Class 10 > Unit # 14: Electrostatics > Coulomb's Law

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Electrostatics - Coulomb’s Law

Q. State and derive expression for Coulomb’s Law.

Statement

"The electric force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."

Figure: Interaction between two point charges $q_1$ and $q_2$ separated by distance $r$.

Mathematical Derivation

Consider two point charges $q_1$ and $q_2$ separated by a distance $r$. According to Coulomb's law, the magnitude of the electrostatic force $F$ exerted by either charge on the other satisfies two conditions:

• Direct proportionality to the product of charges: $F \propto q_1 q_2$
• Inverse square law with distance: $F \propto \frac{1}{r^2}$

Combining these two proportionalities yields:

$F \propto \frac{q_1 q_2}{r^2}$

Introducing the constant of proportionality $k$:

$F = k \frac{q_1 q_2}{r^2} \quad \text{--- (i)}$

Proportionality Constant ($k$) in Free Space

The value of $k$ depends upon the medium present between the charges. For free space (vacuum), $k$ is expressed in terms of the permittivity of free space ($\varepsilon_0$):

$k = \frac{1}{4\pi\varepsilon_0}$

Where the permittivity of free space $\varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$, resulting in the numerical value of $k$:

$k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$

Substituting $k$ into equation (i) for free space gives:

$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$

Effect of an Insulating Medium (Dielectric)

If an insulating medium or dielectric is placed between the charges instead of free space, the electrostatic force decreases by a factor known as the relative permittivity (or dielectric constant) $\varepsilon_r$. The constant $k$ in the medium is given by:

$k = \frac{1}{4\pi\varepsilon}$

Where $\varepsilon$ is the absolute permittivity of the medium, defined as:

$\varepsilon = \varepsilon_0 \varepsilon_r$

Thus, equation (i) for a medium takes the form:

$F = \frac{1}{4\pi\varepsilon_0\varepsilon_r} \frac{q_1 q_2}{r^2}$

Which can also be written as:

$F = \frac{1}{\varepsilon_r} \left( \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \right)$

Nature of Force: The electrostatic force is attractive if the charges are dissimilar (opposite signs) and repulsive if the charges are similar (like signs).

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