Example 16-2 Calculating Electric Force
(a) What is the electric force (magnitude and direction) between two electrons separated by a distance of 10.0 cm = 0.100
m? (b) Suppose you could remove all the electrons from a drop of water 1.00 mm in radius (see Example 16-1 in Section
16-2) and clump them into a ball 1.00 mm in radius. If this ball of electrons is 10.0 cm from the drop of water from which
they were removed, what is the magnitude of the electric force between the drop of water and the ball of electrons?
Set Up
The two electrons repel because both have a
negative charge q = 2e = 21.602 * 10219 C.
From Example 16-1, the combined charge of
all of the electrons in a water drop of this size
is 2224 C; the water drop was initially neutral, so the charge of the water drop after all
of the electrons have been removed is +224 C.
Since the ball of electrons and the water drop
(with electrons removed) have opposite signs
of charge, they attract each other.
We can use Coulomb’s law (Equation
16-1) in both parts of this problem. That’s
because in both cases the charged objects are
much smaller than the distance that separates
them, so we can treat them as point charges.
(Indeed, electrons are very small even compared to the dimensions of an atom.)
Solve
(a) Find the magnitude of the force that each
electron exerts on the other.
Coulomb’s law:
(a)
Fq1 on q2 = Fq2 on q1 =
r
electron
F
k q1   q2 
2
electron
F
0.100 m
(16-1)
(b)
1.00 mm
1.00 mm
F
water
drop
F
0.100 m
electrons
For the two electrons, the charges are
q1 = q2 = 2e = 21.602 * 10219 C
The distance between the electrons is r = 0.100 m. From Equation
16-1, the magnitude of the force that each electron exerts on each
other is
F =
18.99 * 109 N # m2 >C 2 2  -1.602 * 10-19 C   -1.602 * 10-19 C 
= 2.31 * 10
226
(b) Find the magnitude of the force between
the ball of electrons and the water drop from
which they were extracted.
10.100 m2 2
N
For the ball of electrons and the water drop with electrons removed,
the charges are
q1 = 2224 C, q2 = +224 C
The distance between the two objects is r = 0.100 m. From Equation 16-1,
the two objects exert forces on each other of magnitude
F =
18.99 * 109 N # m2 >C 2 2  -224 C   +224 C 
16
= 4.51 * 10
Reflect
N
10.100 m2 2
The repulsive force between the two electrons is tiny because each particle carries only a tiny amount of charge.
By contrast, the attractive force between the ball of electrons and the electron-free water drop is immense. To put
this force into perspective, a solid cube of lead with a weight of 4.51 * 1016 N would be 7.4 kilometers on a side!
This is the magnitude of force that you would have to exert to keep the electrons from flying back into the water
drop. There is no known way to produce a force of this magnitude, which is why you’ll never see an object with all
of its electrons removed. It’s relatively easy to remove a small fraction of an object’s electrons, as for the fur and the
glass rod in Figure 16-2b. But removing all of the electrons from a piece of fur or a glass rod is not a practical thing
to do.