Question

Write Gauss' law of electrostatics and obtain Coulomb's law with its help.

Hint:

Gauss' law is a generalization of Coulomb's law and is particularly useful in situations with spherical symmetry.

Answer 1

Step 1: Gauss' Law of Electrostatics.
Gauss' law states that the electric flux \( \Phi_E \) through any closed surface is proportional to the net charge enclosed within that surface. Mathematically, it is given by: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where: - \( \vec{E} \) is the electric field, - \( d\vec{A} \) is the infinitesimal area vector on the closed surface, - \( Q_{\text{enc}} \) is the total charge enclosed within the surface, - \( \varepsilon_0 \) is the permittivity of free space.
Step 2: Deriving Coulomb's Law from Gauss' Law.
Consider a point charge \( Q \) at the center of a spherical Gaussian surface of radius \( r \). By symmetry, the electric field \( E \) is radial and uniform over the surface, so the electric flux is: \[ \Phi_E = E \cdot 4\pi r^2 \] According to Gauss' law: \[ \Phi_E = \frac{Q}{\varepsilon_0} \] Equating the two expressions for electric flux: \[ E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \] Solving for \( E \): \[ E = \frac{Q}{4 \pi \varepsilon_0 r^2} \] This is the electric field due to a point charge. The force \( F \) on another charge \( q \) due to this electric field is: \[ F = qE = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{r^2} \] This is Coulomb's law, which gives the electrostatic force between two point charges \( Q \) and \( q \) separated by a distance \( r \).
Step 3: Conclusion.
Gauss' law leads directly to Coulomb's law, which describes the force between two point charges.