Download Physics 102 Chapter 19 Homework Solutions

Physics 102 Chapter 19 Homework Solutions
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1.
REASONING When the electron moves from the ground to the cloud, the change in its electric potential
energy is (EPE) = EPEcloud  EPEground. (Remember that the change in any quantity is its final value
minus its initial value.) The change in the electric potential energy is related to the change V in the
potential by EPEq0V (Equation 19.4), where q0 is the charge on the electron. This relation will
allow us to find the change in the electron’s potential energy.
SOLUTION The difference in the electric potentials between the cloud and the ground is V = Vcloud 
Vground = 1.3  108 V, and the charge on an electron is q0 = 1.60  1019 C. Thus, the change in the
electron’s electric potential energy when the electron moves from the ground to the cloud is



  EPE  = q0 V= 1.60 1019 C 1.3 108 V  2.11011 J
3.
SSM REASONING The work WAB done by an electric force in moving a charge q0 from point A to point
WAB
B and the electric potential difference VB  VA are related according to VB  VA 
(Equation 19.4).
q0
This expression can be solved directly for q0.
SOLUTION Solving Equation 19.4 for the charge q0 gives
q0 
6.
WAB
WAB
2.70 103 J


 5.40 105 C
VB  VA VA  VB
50.0 V
REASONING The electric potential difference V experienced by the electron has the same magnitude as
the electric potential difference experienced by the proton. Moreover, the charge q0 on either particle has
the same magnitude. According to EPE = q0V (Equation 19.4), the losses in EPE for the electron and the
proton are the same. Conservation of energy, then, dictates that the electron gains the same amount of
kinetic energy as does the proton.
Both particles start from rest, so the gain in kinetic energy is equal to the final kinetic energies in each case,
which are the same for each particle. However, kinetic energy is 12 mv 2 (Equation 6.2), and the mass of the
electron is much less than the mass of the proton (see the inside of the front cover). Since the kinetic
energies are the same, the speed ve of the electron will be greater than the speed vp of the proton.
SOLUTION Let us assume that the proton accelerates from point A to point B. According to the energyconservation principle (including only kinetic and electric potential energies), we have
KE p, B  EPE p, B  KE p, A  EPE p, A
Final total energy
of proton at point B
Initial total energy
of proton at point A
or
KE p, B  EPE p, A  EPE p, B
Chapter 19 Solutions
2
Here we have used the fact that the initial kinetic energy is zero since the proton starts from rest. According
to Equation 19.3, the electric potential energy of a charge q0 is EPE = q0V, where V is the potential
experienced by the charge. We can, therefore, write the final kinetic energy of the proton as follows:
KEp, B  EPE p, A  EPE p, B  qp VA  VB 
(1)
The electron has a negative charge, so it accelerates in the direction opposite to that of the proton, or from
point B to point A. Energy conservation applied to the electron gives
KEe, A  EPEe, A  KEe, B  EPEe, B
Final total energy
of electron at point A
or
KE e, A  EPE e, B  EPE e, A
Initial total energy
of electron at point B
Here we have used the fact that the initial kinetic energy is zero since the electron starts from rest. Again
using Equation 19.3, we can write the final kinetic energy of the electron as follows:
KEe, A  EPEe, B  EPEe, A  qe VB  VA 
But an electron has a negative charge that is equal in magnitude to the charge on a proton, so qe = –qp.
With this substitution, we can write the kinetic energy of the electron as
KEe, A  EPEe, B  EPEe, A  qp VB  VA   qp VA  VB 
(2)
Comparing Equations (1) and (2), we can see that
KEp, B  KEe, A
or
1 m v2
2 p p
 12 meve2
Solving for the ratio ve/vp and referring to the inside of the front cover for the masses of the electron and
the proton, we obtain
ve
vp

mp
me
1.67 1027 kg
 42.8
9.111031 kg

13. REASONING
The electric potential at a distance r from a point charge q is given by
V = kq/r (Equation 19.6). The total electric potential due to the two charges is the algebraic sum of the
individual potentials.
SOLUTION Using Equation 19.6, we find that the total electric potential due to the two charges is
kq kq
k
V  1  2   q1  q2 
r
r
r
The distance r is one-half the distance between the charges, so r  12 1.20 m  for both charges. Thus, the
total electric potential midway between the charges is
k
8.99 109 N  m2 / C2
V   q1  q2  
3.40 106 C  6.10 106 C  4.05 104 V
1 1.20 m
r

