Like charges repel each other while opposite 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 Coulomb around 1784. Through his observations he was able
to show that the electrostatic force between two point-like
charges is inversely proportional to the square of the distance
between the objects. He also discovered that the force is
proportional to the product of the charges on the two objects. That is:
F
∝
Q
1
Q
2
r
2
,
F
∝
Q
1
Q
2
r
2
,
(1)where Q1Q1 is the charge on the one point-like object, Q2Q2 is
the charge on the second, and rr is the distance between the two.
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 magnitudes of each charge and
inversely proportional to the square of the distance between the
charges:
F
=
k
Q
1
Q
2
r
2
F
=
k
Q
1
Q
2
r
2
(2)The proportionality constant kk is called the
electrostatic constant and has the value:
k
=
8
,
99
×
10
9
N
·
m
2
·
C
-
2
.
k
=
8
,
99
×
10
9
N
·
m
2
·
C
-
2
.
(3)
Notice how similar Coulomb's Law is to the form
of Newton's Universal Law of Gravitation between two point-like
particles:
F
G
=
G
m
1
m
2
r
2
,
F
G
=
G
m
1
m
2
r
2
,
(4)where m1m1 and m2m2 are the masses of the two particles, rr is
the distance between them, and GG is the gravitational constant.
Both laws represent the force exerted by particles (masses or
charges) on each other that interact by means of a field.
Two point-like charges carrying charges of +3×10-9C+3×10-9C and -5×10-9C-5×10-9C are 2m2m apart. Determine the
magnitude of the force between them and state whether it is
attractive or repulsive.
- Step 1. Determine what is required :
We
are required to find the force between two point charges given the
charges and the distance between them.
- Step 2. Determine how to approach the problem :
We can use
Coulomb's Law to find the force.
F
=
k
Q
1
Q
2
r
2
F
=
k
Q
1
Q
2
r
2
(5) - Step 3. Determine what is given :
We are given:
-
Q
1
=
+
3
×
10
-
9
C
Q
1
=
+
3
×
10
-
9
C
-
Q
2
=
-
5
×
10
-
9
C
Q
2
=
-
5
×
10
-
9
C
-
r
=
2
m
r
=
2
m
We know that k=8,99×109N·m2·C-2k=8,99×109N·m2·C-2.
We can draw a diagram of the situation.
- Step 4. Check units :
All quantities are in SI units.
- Step 5. Determine the magnitude of the force :
Using Coulomb's Law
we have
F
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
N
·
m
2
/
C
2
)
(
3
×
10
-
9
C
)
(
5
×
10
-
9
C
)
(
2
m
)
2
=
3
,
37
×
10
-
8
N
F
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
N
·
m
2
/
C
2
)
(
3
×
10
-
9
C
)
(
5
×
10
-
9
C
)
(
2
m
)
2
=
3
,
37
×
10
-
8
N
(6)Thus the magnitude of the force is
3,37×10-8N3,37×10-8N. However since both point charges have
opposite signs, the force will be attractive.
Next is another example that demonstrates the difference in
magnitude between the gravitational force and the electrostatic
force.
Determine the electrostatic force and gravitational force between
two electrons 10-10m10-10m apart (i.e. the forces felt inside an atom).
- Step 1. Determine what is required :
We are required to calculate
the electrostatic and gravitational forces between two electrons,
a given distance apart.
- Step 2. Determine how to approach the problem :
We can use:
F
e
=
k
Q
1
Q
2
r
2
F
e
=
k
Q
1
Q
2
r
2
(7)to calculate the electrostatic
force and
F
g
=
G
m
1
m
2
r
2
F
g
=
G
m
1
m
2
r
2
(8)to calculate the gravitational force.
