Coulomb's law
Like charges repel each other while unlike charges attract each other. If the charges are at
rest then the force between them is known as the electrostatic force. The electrostatic
force between charges increases when the magnitude of the charges increases or the
distance between the charges decreases.
The electrostatic force was first studied in detail by Charles-Augustin de Coulomb around
1784. Through his observations he was able to show that the magnitude of the electrostatic
force between two point-like charges is inversely proportional to the square of the distance
between the charges. He also discovered that the magnitudeof the force is proportional to
the product of the charges. That is:
F∝Q1Q2r2,
where Q1 and Q2 are the magnitudes of the two charges respectively and r is the distance
between them. The magnitude of the electrostatic force between two point-like charges is
given by Coulomb's law.
Definition 1: Coulomb's law
Coulomb's law states that the magnitude of the electrostatic force between two
point charges is directly proportional to the product of the magnitudes of the
charges and inversely proportional to the square of the distance between them.
F=kQ1Q2r2,
The proportionality constant k is called the electrostatic constant and has the value:
9,0×109 N⋅m2⋅C−2 in free space.
Similarity of Coulomb's law to Newton's universal law of gravitation.
Notice how similar in form Coulomb's law is to Newton's universal law of gravitation
between two point-like particles:
FG=Gm1m2d2,
where m1 and m2 are the masses of the two point-like particles, d is the distance between
them, and G is the gravitational constant. Both are inverse-square laws.
Both laws represent the force exerted by particles (point masses or point charges) on each
other that interact by means of a field.
Example 1: Coulomb's law
Question
Two point-like charges carrying charges of +3 × 10−9 C and −5 × 10−9 C are 2 m apart.
Determine the magnitude of the force between them and state whether it is attractive or
repulsive.
Answer
Determine what is required
We are required to determine the force between two point charges given the charges and
the distance between them.
Determine how to approach the problem
We can use Coulomb's law to calculate the magnitude of the force.
F=kQ1Q2r2
Determine what is given
We are given:



Q1 = +3 × 10−9 C
Q2 = −5 × 10−9 C
r=2m
We know that k=9,0×109 N⋅m2⋅C−2.
We can draw a diagram of the situation.
Check units
All quantities are in SI units.
Determine the magnitude of the force
Using Coulomb's law we have
F=kQ1Q2r2=(9,0×109)(3×10−9)(5×10−9)(2)2=3,37×10−8 N
Thus the magnitude of the force is 3,37 × 10−8 N. However since the point charges have
opposite signs, the force will be attractive.
Free body diagram
We can draw a free body diagram to show the forces. Each charge experiences a force with
the same magnitude and the forces are attractive, so we have:
Next is another example that demonstrates the difference in magnitude between the
gravitational force and the electrostatic force.
Example 2: Coulomb's law
Question
Determine the magnitudes of the electrostatic force and gravitational force between two
electrons 10−10 m apart (i.e. the forces felt inside an atom) and state whether the forces are
attractive or repulsive.
Answer
Determine what is required
We are required to calculate the electrostatic and gravitational forces between two
electrons, a given distance apart.
Determine how to approach the problem
We can use:
Fe=kQ1Q2r2
to calculate the electrostatic force and
Fg=Gm1m2d2
to calculate the gravitational force.
Determine what is given



Q1=Q2=1,6×10−19 C (The charge on an electron)
m1=m2=9,1×10−31 kg (The mass of an electron)
r=d=1×10−10 m
We know that:


k=9,0×109 N⋅m2⋅C−2
G=6,67×10−11 N⋅m2⋅kg−2
All quantities are in SI units.
We can draw a diagram of the situation.
Calculate the electrostatic force
Fe=kQ1Q2r2=(9,0×109)(1,60×10−19)(1,60×10−19)(10−10)2=2,30×10−8 N
Hence the magnitude of the electrostatic force between the electrons is 2,30×10−8 N.
Since electrons carry like charges, the force is repulsive.
Calculate the gravitational force
Fg=Gm1m2d2=(6,67×10−11)(9,11×10−31)(9,11×10−31)(10−10)2=5,54×10−51 N
The magnitude of the gravitational force between the electrons is 5,54 × 10−51 N.
Remember that the gravitational force is always an attractive force.
Notice that the gravitational force between the electrons is much smaller than the
electrostatic force.
Tip:
We can apply Newton's third law to charges because two charges exert forces of equal
magnitude on one another in opposite directions.
Tip:
Choosing a positive direction
When substituting into the Coulomb's law equation, one may choose a positive direction
thus making it unnecessary to include the signs of the charges. Instead, select a positive
direction. Those forces that tend to move the charge in this direction are added, while forces
acting in the opposite direction are subtracted.
Example 3: Coulomb's law
Question
Three point charges are in a straight line. Their charges
are Q1=+2×10−9 C, Q2=+1×10−9 C and Q3=−3×10−9 C. The distance
between Q1 and Q2 is 2 × 10−2 m and the distance between Q2 and Q3 is 4 × 10−2 m. What
is the net electrostatic force on Q2 due to the other two charges?
Answer
Determine what is required
We need to calculate the net force on Q2. This force is the sum of the two electrostatic
forces - the forces between Q1 on Q2 and Q3 on Q2.
Determine how to approach the problem

