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# Coulomb's Law

## Any two charged particles any distance from each other will experience repulsion or attraction.

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Coulomb’s Law

Credit: Flickr:ellenm1
Source: http://www.flickr.com/photos/ellenm1/6698266317/

Electric cars are becoming more popular. One large advantage for electric cars is the low cost of operation, which may become an ever bigger advantage as gas prices climb. Energy costs for electric cars average about one-third of the cost for gasoline engine cars, but they can only travel about 200 miles per charge at this point. These cars run using the science of electrical charges and forces.

### Coulomb’s Law

The questions regarding the relationship between the electrical force, the size of the charge, and the separation between the charges were solved by Charles Coulomb in 1785. He determined that electrical force between two charges is directly related to the size of the charges and inversely proportional to the distance between the charges. This is known as Coulomb’s Law.

Fe=Kq1q2d2\begin{align*}F_e=\frac{Kq_1q_2}{d^2}\end{align*}

In this equation, q1 and q2 are the two charges, d is the distance between the two charges, and K is a constant of proportionality. Fe is the electric force, which occurs as a result of interactions between two charged particles. For the purpose of calculating electric forces, we assume all charge is a point charge, in which the entire charge of the particle is located in a massless point.

The SI unit of charge is the coulomb, C, which is the charge of 6.25×1018\begin{align*}6.25 \times 10^{18}\end{align*} electrons. The charge on a single electron is 1.60×1019 C\begin{align*}1.60 \times 10^{-19} \ C\end{align*}. The charge on a single electron is known as the elementary charge. The charge on a proton is the same magnitude but opposite in sign. When the charges are measured in coulombs, the distance in meters, and the force in Newtons, the constant K is 9.0×109 Nm2/C2\begin{align*}9.0 \times 10^9 \ N \cdot m^2/C^2\end{align*}.

The electrical force, like all forces, is a vector quantity. If the two charges being considered are both positive or both negative, the sign of the electrical force is positive and this force is repulsive. If the two charges are opposite in sign, the force will have a negative sign and the force is attractive.

Example Problem: Object A has a positive charge of 6.0×106 C\begin{align*}6.0 \times 10^{-6} \ C\end{align*}. Object B has a positive charge of 3.0×106 C\begin{align*}3.0 \times 10^{-6} \ C\end{align*}. If the distance between A and B is 0.030 m, what is the force on A?

Solution: Fe=Kq1q2d2=(9.0×109 Nm2/C2)(6.0×106 C)(3.0×106 C)(0.030 m)2=180 N\begin{align*}F_e=\frac{Kq_1q_2}{d^2}=\frac{(9.0 \times 10^9 \ N \cdot m^2/C^2)(6.0 \times 10^{-6} \ C)(3.0 \times 10^{-6} \ C)}{(0.030 \ m)^2}=180 \ N\end{align*}

The positive sign of the force indicates the force is repulsive. This makes sense, because both objects have a positive charge.

Example Problem: In the sketch below, the charges are q1=10.0×106 C, q2=2.0×106 C\begin{align*}q_1 = 10.0 \times 10^{-6} \ C, \ q_2 = 2.0 \times 10^{-6} \ C\end{align*}, and q3=6.0×106 C\begin{align*}q_3 = -6.0 \times 10^{-6} \ C\end{align*}. Calculate the total force on q2.

Credit: Samantha Bacic
Source: CK-12 Foundation

Solution: Fe=Kq1q2d2=(9.0×109 Nm2/C2)(10.0×106 C)(2.0×106 C)(2.0 m)2=0.045 N (towards q3)\begin{align*}F_e=\frac{Kq_1q_2}{d^2}=\frac{(9.0 \times 10^9 \ N \cdot m^2/C^2)(10.0 \times 10^{-6} \ C)(2.0 \times 10^{-6} \ C)}{(2.0 \ m)^2}=0.045 \ N \ (\text{towards} \ q_3)\end{align*}

Fe=Kq2q3d2=(9.0×109 Nm2/C2)(2.0×106 C)(6.0×106 C)(4.0 m)2=0.007 N (towards q3)\begin{align*}F_e=\frac{Kq_2q_3}{d^2}=\frac{(9.0 \times 10^9 \ N \cdot m^2/C^2)(2.0 \times 10^{-6} \ C)(-6.0 \times 10^{-6} \ C)}{(4.0 \ m)^2}=-0.007 \ N \ (\text{towards} \ q_3)\end{align*}

Since the two forces act in the same direction, their absolute values can be added together; the total force on q2 is 0.052 N towards q3.

#### Summary

• Coulomb determined that electrical force between two charges is directly related to the size of the charges and inversely proportional to the distance between the charges: Fe=Kq1q2d2\begin{align*}F_e=\frac{Kq_1q_2}{d^2}\end{align*}
• The SI unit of charge is the coulomb, C, which is the charge of 6.25×1018\begin{align*}6.25 \times 10^{18}\end{align*} electrons.
• The charge on a single electron is 1.60×1019 C\begin{align*}1.60 \times 10^{-19} \ C\end{align*} and is known as the elementary charge.
• The electrical force is a vector quantity that is positive in repulsion and negative in attraction.

#### Practice

Questions

The following video covers Coulomb's Law. Use this resource to answer the questions that follow.

1. What happens when like charges are placed near each other?
2. What happens when opposite charged are placed near each other?
3. What happens to the force of attraction if the charges are placed closer together?

Practice problems on Coulomb’s Law: http://physics.info/coulomb/problems.shtml

#### Review

Questions

1. Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m:
1. Will the charges attract or repel?
2. What is the magnitude of the force between them?
3. If the distance between them is doubled, what does the force become?
2. What is the electrical force between two balloons, each having 5.00 C of charge, that are 0.300 m apart?
3. Two spheres are charged with the same charge of -0.0025 C and are separated by a distance of 8.00 m.  What is the electrical force between them?
4. A red foam ball and a blue foam ball are 4.00 m apart. The blue ball has a charge of 0.000337 C and is attracting the red ball with a force of 626 N. What is the charge on the red ball?