2


Chapter 19 Solutions
3
15. SSM REASONING The potential of each charge q at a distance r away is given by Equation 19.6 as
V = kq/r. By applying this expression to each charge, we will be able to find the desired ratio, because the
distances are given for each charge.
SOLUTION According to Equation 19.6, the potentials of each charge are
VA 
kqA
rA
and
VB 
kqB
rB
Since we know that VA = VB, it follows that
kqA kqB

rA
rB
or
qB rB 0.43 m


 2.4
qA rA 0.18 m
32. REASONING The equipotential surfaces that surround a point charge q are spherical. Each has a
kq
potential V that is given by V 
(Equation 19.6), where k = 8.99  109 Nm2/C2 and r is the radius of the
r
sphere. Equation 19.6 can be solved for q, provided that we can obtain a value for the radius r. The radius
can be found from the given surface area A, since the surface area of a sphere is A  4 r 2 .
SOLUTION Solving V 
kq
(Equation 19.6) for q gives
r
Vr
q
k
Since the surface area of a sphere is A  4 r 2 , it follows that r 
A
. Substituting this result into the
4
expression for q gives
q
Vr V

k
k
A
490 V
1.1 m2

 1.6 108 C
4 8.99 109 N  m2 / C2
4
33. SSM REASONING The magnitude E of the electric field is given by Equation 19.7a (without the minus
V
sign) as E 
, where V is the potential difference between the two metal conductors of the spark plug,
s
and s is the distance between the two conductors. We can use this relation to find V.
SOLUTION The potential difference between the conductors is



V  E s  4.7  107 V/m 0.75  103 m  3.5  104 V
Chapter 19 Solutions
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V
35. SSM WWW REASONING AND SOLUTION From Equation 19.7a we know that E  
, where
s
V is the potential difference between the two surfaces of the membrane, and s is the distance between
them. If A is a point on the positive surface and B is a point on the negative surface, then V = VA  VB =
0.070 V. The electric field between the surfaces is
E 
V  VA
V  VB
V
0.070 V
  B
 A

 8.8  106 V/m

9
s
s
s
8.0  10 m
44. REASONING AND SOLUTION The capacitance is given by
C
k 0 A
d



5 8.85  1012 F/m 5  106 m2
8
1  10 m

2  108F
46. REASONING Equation 19.10 gives the capacitance as C = 0A/d, where  is the dielectric constant, and
A and d are, respectively, the plate area and separation. Other things being equal, the capacitor with the
larger plate area has the greater capacitance. The diameter of the circle equals the length of a side of the
square, so the circle fits within the square. The square, therefore, has the larger area, and the capacitor with
the square plates would have the greater capacitance.
To make the capacitors have equal capacitances, the dielectric constant must compensate for the larger area
of the square plates. Therefore, since capacitance is proportional to the dielectric constant, the capacitor
with square plates must contain a dielectric material with a smaller dielectric constant. Thus, the capacitor
with circular plates contains the material with the greater dielectric constant.
SOLUTION The area of the circular plates is Acircle = 
 12 L
2
, while the area of the square plates is
Asquare = L2. Using these areas and applying Equation 19.10 to each capacitor gives
C
 circle 0  12 L 
2
C
and
d
 square 0 L2
d
Since the values for C are the same, we have
 circle 0  12 L 
2
d
 circle 
4 square



 square 0 L2
d
4  3.00 

 3.82
or
 circle 
4 square