We need to calculate the two electrostatic forces on Q2, using Coulomb's law.

We then need to add up the two forces using our rules for adding vector quantities,
because force is a vector quantity.
Determine what is given
We are given all the charges and all the distances.
Calculate the magnitude of the forces.
Force on Q2 due to Q1:
F1=kQ1Q2r2=(9,0×109)(2×10−9)(1×10−9)(2×10−2)2=(9,0×109)(2×10−9)(1×10−9)(4×10−
4)=4,5×10−5 N
Force on Q2 due to Q3:
F3=kQ2Q3r2=(9,0×109)(1×10−9)(3×10−9)(4×10−2)2=(9,0×109)(1×10−9)(3×10−9)(16×10
−4)=1,69×10−5 N
Vector addition of forces
We know the force magnitudes but we need to use the charges to determine whether the
forces are repulsive or attractive. It is helpful to draw the force diagram to help determine
the final direction of the net force on Q2. We choose the positive direction to be to the right
(the positive x-direction).
The force between Q1 and Q2 is repulsive (like charges). This means that it pushes Q2 to
the right, or in the positive direction.
The force between Q2 and Q3 is attractive (unlike charges) and pulls Q2 to the right.
Therefore both forces are acting in the positive direction.
Therefore,
FR=4,5×10−5 N+1,69×10−5 N=6,19×10−5 N
The resultant force acting on Q2 is 6,19 × 10−5 N to the right.
Example 4: Coulomb's law
Question
Three point charges form a right-angled triangle. Their charges
are Q1=4×10−9 C=4 nC, Q2=6×10−9 C=6 nC and Q3=−3×10−9 C=−3 nC. The distance
between Q1 and Q2 is 5 × 10−2 m and the distance between Q1 and Q3 is 3 × 10−2 m. What
is the net electrostatic force on Q1 due to the other two charges if they are arranged as
shown?
Answer
Determine what is required
We need to calculate the net force on Q1. This force is the sum of the two electrostatic
forces - the forces of Q2 on Q1 and Q3 on Q1.
Determine how to approach the problem

We need to calculate, using Coulomb's law, the electrostatic force exerted
on Q1 by Q2, and the electrostatic force exerted on Q1 by Q3.

