The document discusses Coulomb's law, which quantitatively describes the electrical force between two charged objects. It states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Several examples are provided to demonstrate how to use the Coulomb's law equation to calculate electrical force, charge, or distance given two of the three variables.

1. Coulomb's Law Equation
The quantitative expression for the effect of these three variables on electric force is
known as Coulomb's law. Coulomb's law states that the electrical force between two
charged objects is directly proportional to the product of the quantity of charge on the
objects and inversely proportional to the square of the separation distance between the
two objects. In equation form, Coulomb's law can be stated as
where Q1 represents the quantity of charge on object 1 (in Coulombs), Q2 represents the
quantity of charge on object 2 (in Coulombs), and d represents the distance of separation
between the two objects (in meters). The symbol k is a proportionality constant known as
the Coulomb's law constant. The value of this constant is dependent upon the medium
that the charged objects are immersed in. In the case of air, the value is approximately 9.0
x 109 N • m2 / C2.
Example A
Suppose that two point charges, each with a charge of +1.00 Coulomb are separated by a
distance of 1.00 meter. Determine the magnitude of the electrical force of repulsion
between them.
Given:
Q1 = 1.00 C
Q2 = 1.00 C
d = 1.00 m
Find:
Felect = ???
The next and final step of the strategy involves substituting known values into the
Coulomb's law equation and using proper algebraic steps to solve for the unknown
information. This step is shown below.
Felect = k • Q1 • Q2 / d2
Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2
Felect = 9.0 x 109 N
Example B
Two balloons are charged with an identical
quantity and type of charge: -6.25 nC. They
are held apart at a separation distance of

2. 61.7 cm. Determine the magnitude of the
electrical force of repulsion between them.
Given:
Q1 = -6.25 nC = -6.25 x 10-9 C
Q2 = -6.25 nC = -6.25 x 10-9 C
d = 61.7 cm = 0.617 m
Find:
Felect = ???
The final step of the strategy involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for the unknown information. This
substitution and algebra is shown below.
Felect = k • Q1 • Q2 / d2
Felect = (9.0 x 109 N•m2/C2) • (6.25 x 10-9 C) • (6.25 x 10-9 C) / (0.617 m)2
Felect = 9.23 x 10-7 N
Example C
Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of
0.0626 Newton. Determine the separation distance between the two balloons.
The problem states the value of Q1 and Q2. Since these values are in units of
microCoulombs (µC), the conversion to Coulombs will be made. The problem also states
the electrical force (F). The unknown quantity is the separation distance (d). The results
of the first two steps are shown in the table below.
Given:
Q1 = +3.37 µC = +3.37 x 10-6 C
Q2 = -8.21 µC = -8.21 x 10-6 C
Felect = -0.0626 N (use a - force value since
it is attractive)
Find:
d = ???
As mentioned above, the use of the "+" and "-" signs is optional. However, if they are
used, then they have to be used consistently for the Q values and the F values. Their use
in the equation is illustrated in this problem.
The final step of the strategy involves substituting known values into the Coulomb's law
equation and using proper algebraic steps to solve for the unknown information. In this
case, the algebra is done first and the substitution is performed last. This algebra and
substitution is shown below.
Felect = k • Q1 • Q2 / d2
d2 • Felect = k • Q1 • Q2

3. d2 = k • Q1 • Q2 / Felect
d = SQRT(k • Q1 • Q2) / Felect
d = SQRT [(9.0 x 109 N•m2/C2) • (-8.21 x 10-6 C) • (+3.37 x 10-6 C) / (-0.0626 N)]
d = Sqrt [ +3.98 m2 ]
d = +1.99 m
Example:
Find the force on the charge q2 in the diagram below due to the charges q1 and q2
q1 = 1 µC q2 = -2 µC q3 = 3 µC
0.1 m 0.15 m
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|q||q|
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102103
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3
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10998
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
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Example:

4. An example
Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the
top right and bottom left corners are +3.0 x 10-6 C. The charges in the other two corners
are -3.0 x 10-6 C. What is the net force exerted on the charge in the top right corner by the
other three charges?
To solve any problem like this, the simplest thing to do is to draw a good diagram
showing the forces acting on the charge. You should also let your diagram handle your
signs for you. Force is a vector, and any time you have a minus sign associated with a
vector all it does is tell you about the direction of the vector. If you have the arrows
giving you the directionon your diagram, you can just drop any signs that come out of
the equation for Coulomb's law.
Consider the forces exerted on the charge in the top right by the other three:
You have to be very careful to add these forces as vectors to get the net force. In this
problem we can take advantage of the symmetry, and combine the forces from charges 2
and 4 into a force along the diagonal (opposite to the force from charge 3) of magnitude
183.1 N. When this is combined with the 64.7 N force in the opposite direction, the result
is a net force of 118 N pointing along the diagonal of the square.

5. The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have
to do is split the vectors up in to x and y components, add them to find the x and y
components of the net force, and then calculate the magnitude and direction of the net
force from the components. Example 16-4 in the textbook shows this process.
Electric Field
Electric Field
A charged particle exerts a force on particles around it. We can call the influence of this force on surroundings
as electric field. It can be also stated as electrical force per charge. Electric field is represented with E and
Newton per coulomb is the unit of it.
Electric field is a vector quantity. And it decreases with the increasing distance .k=9.109
Nm2
/C2
 Electric field cannot be seen, but you can observe the effects of it on charged particles inside electric field.
 To find the electric field vector of a charge at one point, we assume that as if there is a +1 unit of charge there.
 If you want to find the total electric field of the charges more than one, you should find them one by one and
add them using vector quantities.
Electric Field Lines
Motion path of the “+” charge in an electric field is called electric field line. Intensity of the
lines shows the intensity of the electric field.
Pictures given below show the drawings of field line of the positive charge and negative
charge.

6. Electric field lines;
 Are perpendicular to the surfaces
 Never intercept
 If the electric field lines are parallel to each other, we call this regular electric field and it can be possible
between two oppositely charged plates. E is constant within this plates and zero outside the plates.
Pictures given below show the path of lines of two same charges and two opposite charges.
Example:Find the electric field created by the charges A and B at point C in terms of k.q/d2?

7. Example:If the electric field at point A is zero, find the charge at point D in terms of q.