Coulomb’s law

1. Coulomb’s law:

  • Coulomb’s Law states that “the magnitude of the electrostatic force (F) between two-point charges [math] (Q_1 \, \text{and} \, Q_2) [/math] in a vacuum is directly proportional to the product of the charges and inversely proportional to the square of the distance (r) between them”.
  • Mathematically, this is expressed as:
  • [math] F \propto Q_1 Q_2 \hspace{1cm} (1) \\
    F \propto \frac{1}{r^2} \hspace{1cm} (2) \\
    \text{Combining these two equations:} \\
    F \propto \frac{Q_1 Q_2}{r^2} \\
    F = k \frac{Q_1 Q_2}{r^2} \hspace{1cm} (3)
    [/math]
  • where:
    – F is the electrostatic force
    – k is Coulomb’s constant (approximately [math] 8.99 \times 10^9 Nm^2 .C^{-2} [/math])
  • Or [math] k = \frac{1}{4 \pi \epsilon_0} [/math]. So, the equation 3 becomes
  • [math] F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \hspace{1cm} (4) [/math]
  • Figure 1 Coulomb’s forces applied on charges
  • –  and are the magnitudes of the two-point charges
    – r is the distance between the centers of the two charges
  • The direction of the force is along the line joining the two charges, and the force is:
    – Attractive if the charges are opposite (i.e., one positive and one negative)
    – Repulsive if the charges are like (i.e., both positive or both negative)
  • Coulomb’s Law applies to point charges in a vacuum, but it can also be used to approximate the force between charged objects in other environments, like air or water.

  • ⇒ Examples

  •  Calculate the force of attraction between two-point charges A and B separated by a distance of 0.2m. The charge at A is +2µC and the charge at B is -1µC .
  • Solution:
  • First charge [math] = Q_1 = +2μC = +2 *10^{-6} C [/math]
    Second charge [math] = Q_2 = -1μC= -1 *10^{-6}C [/math]
    Distance between these charges [math] = r = 0.2 [/math]
    Force of attraction = F =?
  • Formula
  • [math] F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2}  [/math]
  • Solution
  • [math] F = \frac{1}{4(3.14)(8.854 * 10^{-12})} \frac{(+2 * 10^{-6})(-1 * 10^{-6})}{(0.2)^2} \\ F = -0.45 \, \text{N} [/math]
  • The significance of the minus sign is to remind us that the force is attractive, but it is not really necessary to include it.

2. Permittivity of free space:

  • Permittivity of Free Space ([math] \varepsilon_0 [/math]) is a fundamental constant in physics, representing the ability of a vacuum to permit electric fields. It’s a crucial concept in understanding electrostatics and electromagnetism.
  • [math] \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} [/math] (Farads per meter)
  • Units:
    F/m (farads per meter)
  • Definition:
    Permittivity of Free Space ([math] \varepsilon_0 [/math]) is the proportionality constant that relates the electric field (E) and electric displacement field (D) in a vacuum:
  • [math] D =  \varepsilon_0 E [/math]
  • Significance:
    1. Electric field: determines the strength of the electric field generated by a charge.
    2. Capacitance: is used to calculate the capacitance of a capacitor.
    3. Electromagnetic waves: plays a role in the propagation of electromagnetic waves, like light.
  • Relationships:
    1. Coulomb’s Law: [math] \varepsilon_0 [/math]  is related to the Coulomb’s constant [math](k= \frac{1}{4 \pi \epsilon_0} ) [/math] by:
    2. Electric constant: is sometimes referred to as the electric constant.
  • Remember, [math] \varepsilon_0 [/math] is a fundamental constant that helps us understand how electric fields behave in a vacuum.

