Note 2 Coulomb’s Law
Using the torsion electrometer, Coulomb showed that the forces between two small, spherical,
charged objects has the following magnitude.
Felectric = k
q1 q 2
r2
Here, the variables q’s are the charges of the two objects. The variable r is the separation
distance between the two point charges. The parameter k is the proportionality constant. It has
the following value.
N ⋅ m2
k = 8.99 ×10
C2
9
The unit of this constant can be derived from the other variables.
The direction of this force depends on the types of charge that are on the objects. If the two
charge types are the same, then the forces point away from each other. If the two charge types
are different, then the forces point toward each other.
If you view one of the charged object as the source of the force and the other object as the target
of the force, then the electric force is more compactly written as this.
!
q
q
Felectric = k source 2 target r̂
r
Here, the direction is indicated by the unit vector that points from the source to the target. This is
the electrostatic force between two point charges. The target charge is also known as the test
charge.
Note that it is usually easier to just write Coulomb’s law this way as you don’t have to calculate the
unit vector.
!
q
q
!
Felectric = k source 3 target r
r
Comparison to Gravity
Coulomb’s law has the same form as the law of universal gravitation. The primary difference is
that there is only one kind if mass and the force is always attractive.
!
m
m
Fgravity = −G source 2 target r̂
r
page 1
Example
Two charges A and B are fixed in the following positions. A third charge is place at position C.
What is the total electrostatic force on the charge at C? The charge q is 1 nC.
B
5m
–4q
–2q C
3m
A
4 m +q
6m
There are two forces on the charge at C. The force from charge at A on the charge at C is this.
!
q q
FA on C = k A2 C r̂A to C
rAC
The distance AC is
rAC = 5 m
The unit vector from A to C is
!
ˆ
rA to C
(−4 m)(iˆ) + (3 m)( j)
r̂A to C =
=
rAC
5m
The force A on C is
!
ˆ
(q)(−2q) (−4)(iˆ) + (3)( j)
8kq 2 ˆ −6kq 2 ˆ
FA on C = k
=
(i ) +
( j)
5
(5 m)2
53
53
The force from charge at B on the charge at C is this.
!
q q !
FB on C = k B3 C rB to C
rBC
The distance BC is
rAC = (−10 m)2 + (−5 m)2 = 125 m
The unit vector from B to C is
!
ˆ
rB to C
(−10 m)(iˆ) + (−5 m)( j)
r̂B to C =
=
rBC
125 m
The force B on C is
!
ˆ
(−4q)(−2q) (−10 m)(iˆ) + (−5 m)( j)
−80kq 2 ˆ −40kq 2 ˆ
FB on C = k
=
(i ) +
( j) 3
3
( 125 m)2
125 m
125
125
page 2
The total force is
!
8kq 2
−6kq 2 ˆ −80kq 2 ˆ −40kq 2 ˆ
ΣF = 3 (iˆ) +
( j) +
(i ) +
( j)
3
3
5
53
125
125
⎛ 8
⎛ 6
!
80 ⎞⎟⎟ ˆ
40 ⎞⎟⎟ ˆ
2⎜
⎜
ΣF = kq 2 ⎜⎜⎜
−
(
i
)
+
kq
−
−
( j)
⎟
⎜⎜ 125
3⎟
3⎟
⎟⎟⎠
⎟
⎜⎝ 125
⎠
⎝
125
125
!
ˆ
ΣF = kq 2(0.00676)(iˆ) + kq 2(−0.0766)( j)
Let’s check the direction of the net force. Graphically, the net force looks like this so our result
above seems good.
B
–2q C
FB on C
–4q
FA on C
A
net force
+q
The final value is
!
ˆ
ΣF = (6.08 ×10−11 N )(iˆ) + (−6.88 ×10−10 N )( j)
page 3
Example (in class exercise)
The following net force is applied on a charge C with a of charge (+1 nC) from two other fixed
charges.
!
ΣF = (1.00 ×10−10 N )(iˆ)
One of the fixed charges (A) is positioned relative to the charge in the following way with a charge
of (+3 nC).
+q C
5m
A
3 m +3q
The second fixed charge (B) is positioned along the horizontal line that passes through the fixed
charge (A) and a distance x from charge (A).
+q C
5m
B
A
3 m +3q
qB
x
a. What is the charge of charge (B)?
b. How far does it need to be placed from charge (A)?
page 4
Solution
We know the net force which is due to two electrostatic forces.
!
!
!
ΣF = (1.00 ×10−10 N )(iˆ) = FA + FB
The force from charge A is
!
3q 2
ˆ ) = kq 2 3 ( (−3)(iˆ) + (5)( j)
ˆ)
FA = k 3/2 ( (−3)(iˆ) + (5)( j)
34
34 3/2
!
ˆ⎤
FA = kq 2 ⎡⎣ (−0.04540)(iˆ) + (0.07566)( j)
⎦
The net force in terms if kq2 is
!
ΣF = (1.00 ×10−10 N )(iˆ) = kq 2(0.01113)(iˆ)
This means that the force from charge B is
!
ˆ
FB = kq 2(0.01113)(iˆ) − kq 2(−0.04540)(iˆ) − kq 2(0.07566)( j)
!
ˆ
F = kq 2(0.05653)(iˆ) − kq 2(0.07566)( j)
B
This can be done by only having a negative charge to the right of charge A. Let the charge of
charge B be written as –nq.
FA
+q C
5m
net force
FB
B
A
3 m +3q
nq
x
The other way to write the force from charge B is using Coulomb’s law which equals to the version
we found above.
!
FB = −k
nq 2
(x + 3)2 + 25
3
ˆ ) = kq 2(0.05653)(iˆ) − kq 2(0.07566)( j)
ˆ
( −(x + 3)(iˆ) + (5)( j)
This splits into two equations, one for each component.
kq 2
−kq
n(x + 3)
(x + 3)2 + 25
n(5)
2
3
(x + 3)2 + 25
= kq 2(0.05653) ⇒ n(x + 3) = (0.05653) (x + 3)2 + 25
3
= −kq 2(0.07566) ⇒ n = (0.01513) (x + 3)2 + 25
3
3
page 5
The ratio of the two equation above is
(x + 3) =
0.05653
0.01513
⇒ x = 0.73579
3
n = 0.01513 (0.73579 + 3)2 + 25 = 3.67877
a. The charge of charge B is
qB = −nq = −3.67877 nC
b. The distance from charge A is
x = 0.73579 m
page 6