D. COULOMB’S LAW
1. A charged sphere behaves as if all the charge were concentrated at the
centre of the sphere; that is, at a point. Whenever we refer to charged
spheres, then, we shall consider them to be charged points regardless of their size.
2. Coulomb discovered several relationships between two charged points placed close to each other.
a) The two points exert an equal but opposite (mutual) force on each other. Electric force is given the symbol Fe.
b) The electric force between the two points is directly proportional to the charge on each point.
Fe µ Q Q
c) The electric force is inversely proportional to the square of the distance separating the charged points.
Fe µ 1/r2 where r = distance separating the two charged points
d) The result of the combination of charge and separation is shown as
Fe µ Q Q
R2
Note that electric charge has as its fundamental unit the elementary charge which is equivalent to the charge on the electron (-1 elch) or the proton (+1 elch). These charges are so small, however, that they are impossible to measure. A practical or measurable unit of charge is a coulomb which is equal to 6.25 x 1018 elementary charges.
1C = 6.25 x 1018 elch
PRACTICE – COULOMB’S LAW
Two point charges, Q1 and Q2, are place 1.0 x 10-20 m away from each other. Q1 has a charge of 3 elch and Q2 has a charge of 4 elch. The two point charges are both negative and the force of repulsion between them is 1.0 x 10-5 N.
a) Draw a diagram illustrating the information in the problem.
b) What force does Q1 exert on Q2?
c) If the separation between the two points were changed as described below, what would be the force between the two points?
i) The separation is doubled.
ii) The separation is halved.
d) The charge on Q1 is changed to 6 elch and Q2 is changed to 8 elch. What is the force between the two points if the separation remains the same?
e) The charge on Q1 is changed to 6 elch and the charge on Q2 is changed to 8 elch. The separation between the two point charges is changed to 2.0 x 10-20 m. What is the force between the two points?
f) Inserting a constant of proportionality (Coulomb’s constant) produces the following equation.
Fe = kQ1 Q2
R2
k = 8.93 x 109 Note: sometimes rounded to 9.0 x 109
Sample
When three points are in a straight line, the vector analysis is easily performed by assigning “+” or “-“ for direction notation and then doing the arithmetic. If the point charges are not arranged in a straight line, however, then two-dimensional vector analysis will be required. Determine the net force that would act on Q2 if the point charges were arranged as shown below.
Q1 = -3.6 x 102 elch Q2 = +4.2 x 102 elch
Q3 = +7.2 x 102 elch Fe12 = 2.0 x 10-16 N
Q1 Q2
o______o
r12 = 4.2 x 10-4 m
r23 = 2.1 x 10-4 m
o
Q3
COULOMB’S LAW
1.
A B C
o ______o ______o
rab = 5.0 x 10-3 m rbc = 2.5 x 10-3 m
a) Determine the mutual force between B and C point charges if:
A = +10 elch B = -7 elch C = +9 elch
FAB = 6.4 x 10-22 N
(2.3 x 10-21 N)
b) Determine Fnet on B.
2.
A B C
o ______o ______o
rab = 5.0 x 10-3 m rbc = 2.5 x 10-3 m
a) Determine the mutual force between B and C point charges if:
A = 1.2 C B = 0.6 C C = 1.1 C
FAB = 2.6 x 1014 N
(9.43 x 1014 N)
b) Determine Fnet on B.
2. Use the supplied information and the diagram to determine the Fnet on B.
o A
A = +3.6 C
rab = 1.414 m B = +4.1 C
C = +6.4 C
FAB = 6.6 x 1010 N
o______o (8.8 x 1010 N 41.6° W of S)
B rbc = 2 m C
3. Use the supplied information and the diagram to determine the Fnet on B.
o A A = +16 C
B = +12 C
C = +8 C
FBC = 1.34 x 1026 N
B o ______o C (1.6 x 1026 N 56.9° W of S)
rab = 1.4 x 10-7 m
rbc = 8.0 x 10-8 m
E. ELECTRIC FIELD
1. Most of the forces we are familiar with in everyday life might be
called contact forces. Contact is required between whatever it is that produces the force and whatever it is that receives the force; as when someone pushes (exerts a force on) a lawnmower. An electric force, however is one that acts at a distance – that is, they act even when the bodies involved are not touching. This idea of a force “acting at a distance” is sometimes difficult to understand so the concept of a field was developed to explain how these forces operate. We can visualize an electric charge giving rise to an electric field that extends out from the charge and permeates space.