- Step 3. Determine what is given :
- Q1=Q2=1,6×10-19CQ1=Q2=1,6×10-19C(The charge on an electron)
- m1=m2=9,1×10-31 kg m1=m2=9,1×10-31 kg (The mass of an electron)
-
r
=
1
×
10
-
10
m
r
=
1
×
10
-
10
m
We know that:
-
k
=
8
,
99
×
10
9
N
·
m
2
·
C
-
2
k
=
8
,
99
×
10
9
N
·
m
2
·
C
-
2
-
G
=
6
,
67
×
10
-
11
N
·
m
2
·
kg
-
2
G
=
6
,
67
×
10
-
11
N
·
m
2
·
kg
-
2
All quantities are in SI units.
We can draw a diagram of the situation.
- Step 4. Calculate the electrostatic force :
F
e
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
)
(
-
1
,
60
×
10
-
19
)
(
-
1
,
60
×
10
-
19
)
(
10
-
10
)
2
=
2
,
30
×
10
-
8
N
F
e
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
)
(
-
1
,
60
×
10
-
19
)
(
-
1
,
60
×
10
-
19
)
(
10
-
10
)
2
=
2
,
30
×
10
-
8
N
(9)Hence the magnitude of the electrostatic force between the
electrons is 2,30×10-8N2,30×10-8N. Since electrons carry the
same charge, the force is repulsive.
- Step 5. Calculate the gravitational force :
F
g
=
G
m
1
m
2
r
2
=
(
6
,
67
×
10
-
11
N
·
m
2
/
kg
2
)
(
9
.
11
×
10
-
31
C
)
(
9
.
11
×
10
-
31
kg
)
(
10
-
10
m
)
2
=
5
,
54
×
10
-
51
N
F
g
=
G
m
1
m
2
r
2
=
(
6
,
67
×
10
-
11
N
·
m
2
/
kg
2
)
(
9
.
11
×
10
-
31
C
)
(
9
.
11
×
10
-
31
kg
)
(
10
-
10
m
)
2
=
5
,
54
×
10
-
51
N
(10)The magnitude of the gravitational force between the electrons is
5,54×10-51N5,54×10-51N. This is an attractive force.
Notice that the gravitational force between the electrons is much
smaller than the electrostatic force. For this reason, the
gravitational force is usually neglected when determining the
force between two charged objects.
We can apply Newton's Third Law to charges because, two
charges exert forces of equal magnitude on one another in opposite
directions.
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.
Three point charges are in a straight line.
Their charges are Q1Q1 = +2×10-9C+2×10-9C, Q2Q2 = +1×10-9C+1×10-9C and Q3Q3 = -3×10-9C-3×10-9C. The distance between
Q1Q1 and Q2Q2 is 2×10-2m2×10-2m and the distance between Q2Q2 and
Q3Q3 is 4×10-2m4×10-2m. What is the net electrostatic force on Q2Q2
from the other two charges?
- Step 1. Determine what is required :
We are needed to calculate the net force on Q2Q2. This force is the sum of the two electrostatic forces - the forces between Q1Q1 on Q2Q2 and Q3Q3 on Q2Q2.
- Step 2. Determine how to approach the problem :
- We need to calculate the two electrostatic forces on Q2Q2, 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.
- Step 3. Determine what is given :
We are given all the charges and all the distances.
- Step 4. Calculate the forces. :
Force of Q1Q1 on Q2Q2:
F
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
)
(
2
×
10
-
9
)
(
1
×
10
-
9
)
(
2
×
10
-
4
)
=
4
,
5
×
10
-
5
N
F
=
k
Q
1
Q
2
r
2
=
(
8
,
99
×
10
9
)
(
2
×
10
-
9
)
(
1
×
10
-
9
)
(
2
×
10
-
4
)
=
4
,
5
×
10
-
5
N
(11)Force of Q3Q3 on Q2Q2:
F
=
k
Q
2
Q
3
r
2
=
(
8
,
99
×
10
9
)
(
1
×
10
-
9
)
(
3
×
10
-
9
)
(
4
×
10
-
4
=
1
,
69
×
10
-
5
N
F
=
k
Q
2
Q
3
r
2
=
(
8
,
99
×
10
9
)
(
1
×
10
-
9
)
(
3
×
10
-
9
)
(
4
×
10
-
4
=
1
,
69
×
10
-
5
N
(12)Both forces act in the same direction because the force between Q1Q1 and Q2Q2
is repulsive (like charges) and the force between Q2Q2 and Q3Q3 is attractive
(unlike charges).