We then need to add up the two forces using our rules for adding vector quantities,
because force is a vector quantity.
Determine what is given
We are given all the charges and two of the distances.
Calculate the magnitude of the forces.
The magnitude of the force exerted by Q2 on Q1, which we will call F2, is:
F2=kQ1Q2r2=(9,0×109)(4×10−9)(6×10−9)(5×10−2)2=(9,0×109)(4×10−9)(6×10−9)(25×10
−4)=8,630×10−5 N
The magnitude of the force exerted by Q3 on Q1, which we will call F3, is:
F3=kQ1Q3r2=(9,0×109)(4×10−9)(3×10−9)(3×10−2)2=(9,0×109)(4×10−9)(3×10−9)(9×10−
4)=1,199×10−4 N
Vector addition of forces
This is a two-dimensional problem involving vectors. We have already solved many twodimensional force problems and will use precisely the same procedure as before.
Determine the vectors on the Cartesian plane, break them into components in the x- and ydirections, and then sum components in each direction to get the components of the
resultant.
We choose the positive directions to be to the right (the positive x-direction) and up (the
positive y-direction). We know the magnitudes of the forces but we need to use the signs of
the charges to determine whether the forces are repulsive or attractive. Then we can use a
diagram to determine the directions.
The force between Q1 and Q2 is repulsive (like charges). This means that it pushes Q1 to
the left, or in the negative x-direction.
The force between Q1 and Q3 is attractive (unlike charges) and pulls Q1 in the positive ydirection.
We can redraw the diagram as a free-body diagram illustrating the forces to make sure we
can visualise the situation:
Resultant force
The magnitude of the resultant force acting on Q1 can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the x- and ydirections:
F2RFRFR=F22+F23 by Pythagoras'
theorem=(8,630×10−5)2+(1,199×10−4)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√=1,48×
10−4 N
and the angle, θR made with the x-axis can be found using trigonometry.
tan(θR)tan(θR)θRθR=y-componentxcomponent=1,199×10−48,630×10−5=tan−1(1,199×10−48,630×10−5)=54,25° to 2
decimal places
The final resultant force acting on Q1 is 1,48 × 10−4 N acting at an angle of 54,25° to the
negative x-axis or 125,75°to the positive x-axis.
We mentioned in grade 10 that charge placed on a spherical conductor spreads evenly
along the surface. As a result, if we are far enough from the charged sphere,
electrostatically, it behaves as a point-like charge. Thus we can treat spherical conductors
(e.g. metallic balls) as point-like charges, with all the charge acting at the centre.
Exercise 1: Electrostatic forces
Problem 1:
Calculate the electrostatic force between two charges of +6 nC and +1 nC if they are
separated by a distance of 2 mm.
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Answer 1:
Fe=kQ1Q2r2=(9,0×109)(6×10−9)(1×10−9)(2×10−3)2=1,35×10−2 N
This force is repulsive since it is between two like charges.
Problem 2:
What is the magnitude of the repulsive force between two pith balls (a pith ball is a small,
light ball that can easily be charged) that are 8 cm apart and have equal charges of −30
nC?
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Answer 2:
Fe=kQ1Q2r2=(9,0×109)(30×10−9)(30×10−9)(8×10−2)2=1,27×10−3 N
This force is repulsive since it is between two like charges.
Problem 3:
How strong is the attractive force between a glass rod with a 0,7 μC charge and a silk cloth
with a −0,6 μC charge, which are 12 cm apart, using the approximation that they act like
point charges?
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Answer 3:
Fe=kQ1Q2r2=(9,0×109)(0,6×10−6)(0,7×10−6)(12×10−2)2=0,26 N
This force is attractive since it is between two unlike charges.
Problem 4:
Two point charges exert a 5 N force on each other. What will the resulting force be if the
distance between them is increased by a factor of three?
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Answer 4:
Let the charges be Q1 and Q2. For the first situation we have:
Fe1=kQ1Q2r2
Now we increase the distance by a factor of three so we get:
Fe2=kQ1Q29r2
We now note the following:
(Fe1)(r2)=kQ1Q2
and:
(Fe2)(9r2)=kQ1Q2
Therefore:
(Fe1)(r2)Fe1∴Fe2=(Fe2)(9r2)=(Fe2)(9)=0,56 N
Problem 5:
Two point charges are brought closer together, increasing the force between them by a
factor of 25. By what factor was their separation decreased?
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Answer 5:
Let the charges be Q1 and Q2. For the first situation we have:
Fe1=kQ1Q2r2
Now we decrease by an unknown factor, x, the distance and we get:
Fe2=kQ1Q2(xr)2
We now note the following:
and:
(Fe2)=25kQ1Q2r2
Therefore:
25(kQ1Q2r2)25∴x=kQ1Q2(xr)2=1x2=0,2
Problem 6:
If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is
the magnitude of the force acting between them?
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Answer 6:
Fe=kQ1Q2r2=(9,0×109)(1)(1)(1×103)2=9000 N
This force is repulsive since it is between two like charges.
You should notice that even at a distance as large as 1 km, the repulsive force is substantial
because 1 C is a very significant amount of charge.
Problem 7:
Calculate the distance between two charges of +4 nC and −3 nC if the electrostatic force
between them is 0,005 N.
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Answer 7:
Fe0,0050,005d2d2=kQ1Q2r2=(9,0×109)(4×10−9)(3×10−9)(d)2=1,07×10−9=2,1576×10−5
=4,6×10−3 m
Problem 8:
For the charge configuration shown, calculate the resultant force on Q2 if:





Q1=2,3 × 10−7 C
Q2=4 × 10−6 C
Q3=3,3 × 10−7 C
r1=2,5 × 10−1 m
r2=3,7 × 10−2 m
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Answer 8:
We first calculate the force on Q2 from Q3:
Fe1=kQ1Q2r2=(9,0×109)(2,3×10−7)(4×10−6)(3,7×10−2+2,5×10−1)2=1,00×10−1 N
And then we calculate the force of Q1 on Q2. Note that for this force we must add r1 and r2.
Fe2=kQ2Q3r2=(9,0×109)(4×10−6)(3,3×10−7)(2,5×10−1+3,7×10−2)2=8,67 N
Next we note that the force of Q3 on Q2 is repulsive and the force of Q1 on Q2 is also
repulsive. So these two forces act in the same direction (towards the right). The resultant
force is:
FeR=Fe1+Fe2=8,67 N+0,1 N=8,77 N to the right.
Problem 9:
For the charge configuration shown, calculate the charge on Q3 if the resultant force
on Q2 is 6,3 × 10−1 N to the right and:




Q1=4,36 × 10−6 C
Q2=−7 × 10−7 C
r1=1,85 × 10−1 m
r2=4,7 × 10−2 m
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Answer 9:
We are told that the resultant force is 6,3 × 10−5 N to the right. Since the force
of Q1 on Q2 is attractive, the force of Q3 on Q2 must be repulsive to cause a resultant force
to the right (if it was also attractive, the resultant force would be to the left). So we know
that Q3 must be negative.
We first calculate the force on Q2 from Q1:
Fe1=kQ1Q2r2=(9,0×109)(4,36×10−6)(7×10−7)(1,85×10−1+4,7×10−2)2=0,51 N
Next we use this and the resultant force to find the force on Q2 from Q3
FeRFe2=Fe1+Fe2=6,3×10−1 N−0,51 N=0,12 N
And then we calculate the charge on Q3:
Fe20,122,6×10−4Q3=kQ2Q3r2=(9,0×109)(7×10−7)(Q3)(4,7×10−2)2=(6,293×103)(Q3)=4,
2×10−8 C
Problem 10:
Calculate the resultant force on Q1 given this charge configuration:
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Answer 10:
We first calculate the force on Q1 from Q2:
Fe=kQ1Q2r2=(9,0×109)(1×10−9)(2×10−9)(0,07)2=3,7×10−6 N
And then we calculate the force of Q3 on Q1:
Fe=kQ1Q3r2=(9,0×109)(3×10−9)(2×10−9)(0,04)2=3,4×10−5 N
The magnitude of the resultant force acting on Q1 can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the x- and ydirections:
F2RFR=F2x+F2y=(3,7×10−6)2+(3,4×10−5)2−−−−−−−−−−−−−−−−−−−−−−−√=3,42×10−
5N
We can find the angle using trigonometry:
tanθRθR=y-componentx-component=3,42×10−53,7×10−6=9,2432…=83,8°
The final resultant force acting on Q1 is 3,42 × 10−5 N acting at an angle of 83,8° to the
negative x-axis.
Problem 11:
Calculate the resultant force on Q1 given this charge configuration:
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Answer 11:
We first calculate the force on Q1 from Q2:
Fe=kQ1Q2r2=(9,0×109)(1×10−9)(9×10−9)(0,65)2=1,92×10−7 N
And then we calculate the force of Q3 on Q1:
Fe=kQ1Q3r2=(9,0×109)(3×10−9)(9×10−9)(0,4)2=1,52×10−6 N
The magnitude of the resultant force acting on Q1 can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the x- and ydirections:
F2RFR=F2x+F2y=(1,92×10−7)2+(1,52×10−6)2−−−−−−−−−−−−−−−−−−−−−−−−−√=1,52
×10−6 N
We can find the angle using trigonometry:
tanθRθR=y-componentx-component=1,52×10−61,92×10−7=7,91666…=82,80°
The final resultant force acting on Q1 is 1,52 × 10−6 N acting at an angle of 82,80° to the
positive x-axis.
Problem 12:
Calculate the resultant force on Q2 given this charge configuration:
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Answer 12:
We first calculate the force on Q2 from Q1:
Fe=kQ1Q2r2=(9,0×109)(8×10−9)(3×10−9)(0,05)2=8,63×10−5 N
And then we calculate the force of Q3 on Q2:
Fe=kQ1Q3r2=(9,0×109)(3×10−9)(2×10−9)(0,03)2=5,99×10−5 N
The magnitude of the resultant force acting on Q2 can be calculated from the forces using
Pythagoras' theorem because there are only two forces and they act in the x- and ydirections:
F2RFR=F2x+F2y=(8,63×10−5)2+(5,99×10−5)2−−−−−−−−−−−−−−−−−−−−−−−−−√=1,05
×10−4 N
We can find the angle using trigonometry:
tanθRθR=y-componentx-component=5,99×10−58,63×10−5=0,694…=34,76°
The final resultant force acting on Q1 is 1,05 × 10−4 N acting at an angle of 34,76° to the
positive x-axis.