The diagram at left shows the electric field
coming out from the positive charge Q1 in
Å all directions. A second charge Q2 would
experience a force from Q1, because of the
. Q2 electric field.
An electric field is detected through the use of a positive test charge. This elementary positive test charge is placed in proximity to the charged body whose electric field we want to determine. The magnitude and direction of the electric force acting on the positive test charge is the electric field. It is important at this point to remember that all forces are vectors and must have direction associated with them whenever a calculation or description involving a force takes place. Since a field is a special case involving a force, then it too must always be described as a vector. Note that the electric field is directed away from a positively charged point. As the positive test charge moves closer to the positively charged point, the electric field becomes larger.
Note the electric field points toward the negatively charged object.
In any case, the electric field is described as the electric force acting on a single positive test charge. This means we can use Coulomb’s Law to determine the electric field around a charged point, since there are two points – the charged object around which we are trying to find the electric field and the positive test charge. Since the positive test charge is only one elementary charge, one “Q” can be eliminated from the equation.
E = kQ
R2 E = electric field in N/elch or N/C
k = coulomb’s constant (8.93 x 109)
r = separation between the charged point and the
positive test charge
Q = the electric charge on the charged point
expressed either as elementary charges or
coulombs
Another equation can be used to determine the electric field when there are two charged objects.
E = Fe
q E = electric field in N/elch or N/C
Fe = electric force between the two charged points
q = electric charge on the charged point where the
electric field is to be determined in elch or
coulombs
It is very important that you remember the difference between the “Q” and “q” in the two formulas.
PRACTICE
Three point charges are placed in a straight horizontal line. Q1 = 5.6 x 10-12 C,
Q2 = 7.2 x 10-12 C, Q3 = 9.4 x 10-12 C. The charge separation r12 = 4.8 x 10-2 m and the charge separation r23 = 2.4 x 10-2 m. The three points are all negatively charged..
a) Draw a diagram to represent the information above.
b) Determine the force that Q1 exerts on Q2.
c) Determine the force that Q3 exerts on Q2.
d) Determine the net force that is exerted on Q2.
e) Determine the electric field that:
i) Q1 exerts at Q2.
ii) Q1 exerts at Q3.
ELECTRIC FIELDS
1. One point has a charge of +3.6 x 104 elch and the other point charge 1.0 cm away has a charge of –2.4 x 104 elch. The electric force between the particles is +2.0 x 10-15 N. A third point having a charge of +1.2 x 104 elch is located
0.5 cm from the second point as shown below.
Q1 Q2 Q3
· · ·
1.0 cm 0.5 cm
Determine:
a) the electric force between Q2 and Q3. (2.63 x 10-15 N)
b) the net force that Q1 and Q3 exert on Q2. (6.3 x 10-16 N R)
c) the electric field that Q1 exerts at Q2. (8.3 x 10-20 N/elch R)
d) the electric field that Q2 exerts at Q3. (2.19 x 10-19 N/elch L)
2. The following diagram shows three point charges where:
Q1 = +1.8 x 10-6 C r12 = 3.2 x 10-3 m F12 = 3.8 x 103 N
Q2 = - 2.4 x 10-6 C r23 = 1.4 x 10-3 m
Q3 = +3.2 x 10-6 C
Q1
o o Q2
o Q3
Determine:
a) the electric force between Q2 and Q3.
(3.5 x 104 N)
b) the net force that Q1 and Q3 exert on Q2.
(3.5 x 104 N 83.8° S of W)
c) the electric field that Q2 exerts at Q1.
(2.1 x 109 N/C right)
d) the electric field that Q2 exerts at Q3.
(1.1 x 1010 N/C up)