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

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

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

)righttheto(N.N.N.FFF

)righttheto(N.)m.(

)Cx)(Cx()C/mNx.(

r

|q||q|kF

)lefttheto(N.)m.(

)Cx)(Cx()C/mNx.(

r

|q||q|kF

e

e

604281

42150

10210310998

3

8110

10210110998

32122

2

66229

232

232

2

66229

212

2112

Example:

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 direction on 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.

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=9.109Nm2/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.

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?

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