Question

State Gauss’ theorem. Obtain the expression for the intensity of the electric field at a point due to a thin charged wire of infinite length with its help.

Hint:

The electric field due to an infinitely long charged wire decreases with the distance from the wire, and it is inversely proportional to the distance from the wire.

Answer 1

Step 1: Gauss’ Theorem.
Gauss’ theorem states that the total electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed \( Q_{\text{enc}} \) divided by the permittivity of free space \( \varepsilon_0 \): \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]
Step 2: Electric Field Due to a Thin Charged Wire.
Consider an infinitely long, straight charged wire with linear charge density \( \lambda \) (charge per unit length). To find the electric field at a distance \( r \) from the wire, we use a cylindrical Gaussian surface with radius \( r \) and length \( L \). The electric flux through the surface is given by: \[ \Phi_E = E \cdot 2\pi r L \] Since the charge enclosed is \( Q_{\text{enc}} = \lambda L \), applying Gauss' law: \[ E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \] Solving for \( E \): \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]
Final Answer:
The intensity of the electric field at a distance \( r \) from a thin charged wire of infinite length is: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]