Question

State Gauss’s law. Determine the electric field intensity at a point due to an infinitely long uniformly charged straight wire.

Hint:

Remember standard Gauss law results: Line charge: \(E \propto \frac{1}{r}\) Plane sheet: Constant field Point charge: \(E \propto \frac{1}{r^2}\)

Answer 1

Concept:
Gauss’s law relates electric flux through a closed surface to the charge enclosed: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] For highly symmetric charge distributions (spherical, cylindrical, planar), Gauss’s law helps directly calculate electric field.
Step 1: Statement of Gauss’s law
The total electric flux through any closed surface is equal to \(1/\varepsilon_0\) times the total charge enclosed inside the surface.
Step 2: Symmetry of an infinite line charge
For an infinitely long straight wire:
- Electric field is radial (perpendicular to wire)
- Same magnitude at equal distance \(r\)
- Cylindrical symmetry exists
Step 3: Choose Gaussian surface
Take a cylindrical Gaussian surface:
- Radius = \(r\)
- Length = \(L\)
- Wire along axis of cylinder
Step 4: Electric flux through surface
Electric field is:
- Perpendicular to curved surface
- Parallel to flat ends
So flux through flat ends = 0.
Flux only through curved surface: \[ \Phi = E \times (2\pi r L) \]
Step 5: Charge enclosed
If linear charge density = \(\lambda\), then: \[ Q_{\text{enc}} = \lambda L \]
Step 6: Apply Gauss’s law \[ E(2\pi r L) = \frac{\lambda L}{\varepsilon_0} \] Cancel \(L\): \[ E(2\pi r) = \frac{\lambda}{\varepsilon_0} \] \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
Step 7: Nature of electric field
- Decreases as \(1/r\) - Radially outward for positive charge - Radially inward for negative charge