE + PE]final − [KE + PE]initial
1
= [ mv2 + (−1.60 × 10−17 J )] − [0 J + 1.60 × 10−17 J ]
2
=⇒
1 2
mv = 3.20 × 10−17 J .
2
649
17.44.
If you prefer to work in units of eV , the answer would be
Wnonconservative = ∆KE + ∆PE
=⇒ 0 J = [KE + PE]final − [KE + PE]initial
1
= [ mv2 + (−100 eV )] − [0 J + 100 eV ]
2
1 2
mv = 200 eV .
2
Solve for the speed v in part e),
s
2(3.20 × 10−17 J )
= 8.38 × 106 m/s .
v=
9.11 × 10−31 kg
=⇒
f)
Note that since we usually express speeds in the SI units of meters per second, we start this part of the
problem with kinetic energy expressed in the SI units of joules.
17.44 The work done by the total force on the electron is equal to the change in its kinetic energy. The
electric force is the only force acting and, in this particular problem, it is constant. Hence using the CWE
theorem,
Wtotal = ∆KE
=⇒ ~
F • ∆~r = KEfinal − KEinitial .
Let î be in the direction of the initial motion of the electron, and choose the origin at the place where the
electron enters the field. The electric force is opposite in direction to the change in electron’s position vector.
When the electron has zero speed, let the distance into the field it has traveled be x. With these choices,
we have
1
(−eE î) • (xî) = 0 J − mv02
2
=⇒ x =
mv02
(9.11 × 10−31 kg )(4.00 × 106 m/s )2
=
2eE
2(1.602 × 10−19 C )(300 N/C )
=⇒ x = 0.152 m .
17.45
a) The electric field is directed from regions of higher to regions of lower electric potential, so it is directed
from the 0 V sheet to the −200 V sheet. Since the positron has positive charge, the electric force on it is
~ Therefore, release the positron near the 0 V sheet.
parallel to the field (~
F = q E).
b) The change in the kinetic energy of the positron expressed in electron-volts is numerically the same as
the potential difference, expressed in volts, through which the particle was accelerated. Hence, the change
in the kinetic energy of the positron is
|∆KE| = |qV | = |(−1e)(200V )| = 200 eV .
Converted to joules, this is
∆KE = 200 eV = (200 eV )
1.602 × 10−19 J
eV
= 3.20 × 10−17 J .