Question

Derive the formula for electric field due to a uniformly charged straight wire of infinite length using Gauss's law.

Hint:

The key to using Gauss's law is choosing a "Gaussian surface" that matches the symmetry of the charge distribution. For a line of charge, a cylinder is the ideal choice. For a point charge, a sphere. For a plane of charge, a cylinder or a cuboid (pillbox).

Answer 1

Step 1: Understanding the Concept and Choosing a Gaussian Surface:
We want to find the electric field (\(E\)) at a distance \(r\) from an infinitely long straight wire with a uniform linear charge density \(\lambda\) (charge per unit length). Gauss's law is effective for symmetric charge distributions. Due to the cylindrical symmetry of the infinite wire, the electric field must be directed radially outwards (for \(\lambda>0\)) and its magnitude can only depend on the radial distance \(r\). We choose a cylindrical Gaussian surface of radius \(r\) and length \(l\), coaxial with the wire.
Step 2: Applying Gauss's Law:
Gauss's law states that the total electric flux (\(\Phi_E\)) through a closed surface is equal to the net charge enclosed (\(q_{in}\)) divided by the permittivity of free space (\(\epsilon_0\)). \[ \Phi_E = \oint \vec{E} . d\vec{A} = \frac{q_{in}}{\epsilon_0} \] The total flux through our cylindrical surface can be split into the flux through the two flat end caps (top and bottom) and the flux through the curved side wall. \[ \oint \vec{E} . d\vec{A} = \int_{\text{top}} \vec{E} . d\vec{A} + \int_{\text{bottom}} \vec{E} . d\vec{A} + \int_{\text{curved}} \vec{E} . d\vec{A} \] Step 3: Calculating the Electric Flux:
Flux through the end caps: On the top and bottom circular surfaces, the electric field vector \(\vec{E}\) is radial and perpendicular to the wire, while the area vector \(d\vec{A}\) is directed along the axis of the cylinder (upwards for the top cap, downwards for the bottom). Thus, \(\vec{E}\) is perpendicular to \(d\vec{A}\), and the angle between them is 90\(^{\circ}\). \[ \vec{E} . d\vec{A} = E \, dA \, \cos(90^\circ) = 0 \] So, the flux through the top and bottom caps is zero.
Flux through the curved surface: On the curved side wall, the electric field vector \(\vec{E}\) is always parallel to the area vector \(d\vec{A}\) (both point radially outwards). The angle between them is 0\(^{\circ}\). \[ \int_{\text{curved}} \vec{E} . d\vec{A} = \int_{\text{curved}} E \, dA \, \cos(0^\circ) = \int_{\text{curved}} E \, dA \] Since the magnitude of the electric field E is constant at a fixed distance r from the wire, we can take it out of the integral. \[ \Phi_E = E \int_{\text{curved}} dA = E \times (\text{Area of curved surface}) = E(2\pi r l) \] Step 4: Calculating the Enclosed Charge and Final Formula:
The charge enclosed by the Gaussian surface of length \(l\) is the linear charge density \(\lambda\) multiplied by the length \(l\). \[ q_{in} = \lambda l \] Now, we substitute the flux and enclosed charge back into Gauss's law: \[ E(2\pi r l) = \frac{\lambda l}{\epsilon_0} \] The length \(l\) cancels out from both sides, which is expected as the wire is infinitely long. \[ E = \frac{\lambda}{2\pi\epsilon_0 r} \] This is the formula for the electric field due to an infinitely long charged wire. The field is inversely proportional to the distance from the wire.