Talha's Physics Academy
Coulomb's Law and Electrostatic Force
Video Lecture
Watch the complete video lecture below to understand Coulomb's law, electrostatic force between point charges, and the effect of dielectric media on electric force.
Coulomb's Law
Statement
Consider two point charges $q_1$ and $q_2$ separated by a distance $r$ from each other. By this law, the magnitude of the force $F$ exerted by either of the charges on the other is given by:
$$F \propto \frac{q_1 q_2}{r^2}$$
$$F = k \frac{q_1 q_2}{r^2} \quad \text{--- (i)}$$
Where $k$ is a constant of proportionality, and its value depends on the medium between the charges. For free space (vacuum), the value of $k$ is:
$$k = \frac{1}{4\pi \varepsilon_0}$$
$$k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$$
Substituting the value of $k$ in equation (i) for free space:
$$F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$$
Where $\varepsilon_0$ = permittivity of free space.
Effect of Medium
If instead of free space there is some insulating material or dielectric medium between the charges, then the proportionality constant becomes:
$$k = \frac{1}{4\pi \varepsilon}$$
Where $\varepsilon$ is the absolute permittivity of the medium, which can be expressed as:
$$\varepsilon = \varepsilon_0 \varepsilon_r$$
Here, $\varepsilon_r$ represents the relative permittivity (or dielectric constant) of the medium.
Thus, equation (i) for a material medium becomes:
$$F = \frac{1}{4\pi \varepsilon} \frac{q_1 q_2}{r^2}$$
$$F = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{q_1 q_2}{r^2}$$
$$F = \frac{1}{\varepsilon_r} \left( \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2} \right)$$
Nature of Force: The electrostatic force will be attractive if the charges are dissimilar (unlike charges) and repulsive if the charges are similar (like charges).
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