Lesson 1, Topic 1
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Coulomb’s law-force between two point charges

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Like charges repel each other while unlike 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-Augustin de Coulomb around 1784. Through his observations he was able to show that the magnitude of the electrostatic force between two point-like charges is inversely proportional to the square of the distance between the charges. He also discovered that the magnitude of the force is proportional to the product of the charges. That is: \[F \propto \frac{Q_1 Q_2}{r^2},\] where \(Q_1\) and \(Q_2\) are the magnitudes of the two charges respectively and r is the distance between them. The magnitude of the electrostatic force between two point-like charges is given by Coulomb’s law.Coulomb’s law

Coulomb’s law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. \[F = \frac{k Q_1 Q_2}{r^2},\]

The proportionality constant \(k\) is called the electrostatic constant and has the value:

\(\text{9,0} \times \text{10}^{\text{9}}\text{ N·m$^{2}$·C$^{-2}$}\) in free space.

Similarity of Coulomb’s law to Newton’s universal law of gravitation.

Notice how similar in form Coulomb’s law is to Newton’s universal law of gravitation between two point-like particles: \[F_G = \frac{G m_1 m_2}{d^2},\]

where \({m}_{1}\) and \({m}_{2}\) are the masses of the two point-like particles, \(d\) is the distance between them, and \(G\) is the gravitational constant. Both are inverse-square laws.

Both laws represent the force exerted by particles (point masses or point charges) on each other that interact by means of a field.

Two point-like charges carrying charges of \(\text{+3} \times \text{10}^{-\text{9}}\) \(\text{C}\) and \(-\text{5} \times \text{10}^{-\text{9}}\) \(\text{C}\) are \(\text{2}\) \(\text{m}\) apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

Determine what is required

We are required to determine the force between two point charges given the charges and the distance between them.

Determine how to approach the problem

We can use Coulomb’s law to calculate the magnitude of the force.\(F=k\frac{{Q}_{1}{Q}_{2}}{{r}^{2}}\)

Determine what is given

We are given:

  • \(Q_1\) = \(\text{+3} \times \text{10}^{-\text{9}}\) \(\text{C}\)
  • \(Q_2\) = \(-\text{5} \times \text{10}^{-\text{9}}\) \(\text{C}\)
  • \(r\) = \(\text{2}\) \(\text{m}\)

We know that \(k=\text{9,0} \times \text{10}^{\text{9}}\text{ N·m$^{2}$·C$^{-2}$}\).

We can draw a diagram of the situation.f8dda715f91ab9cdc2199deeba05ca58.png

Check units

All quantities are in SI units.

Determine the magnitude of the force

Using Coulomb’s law we have

\begin{align*} F &= \frac{k Q_1 Q_2}{r^2} \\ &= \frac{(\text{9,0}\times 10^{9})(\text{3}\times 10^{-9})(\text{5}\times 10^{-9})}{(2)^2} \\ &= \text{3,37} \times \text{10}^{-\text{8}}\text{ N} \end{align*}

Thus the magnitude of the force is \(\text{3,37} \times \text{10}^{-\text{8}}\) \(\text{N}\). However since the point charges have opposite signs, the force will be attractive.

Free body diagram

We can draw a free body diagram to show the forces. Each charge experiences a force with the same magnitude and the forces are attractive, so we have:

b4cc147ebe777bd5b4334e2943633a2a.png

WORKED EXAMPLE 2: COULOMB’S LAW

Determine the magnitudes of the electrostatic force and gravitational force between two electrons \(\text{10}^{-\text{10}}\) \(\text{m}\) apart (i.e. the forces felt inside an atom) and state whether the forces are attractive or repulsive.

Determine what is required

We are required to calculate the electrostatic and gravitational forces between two electrons, a given distance apart.

Determine how to approach the problem

We can use: \[F_e = \frac{k Q_1 Q_2}{r^2}\]

to calculate the electrostatic force and \[F_g = \frac{G m_1 m_2}{d^2}\]

to calculate the gravitational force.

Determine what is given

  • \({Q}_{1}={Q}_{2} = \text{1,6} \times \text{10}^{-\text{19}}\text{ C}\) (The charge on an electron)
  • \({m}_{1}={m}_{2}= \text{9,1} \times \text{10}^{-\text{31}}\text{ kg}\) (The mass of an electron)
  • \(r = d = \text{1} \times \text{10}^{-\text{10}}\text{ m}\)

We know that:

  • \(k=\text{9,0} \times \text{10}^{\text{9}}\text{ N·m$^{2}$·C$^{-2}$}\)
  • \(G=\text{6,67} \times \text{10}^{-\text{11}}\text{ N·m$^{2}$·kg$^{-2}$}\)

All quantities are in SI units.

We can draw a diagram of the situation.5f998ff81fd359913c3c0b7a10f5d991.png

Calculate the electrostatic force

\begin{align*} F_e &= \frac{k Q_1 Q_2}{r^2} \\ &= \frac{(\text{9,0}\times 10^9)(\text{1,60}\times 10^{-19})(\text{1,60}\times 10^{-19})}{(10^{-10})^2} \\ &= \text{2,30} \times \text{10}^{-\text{8}}\text{ N} \end{align*} Hence the magnitude of the electrostatic force between the electrons is \(\text{2,30} \times \text{10}^{-\text{8}}\text{ N}\). Since electrons carry like charges, the force is repulsive.

Calculate the gravitational force

\begin{align*} F_g &= \frac{G m_1 m_2}{d^2} \\ &= \frac{(\text{6,67}\times 10^{-11})(\text{9,11}\times 10^{-31})(\text{9,11}\times 10^{-31})}{(10^{-10})^2} \\ &= \text{5,54} \times \text{10}^{-\text{51}}\text{ N} \end{align*} The magnitude of the gravitational force between the electrons is \(\text{5,54} \times \text{10}^{-\text{51}}\) \(\text{N}\). Remember that the gravitational force is always an attractive force.

Notice that the gravitational force between the electrons is much smaller than the electrostatic force.