Physics 107 HOMEWORK ASSIGNMENT #15
Cutnell & Johnson, 7th edition
Chapter 18: Problem 43
Chapter 19: Problems 21, 23, 35, 54
**43 A small plastic ball with a mass of 6.50 x 10-3 kg and with a
charge of +0.150 µC is suspended from an insulating thread and
hangs between the plates of a capacitor (see the drawing). The ball is
in equilibrium, with the thread making an angle of 30.0° with respect
to the vertical. The area of each plate is 0.0150 m2. What is the
magnitude of the charge on each plate?
*21 Refer to Interactive Solution 19.21 to review one way in which this problem can be
solved. Two protons are moving directly toward one another. When they are very far apart, their
initial speeds are 3.00 x 106 m/s. What is the distance of closest approach?
*23 ssm Determine the electric potential energy for the array of three
charges shown in the drawing, relative to its value when the charges are
infinitely far away.
*35 ssm Equipotential surface A has a potential of 5650 V, while equipotential surface B has a
potential of 7850 V. A particle has a mass of 5.00 x 10-2 kg and a charge of +4.00 x 10-5 C. The
particle has a speed of 2.00 m/s on surface A. An outside force is applied to the particle, and it
moves to surface B, arriving there with a speed of 3.00 m/s. How much work is done by the
outside force in moving the particle from A to B?
54 The drawing shows four point charges. The value of q is 2.0 µC,
and the distance d is 0.96 m. Find the total potential at the location P.
Assume that the potential of a point charge is zero at infinity.
43. REASONING AND SOLUTION Since the thread makes an angle of 30.0° with the
vertical, it can be seen that the electric force on the ball, Fe, and the gravitational force, mg,
are related by
Fe = mg tan 30.0°
The force Fe is due to the charged ball being in the electric field of the parallel plate
capacitor. That is,
Fe  E qball
(1)
where qball is the magnitude of the ball's charge and E is the magnitude of the field due to
the plates. According to Equation 18.4 E is
E
q
0 A
(18.4)
where q is the magnitude of the charge on each plate and A is the area of each plate.
Substituting Equation 18.4 into Equation (1) gives
Fe  mg tan 30.0 
q qball
0 A
Solving for q yields
q
 0 Amg tan 30.0
qball





8.85  10 –12 C2 / N  m 2  0.0150 m 2 6.50  10 –3 kg 9.80 m/s 2 tan 30.0


0.150  10 –6 C
 3.25  10 –8 C
21. REASONING The only force acting on each proton is the conservative electric force.
Therefore, the total energy (kinetic energies plus electric potential energy) is conserved at all
points along the motion. For two points, A and B, the conservation of energy is expressed as
follows:
1 mv 2
A
2
 12 mv 2A 
Initial kinetic energies
of the two protons
EPE A
Initial electric
potential
energy

1 mv 2
B
2
 12 mvB2 
Final kinetic energies
of the two protons
EPE B
Final electric
potential
energy
The electric potential energy of two protons (charge on each proton = +e) that are separated
by a distance r can be found by combining the relations EPE = eV (Equation 19.3) and
V = ke/r (Equation 19.6) to give EPE = ke2/r. By using this expression for EPE in the
conservation of energy relation, we will be able to determine the distance of closest
approach.
SOLUTION When the protons are very far apart (rA = ), so that EPEA = 0 J. At the
distance rB of closest approach, the speed of each proton is momentarily zero (vB = 0 m/s).
With these substitutions, the conservation of energy equation reduces to
1 mv 2
A
2
 12 mv 2A 
ke2
rB
Solving for rB, the distance of closest approach, gives


8.99  109 N  m2 / C2 1.60  1019 C
ke2
rB 

2
mv 2A
1.67  1027 kg 3.00  106 m/s




2
 1.53  1014 m
We have taken the mass of a proton from the inside of the front cover of the text.
23. SSM REASONING Initially, the three charges are infinitely far apart. We will proceed as
in Example 8 by adding charges to the triangle, one at a time, and determining the electric
potential energy at each step. According to Equation 19.3, the electric potential energy EPE
is the product of the charge q and the electric potential V at the spot where the charge is
placed, EPE = qV. The total electric potential energy of the group is the sum of the energies
of each step in assembling the group.
SOLUTION Let the corners of the triangle be numbered clockwise as 1, 2 and 3, starting
with the top corner. When the first charge (q1 = 8.00 µC) is placed at a corner 1, the charge
has no electric potential energy, EPE1 = 0. This is because the electric potential V1 produced
by the other two charges at corner 1 is zero, since they are infinitely far away.
Once the 8.00-C charge is in place, the electric potential V2 that it creates at corner 2 is
V2 
kq1
r21
where r21 = 5.00 m is the distance between corners 1 and 2, and q1 = 8.00 C. When the
20.0-C charge is placed at corner 2, its electric potential energy EPE2 is
 kq 
EPE 2  q2V2  q2  1 
r 
 21 

 20.0  10
6


 8.99  109 N  m 2 /C2 8.00  106 C
C 

5.00 m


   0.288 J


The electric potential V3 at the remaining empty corner is the sum of the potentials due to the
two charges that are already in place on corners 1 and 2:
V3 
kq1 kq2

r31 r32
where q1 = 8.00 C, r31 = 3.00 m, q2 = 20.0 C, and r32 = 4.00 m. When the third charge
(q3 = 15.0 C) is placed at corner 3, its electric potential energy EPE3 is
 kq
q
kq 
q 
EPE3  q3V3  q3  1  2   q3 k  1  2 
r

r

 31 r32 
 31 r32 
 8.00  106 C 20.0  106 C 
 15.0  106 C 8.99  109 N  m 2 /C2 

   1.034 J
4.00 m 
 3.00 m



The electric potential energy of the entire array is given by
EPE  EPE1  EPE2  EPE3  0  0.288 J + ( 1.034 J) = 0.746 J
35. SSM REASONING The total energy of the particle on the equipotential surface A is
E A  12 mv A2  qVA . Similarly, the total energy of the particle when it reaches equipotential
surface B is EB  12 mvB2  qVB . According to Equation 6.8, the work W done by the outside
force is equal to the final total energy E B minus the initial total energy E A ; W  EB  E A .
SOLUTION The work done by the outside force in moving the particle from A to B is,
W  EB  E A 
 12 mvB2  qVB  –  12 mvA2  qVA   12 m vB2  vA2   q(VB  VA )
 12 (5.00 102 kg) (3.00 m/s) 2  (2.00 m/s) 2 
 (4.00 105 C) 7850 V  5650 V  0.213 J
54. REASONING The electric potential at a distance r from
a point charge q is given by Equation 19.6 as V  kq / r .
The total electric potential at location P due to the four
point charges is the algebraic sum of the individual
potentials.
q
d
SOLUTION The total electric potential at P is (see the
drawing)
V
k  q 
d

k  q 
2d

k  q 
d

k  q 
d

d
d
P
kq
2d
d
q
Substituting in the numbers gives

N  m2 
  8.99 109
2.0 106 C
2 
C 
kq
V
 
 9.4 103 V
2d
2  0.96 m 

q
q
