QUICK QUIZ 23.1
(For the end of section 23.2)
You bring a positively charged rod toward a hanging sphere on a thread.
At first, as you bring the rod towards the sphere, the sphere is repelled.
Eventually, however, when the rod is very close to the sphere, the sphere is
attracted to the rod. Assuming that no charges leak from the rod or sphere, the
best explanation for these results is a) the sphere is positively charged, b) the
sphere is negatively charged, c) the sphere is neutral, or d) a, b, or c are possible.
QUICK QUIZ 23.1 ANSWER
(a) Recall that attraction will occur between objects that are oppositely charged or due
to charge realignment. Repulsion will occur between two objects that have the same
charge. However, charge realignment will not cause repulsion. If the sphere had a
net negative charge or if the sphere were neutral, the rod would never repel the
sphere. However, if the sphere had a net positive charge, at large distances the sphere
would appear as a positive point charge and repel the rod. At short distances,
significant charge realignment could occur and, due to the force dependence on
distance, the attraction of the negative charges on the left side of the sphere to the rod
could overcome the repulsive forces of the positive charges on the right side of the
sphere, even if there were
slightly more positive than
negative charges. You may
verify such an experiment
with a pith ball and charged
rod.
QUICK QUIZ 23.2
(For the end of section 23.3)
Two point charges, q1 = +q and q2 = –2q,
are fixed at positions on the x-axis, as shown
below. If you were to place a third point
charge somewhere so that the force on it
would be zero, you should place that charge
somewhere a) on the x-axis to the right of the
two charges, b) on the x-axis in between the
two charges, c) on the x-axis to the left of the
two charges, or d) off of the x-axis.
QUICK QUIZ 23.2 ANSWER
(c). Consider a positive point charge placed to the right of the two charges. From
Coulomb’s Law, the net force on this third charge is given by:
Fnet 
keq1qo keq2qo

.
2
2
r1o
r2o
Since the magnitude of q2 is greater than q1 and the distance from qo to q2 is less
than to q1, the force from q2 on qo will always be greater than from q1 on qo and the
net force won’t be zero. At positions in between the two charges, the force on qo
will be to the right from both charges. Only at some position to the left of both
charges will it be possible to balance the attractive force from q2 with the repulsive
force from q1.
QUICK QUIZ 23.3
(For the end of section 23.4)
Many types of molecules may be modeled as dipoles. For
example, two molecules modeled as simple dipoles are aligned on
the x-axis, as shown. Each dipole consists of a positive charge q
and negative charge –q, separated by a distance d. The distance
between the centers of the two dipoles is x. Considering the forces
due to all four charges, the net force of one dipole on the other will
be a) attractive, b) repulsive, c) zero, or d) it depends on the relative
distances d and x.
QUICK QUIZ 23.3 ANSWER
(a). You can analyze the situation by considering the forces on the left dipole. If the net
force on the left dipole is to the right, there is an attractive interaction. If the net force is to
the left, there is a repulsive interaction. The charges are numbered 1 through 4 with the
various forces on each charge for the left dipole shown on the diagram (the internal forces,
that is, the forces between the positive and negative charge within the dipole, are not shown
since they are not relevant to the forces between the two dipoles). Considering forces to the
right as positive and forces to the left as negative, the net force on the left dipole is:
F  F41  F32 - F42 - F31
keq 2
k eq 2
k eq 2


2 2
2
2
( x  d ) (x  d )
x
 x 2 (x  d )2  x 2 (x  d )2  2(x  d )2 (x  d )2 
 keq 

2
2 2
(
x

d
)
(
x

d
)
x


2
The denominator of the fraction will always be positive. The direction of the
force is therefore determined completely by the numerator. Looking at the
numerator only, we obtain
x 2 (x  d )2  x 2 (x  d )2  2(x  d )2 (x  d )2
 x 2 (2x 2  2d 2 )  2(x 2  d 2 )2  6x 2d 2  2d 4  2d 2 (3x 2  d 2 )
The minimum possible value for x will be exactly equal to d when the two
dipoles are touching. Even in this case, the quantity (3x2 – d2) will be positive.
For all values of x and d, therefore, the numerator will be positive and the net
force on the left dipole will be to the right. A similar analysis of the net force on
the right dipole will yield the same net force to the left, indicating an attractive
interaction between the two dipoles.
QUICK QUIZ 23.4
(For the end of section 23.5)
As illustrated in Figure 23.18 and below, an electric field is
produced on the axis of a uniformly charged ring of radius
a and total charge Q, at point P a distance x away from the
center of the ring. If the charge Q is then spread uniformly
into a disk of the same radius at the same location, the
electric field at point P a distance x away from the center of
this disk will be a) less than for the ring, b) greater than for
the ring, c) the same as for the ring, or d) impossible to
determine.
QUICK QUIZ 23.4 ANSWER
(b). As demonstrated in Example 23.8, the electric field on the axis
of the ring is due to the charge elements of the ring, all a distance r
from point P. In order to visualize the situation, one could
approximate the infinite number of infinitesimal charge elements
as a finite number of charge elements that add up to the total
charge Q. For example, if there were 1000 charge elements, each
element would have the same charge of Q/1000. If one were to
flatten the ring into a disk and in essence fill up the hole in the
ring, except for the outer elements, these charge elements would
now be distributed closer to point P than for the ring. In addition,
the component of the electric field along the axis would be greater
for charge elements closer to the center of the disk. Both of these
conditions would produce a larger electric field at point P.
QUICK QUIZ 23.5
(For the end of section 23.7)
In an experiment similar to the one pictured below, an electron is projected
horizontally at a speed vi into a uniform electric field pointing up. The magnitude
of the total vertical deflection, ye, of the electron is measured to be 1 mm. The
same experiment is repeated with a proton (whose mass is 1840 times that of the
electron) that is also projected horizontally at a speed vi into the same uniform
electric field. The magnitude of the total vertical deflection, yp, for the proton will
be a) 1840 mm, b) 1/1840 mm, c) 18402 mm, d) 1/(1840)2 mm, or e) the same as
the electron, 1 mm, since the gravitational forces and therefore mass make
negligible contributions to the situation.
QUICK QUIZ 23.5 ANSWER
(b). From Equation 23.17, y = -eEt2/2m, the vertical deflection is
inversely proportional to the mass. The time spent in the electric
field is completely determined by the initial horizontal speed,
which is the same for each particle. Therefore, the deflection is
completely determined by the mass and for the proton will be
1/1840 times as much as for the electron.