Question

State Gauss's Law in electrostatics. Using it (i) find electric field due to a point source charge \( q \) and (ii) deduce Coulomb's law between source charge \( q \) and test charge \( q_0 \). 
 

Hint:

Gauss's law provides a direct method for deriving Coulomb’s law by calculating the electric field due to a point charge.

Answer 1

- Gauss's Law in Electrostatics: Gauss's law states that the electric flux through a closed surface is proportional to the enclosed electric charge. It is given by: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}, \] where:
- \( \Phi_E \) is the electric flux,
- \( \vec{E} \) is the electric field,
- \( d\vec{A} \) is the differential area element on the closed surface,
- \( q_{\text{enc}} \) is the total charge enclosed within the surface,
- \( \epsilon_0 \) is the permittivity of free space. - (A) Electric Field due to a Point Source Charge \( q \): Consider a spherical Gaussian surface with a radius \( r \) centered at a point charge \( q \). Applying Gauss's law, we get: \[ E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}, \] where \( E \) is the electric field at distance \( r \) from the charge. Solving for \( E \), we get: \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2}. \] - (B) Deduce Coulomb’s Law: Using the expression for \( E \), the force on a test charge \( q_0 \) placed in the electric field \( E \) is given by: \[ F = q_0 \cdot E = q_0 \cdot \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2}. \] This is Coulomb’s law, which states that the force between two point charges \( q \) and \( q_0 \) separated by a distance \( r \) is: \[ F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q \cdot q_0}{r^2}. \]