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Solutions to Homework for May 10
3. When the charged insulating rod is brought near to (but not touching) the sphere, the free electrons in the
sphere will move. If the rod is negatively charged, the free electrons will move to the side of the sphere that
is opposite to the side where the rod is; if the rod is positively charged, the free electrons will migrate to the
side of the sphere where the rod is. In either case, the region of the sphere near the vicinity of the rod will
acquire a charge that has the opposite sign as the charge on the rod.
a. Since oppositely charged objects always attract each other, the rod and sphere will always experience a
mutual attraction.
b. Since the side of the sphere in the vicinity of the rod will always have charge that is opposite in sign to the
charge on the rod, the rod and the sphere will always attract each other. They never repel each other.
8.
Identical point charges are fixed to opposite corners of a 1
square, as shown in the figure at the right.
There exists an electric field at each point in space in the
vicinity of this configuration. The electric field at a given
point is the resultant of the electric field at that point due to
the charge at corner 1 and the charge at corner 3.
2
C
The magnitude of the electric field due to a single point
charge q is given by Equation 18.3: E  k q / r 2 ,
4
3
where r is the distance from the charge to the point in question.
Since the point charges at corners 1 and 3 are identical, they each have the same value of q. Furthermore,
they are equidistant from the center point C. Therefore, at the center of the square, each charge gives rise to
an electric field that is equal in magnitude and opposite in direction. The resultant electric field at the center
of the square is, therefore, zero. Note that this result is independent of the polarity of the charges.
The distance from either empty corner, 2 or 4, to either of the charges is the same. Since the charges are
equidistant from either empty corner, each charge gives rise to an electric field that is equal in magnitude.
Since the direction of the electric field due to a point charge is radial (radially inward for negative charges
and radially outward for positive charges), we see that the electric fields due to each of the two point charges
will be mutually perpendicular. Their resultant can be found by using the Pythagorean theorem.
The magnitude of the force experienced by a third point charge placed in this system is, from
Equation 18.2: F  q0 E , where q0 is the magnitude of the charge and E is the magnitude of the electric
field at the location of the charge. Since the electric field is zero at the center of the square, a third point
charge will experience no force there. Thus, a third point charge will experience the greater force at one of
the empty corners of the square.
10. The magnitude of the electrostatic force that acts on particle 1 is given by Coulomb’s law as
F  k q1 q2 / r 2 . This equation can be used to find the magnitude q2 of the charge.
SOLUTION Solving Coulomb’s law for the magnitude q2 of the charge gives
 3.4 N  0.26 m 
F r2
q2 

 7.3  106 C
9
2
2

6
k q1
9.0  10 N  m /C 3.5  10 C
2



(18.1)
Since q1 is positive and experiences an attractive force, the charge q2 must be negative
25. Knowing the electric field at a spot allows us to calculate the force that acts on a charge placed at that spot,
without knowing the nature of the object producing the field. This is possible because the electric field is
defined as E = F/q0, according to Equation 18.2. This equation can be solved directly for the force F, if the
field E and charge q0 are known.
SOLUTION Using Equation 18.2, we find that the force has a magnitude of


F  E q0   260 000 N/C  7.0  10–6 C  1.8 N
If the charge were positive, the direction of the force would be due west, the same as the direction of the
field. But the charge is negative, so the force points in the opposite direction or due east. Thus, the force on
the charge is 1.8 N due east.
27. Two forces act on the charged ball (charge q); they are the downward force of gravity mg and the electric
force F due to the presence of the charge q in the electric field E. In order for the ball to float, these two
forces must be equal in magnitude and opposite in direction, so that the net force on the ball is zero (Newton's
second law). Therefore, F must point upward, which we will take as the positive direction. According to
Equation 18.2, F = qE. Since the charge q is negative, the electric field E must point downward, as the
product qE in the expression F = qE must be positive, since the force F points upward. The magnitudes of
the two forces must be equal, so that mg  q E . This expression can be solved for E.
SOLUTION The magnitude of the electric field E is
mg (0.012 kg)(9.80 m/s2 )
E

 6.5 103 N/C
–6
q
18 10 C
As discussed in the reasoning, this electric field points downward .
29. a. The drawing shows the two point charges q1 and q2. Point A is located at x = 0 cm, and point B is at x =
+6.0 cm.
E1
A
3.0 cm
3.0 cm
B
3.0 cm
q2
q1
E2
Since q1 is positive, the electric field points away from it. At point A, the electric field E1 points to the left,
in the x direction. Since q2 is negative, the electric field points toward it. At point A, the electric field E2
points to the right, in the +x direction. The net electric field is E = E1 + E2. We can use Equation 18.3,
E  k q / r 2 , to find the magnitude of the electric field due to each point charge.
b. The drawing shows the electric field produced by the charges q1 and q2 at point B, which is located at x =
+6.0 cm.
A
3.0 cm
3.0 cm
q1
B
q2
3.0 cm
E1
E2
Since q1 is positive, the electric field points away from it. At point B, the electric field points to the right, in the
+x direction. Since q2 is negative, the electric field points toward it. At point B, the electric field points to the
right, in the +x direction. The net electric field is E = +E1 + E2.
SOLUTION
a.
The net electric field at the origin (point A) is E = E1 + E2:
E   E1  E2 


k q1
r12

k q2
r22

2
3.0  102 m 
 8.99  109 N  m 2 /C2 8.5  106 C
  8.99  109 N  m2 /C2  21  106 C 
2
9.0  102 m 
 6.2  107 N/C
The minus sign tells us that the net electric field points along the x axis.
b.
The net electric field at x = +6.0 cm (point B) is E = E1 + E2:
E  E1  E2 
k q1
r12

k q2
r22
8.99  109 N  m 2 /C2 8.5  106 C  8.99  109 N  m 2 /C2  21  106 C 



2
2
2
3.0  10 m 
3.0  102 m 
 2.9  108 N/C
The plus sign tells us that the net electric field points along the +x axis.
33. Since the charged droplet (charge = q) is suspended motionless in the electric field E, the net force on the
droplet must be zero. There are two forces that act on the droplet, the force of gravity W  mg , and the
electric force F = qE due to the electric field. Since the net force on the droplet is zero, we conclude that
mg  q E . We can use this reasoning to determine the sign and the magnitude of the charge on the droplet.
SOLUTION
a. Since the net force on the droplet is zero, and the weight of magnitude W
points downward, the electric force of magnitude F  q E must point
upward. Since the electric field points upward, the excess charge on the
droplet must be positive in order for the force F to point upward.
b. Using the expression mg  q E , we find that the magnitude of the excess charge on the droplet is
q 
mg (3.50 10 –9 kg)(9.80 m/s 2 )

 4.04  10 –12 C
E
8480 N/C
The charge on a proton is 1.60  10–19 C, so the excess number of protons is

 4.04 10–12 C  1.601 proton

–19
10
C
2.53 107 protons