3. Air can be treated as a vacuum:

  • In most cases, air can be considered a vacuum for the purpose of calculating the force between charges. This is because the presence of air has a negligible effect on the electric field and the force between charges.
  • Here’s why:
    1. Air is a poor conductor: Air is a poor conductor of electricity, which means it doesn’t easily conduct electric charge. As a result, the electric field and the force between charges are not significantly affected by the presence of air.
    2. Permittivity of air: The permittivity of air ([math] \varepsilon_{air} [/math]) is very close to the permittivity of free space ([math] \varepsilon_0 [/math]). In fact, the relative permittivity of air [math](\varepsilon_r = \frac{\varepsilon_{\text{air}}}{\varepsilon_0} )[/math] is approximately 1.0006, which is very close to 1.
  • Given this, we can assume that the force between charges in air is approximately the same as in a vacuum. This simplifies calculations and allows us to use Coulomb’s Law without worrying about the effects of air.
  • However, keep in mind that this assumption breaks down in certain situations, such as:
    1. High voltages: At high voltages, air can become ionized and conduct electricity, affecting the electric field and force between charges.
    2. High pressures: At high pressures, the permittivity of air can deviate significantly from the permittivity of free space.
  • In such cases, the effects of air must be taken into account, and more advanced calculations are required.
  • Remember, treating air as a vacuum simplifies calculations, but it’s essential to consider the limitations of this assumption.

4. Charged sphere:

  • According to Coulomb’s Law, the electric field (E) due to a charge (Q) at a distance (r) from the charge is given by:
  • [math] E = k \frac{Q}{r^{2}} [/math]
  • For a charged sphere, we can consider the charge (Q) to be concentrated at the center of the sphere. Then, using Coulomb’s Law, we can calculate the electric field (E) at any point outside the sphere.


    Figure 2 Charged Sphere

  • By considering the charge to be at the center, we can treat the sphere as a point charge, and Coulomb’s Law simplifies the calculation of the electric field.
    1. The sphere is spherical symmetric (i.e., it’s a perfect sphere).
    2. The charge is uniformly distributed throughout the sphere.
    3. The distance from the sphere is much greater than the sphere’s radius.
  • Note that this assumption is only valid outside the sphere (r > radius of the sphere). Inside the sphere, the electric field is zero (E = 0) due to the spherical symmetry of the charge distribution.
  • Remember, Coulomb’s Law is a fundamental principle in electrostatics, and this assumption helps us apply it to charged spheres.

5. Gravitational and electrostatic forces:

  • Let’s compare the magnitude of gravitational and electrostatic forces between subatomic particles using Coulomb’s Law.
  • Gravitational Force:
  • The gravitational force between two objects is given by:
  • [math] F_{\text{grav}} = G \frac{m_1 m_2}{r^2} [/math]
  • where:
    G = [math] 6.674 \times 10^{-11} Nm^2kg^{-2}[/math] (gravitational constant)
    [math] m_1 \, \text{and} \, m_2 [/math] = masses of the objects
    r = distance between the objects
  • Figure 3 Gravitational force between earth and moon
  • Electrostatic Force (Coulomb’s Law):
  • The electrostatic force between two charged particles is given by:
  • [math] F_{\text{electrostatic}} = k \frac{Q_1 Q_2}{r^2} [/math]
  • where:
    [math] k = 8.987 \times 10^9 Nm^2 C^{-2} [/math](Coulomb’s constant)
    [math] Q_1 \, \text{and} \, Q_2 [/math] = charges of the particles
    r = distance between the particles
  • Figure 4 Coulomb’s force between two charges
  • Now, let’s compare the magnitude of these forces for subatomic particles like protons and electrons:
  • – Gravitational force between a proton and an electron:
  • [math] F_{\text{grav}} = G \frac{m_{\text{proton}} m_{\text{electron}}}{r^2} \approx 10^{-67} [/math]
  • – Electrostatic force between a proton and an electron:
  • [math] F_{\text{grav}} = G \frac{Q_{\text{proton}} Q_{\text{electron}}}{r^2} \approx 10^{-28} [/math]
  • Notice that the electrostatic force is much stronger than the gravitational force for subatomic particles!
  • In fact, the ratio of the electrostatic force to the gravitational force is approximately:
  • [math] \frac{F_{\text{electrostatic}}}{F_{\text{grav}}} \approx 10^{39} [/math]
  • This is why electrostatic forces dominate at the atomic and subatomic level, while gravitational forces are more significant at larger scales, like planetary and celestial objects.
  • Remember, Coulomb’s Law helps us understand the electrostatic forces between charged particles, while the gravitational force is governed by the laws of gravity.
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