614
CHAPTE R 21 COU LOM B’S L AW
axis extending through the two particles, radially away from particle 2. (As with
other unit vectors, has a magnitude of exactly 1 and no unit; its purpose is to
point, like a direction arrow on a street sign.) With these decisions, we write the
electrostatic force as
(Coulomb’s law),
(21-1)
where r is the separation between the particles and k is a positive constant called
the electrostatic constant or the Coulomb constant. (We’ll discuss k below.)
Let’s first check the direction of the force on particle 1 as given by Eq. 21-1. If
q 1 and q 2 have the same sign, then the product q 1 q 2 gives us a positive result. So,
Eq. 21-1 tells us that the force on particle 1 is in the direction of . That checks, because particle 1 is being repelled from particle 2. Next, if q 1 and q 2 have opposite
signs, the product q 1 q 2 gives us a negative result. So, now Eq. 21-1 tells us that the
force on particle 1 is in the direction opposite . That checks because particle 1 is
being attracted toward particle 2.
An Aside. Here is something that is very curious. The form of Eq. 21-1 is the
same as that of Newton’s equation (Eq. 13-3) for the gravitational force between
two particles with masses m 1 and m 2 and separation r:
(Newton’s law),
(21-2)
where G is the gravitational constant. Although the two types of forces are wildly
different, both equations describe inverse square laws (the 1/r
2
dependences)
that involve a product of a property of the interacting particles—the charge in
one case and the mass in the other. However, the laws differ in that gravitational
forces are always attractive but electrostatic forces may be either attractive or repulsive, depending on the signs of the charges. This difference arises from the fact
that there is only one type of mass but two types of charge.
Unit. The SI unit of charge is the coulomb. For practical reasons having to do
with the accuracy of measurements, the coulomb unit is derived from the SI unit ampere for electric current i. We shall discuss current in detail in Chapter 26, but here
let’s just note that current i is the rate dq/dt at which charge moves past a point or
through a region:
(electric current).
( 2 1 - 3 )
Rearranging Eq. 21-3 and replacing the symbols with their units (coulombs C,
amperes A, and seconds s) we see that
1 C ϭ (1 A)(1 s).
Force Magnitude. For historical reasons (and because doing so simplifies
many other formulas), the electrostatic constant k in Eq. 21-1 is often written as
1/4p´ 0 . Then the magnitude of the electrostatic force in Coulomb’s law becomes
(Coulomb’s law).
( 2 1 - 4 )
The constants in Eqs. 21-1 and 21-4 have the value
(21-5)
The quantity ´ 0 , called the permittivity constant, sometimes appears separately in
equations and is
´ 0 ϭ 8.85 ϫ 10
Ϫ12
C
2
/N и m
2
.
( 2 1 - 6 )
Working a Problem. Note that the charge magnitudes appear in Eq. 21-4,
which gives us the force magnitude. So, in working problems in this chapter, we
use Eq. 21-4 to find the magnitude of a force on a chosen particle due to a second
k ϭ
1
4p´ 0
ϭ 8.99 ϫ 10
9
Nиm
2
/C
2
.
F ϭ
1
4p´ 0
͉q 1 ͉͉q 2 ͉
r
2
i ϭ
dq
dt
F
: ϭ G
m 1 m 2
r
2
r ˆ
r ˆ
r ˆ
F
: ϭ k
q 1 q 2
r
2
r ˆ
r ˆ
CHAPTE R 21 COU LOM B’S L AW
axis extending through the two particles, radially away from particle 2. (As with
other unit vectors, has a magnitude of exactly 1 and no unit; its purpose is to
point, like a direction arrow on a street sign.) With these decisions, we write the
electrostatic force as
(Coulomb’s law),
(21-1)
where r is the separation between the particles and k is a positive constant called
the electrostatic constant or the Coulomb constant. (We’ll discuss k below.)
Let’s first check the direction of the force on particle 1 as given by Eq. 21-1. If
q 1 and q 2 have the same sign, then the product q 1 q 2 gives us a positive result. So,
Eq. 21-1 tells us that the force on particle 1 is in the direction of . That checks, because particle 1 is being repelled from particle 2. Next, if q 1 and q 2 have opposite
signs, the product q 1 q 2 gives us a negative result. So, now Eq. 21-1 tells us that the
force on particle 1 is in the direction opposite . That checks because particle 1 is
being attracted toward particle 2.
An Aside. Here is something that is very curious. The form of Eq. 21-1 is the
same as that of Newton’s equation (Eq. 13-3) for the gravitational force between
two particles with masses m 1 and m 2 and separation r:
(Newton’s law),
(21-2)
where G is the gravitational constant. Although the two types of forces are wildly
different, both equations describe inverse square laws (the 1/r
2
dependences)
that involve a product of a property of the interacting particles—the charge in
one case and the mass in the other. However, the laws differ in that gravitational
forces are always attractive but electrostatic forces may be either attractive or repulsive, depending on the signs of the charges. This difference arises from the fact
that there is only one type of mass but two types of charge.
Unit. The SI unit of charge is the coulomb. For practical reasons having to do
with the accuracy of measurements, the coulomb unit is derived from the SI unit ampere for electric current i. We shall discuss current in detail in Chapter 26, but here
let’s just note that current i is the rate dq/dt at which charge moves past a point or
through a region:
(electric current).
( 2 1 - 3 )
Rearranging Eq. 21-3 and replacing the symbols with their units (coulombs C,
amperes A, and seconds s) we see that
1 C ϭ (1 A)(1 s).
Force Magnitude. For historical reasons (and because doing so simplifies
many other formulas), the electrostatic constant k in Eq. 21-1 is often written as
1/4p´ 0 . Then the magnitude of the electrostatic force in Coulomb’s law becomes
(Coulomb’s law).
( 2 1 - 4 )
The constants in Eqs. 21-1 and 21-4 have the value
(21-5)
The quantity ´ 0 , called the permittivity constant, sometimes appears separately in
equations and is
´ 0 ϭ 8.85 ϫ 10
Ϫ12
C
2
/N и m
2
.
( 2 1 - 6 )
Working a Problem. Note that the charge magnitudes appear in Eq. 21-4,
which gives us the force magnitude. So, in working problems in this chapter, we
use Eq. 21-4 to find the magnitude of a force on a chosen particle due to a second
k ϭ
1
4p´ 0
ϭ 8.99 ϫ 10
9
Nиm
2
/C
2
.
F ϭ
1
4p´ 0
͉q 1 ͉͉q 2 ͉
r
2
i ϭ
dq
dt
F
: ϭ G
m 1 m 2
r
2
r ˆ
r ˆ
r ˆ
F
: ϭ k
q 1 q 2
r
2
r ˆ
r ˆ