Therefore,
F
t
o
t
=
4
,
50
×
10
-
5
+
4
,
50
×
10
-
5
=
6
,
19
×
10
-
5
N
F
t
o
t
=
4
,
50
×
10
-
5
+
4
,
50
×
10
-
5
=
6
,
19
×
10
-
5
N
(13)
We mentioned in Chapter (Reference) 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.
In the picture below, X is a small negatively charged sphere with
a mass of 10kg. It is suspended from the roof by an insulating
rope which makes an angle of 60∘60∘ with the roof. Y is a
small positively charged sphere which has the same magnitude of
charge as X. Y is fixed to the wall by means of an insulating
bracket. Assuming the system is in equilibrium, what is the
magnitude of the charge on X?
How are we going to determine the charge on X? Well, if we know
the force between X and Y we can use Coulomb's Law to determine
their charges as we know the distance between them. So, firstly,
we need to determine the magnitude of the electrostatic force
between X and Y.
- Step 1. :
Is everything in S.I. units? The
distance between X and Y is 50 cm =0,5m50 cm =0,5m, and the mass of
X is 10kg.
- Step 2. Draw a force diagram :
Draw the forces on X (with
directions) and label.
- Step 3. Calculate the magnitude of the electrostatic force, FEFE :
Since nothing is moving (system is in equilibrium) the vertical
and horizontal components of the gravitational force must cancel the vertical and horizontal components of the electrostatic force. Thus
F
E
=
T
cos
(
60
∘
)
;
F
g
=
T
sin
(
60
∘
)
.
F
E
=
T
cos
(
60
∘
)
;
F
g
=
T
sin
(
60
∘
)
.
(14)The only force we know is the gravitational force Fg=mgFg=mg. Now we
can calculate the magnitude of TT from above:
T
=
F
g
sin
(
60
∘
)
=
(
10
)
(
10
)
sin
(
60
∘
)
=
115
,
5
N
.
T
=
F
g
sin
(
60
∘
)
=
(
10
)
(
10
)
sin
(
60
∘
)
=
115
,
5
N
.
(15)Which means that FEFE is:
F
E
=
T
cos
(
60
∘
)
=
115
,
5
·
cos
(
60
∘
)
=
57
,
75
N
F
E
=
T
cos
(
60
∘
)
=
115
,
5
·
cos
(
60
∘
)
=
57
,
75
N
(16) - Step 4. :
Now that we know the magnitude of the electrostatic
force between X and Y, we can calculate their charges using
Coulomb's Law. Don't forget that the magnitudes of the charges on
X and Y are the same: QX=QYQX=QY. The magnitude of
the electrostatic force is
F
E
=
k
Q
X
Q
Y
r
2
=
k
Q
X
2
r
2
Q
X
=
F
E
r
2
k
=
(
57
.
75
)
(
0
.
5
)
2
8
.
99
×
10
9
=
5
.
66
×
10
-
5
C
F
E
=
k
Q
X
Q
Y
r
2
=
k
Q
X
2
r
2
Q
X
=
F
E
r
2
k
=
(
57
.
75
)
(
0
.
5
)
2
8
.
99
×
10
9
=
5
.
66
×
10
-
5
C
(17)Thus the charge on X is -5.66×10-5C-5.66×10-5C.
- Calculate the electrostatic force between two charges of +6 nC +6 nC and +1 nC +1 nC if they are separated by a distance of 2 mm 2 mm .
- Calculate the distance between two charges of +4 nC +4 nC and -3 nC -3 nC if the electrostaticforce between them is 0,005N0,005N.
- Calculate the charge on two identical spheres that are similiarly charged if they are separated by 20 cm 20 cm and the electrostatic force between them is 0,06N0,06N.
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