Question

State Gauss’s Theorem. Use it to derive the electric field intensity due to an infinitely long straight charged wire.

Hint:

Use Gauss’s law when symmetry exists: Spherical $\rightarrow$ sphere, Cylindrical $\rightarrow$ cylinder, Planar $\rightarrow$ Gaussian pillbox.

Answer 1

Concept: Gauss’s theorem helps calculate electric fields for highly symmetric charge distributions such as spherical, cylindrical, or planar symmetry. Statement of Gauss’s Theorem: The total electric flux through a closed surface is equal to \( \frac{1}{\varepsilon_0} \) times the total charge enclosed within the surface. \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where:
\( \vec{E} \) = electric field
\( d\vec{A} \) = area vector
\( \varepsilon_0 \) = permittivity of free space
Electric field due to an infinitely long straight charged wire Let:
Linear charge density = \( \lambda \) (charge per unit length)
Distance from wire = \( r \)
Step 1: Choose Gaussian surface Due to cylindrical symmetry, choose a cylindrical Gaussian surface:
Radius = \( r \)
Length = \( L \)
The electric field is:
Radial
Same magnitude everywhere on curved surface
Step 2: Calculate electric flux Flux through cylinder:
Curved surface contributes flux
Flat ends give zero flux (field parallel to surface)
\[ \text{Flux} = E \times \text{Curved surface area} \] \[ \text{Curved area} = 2\pi rL \] \[ \Rightarrow \Phi = E(2\pi rL) \] Step 3: Charge enclosed \[ Q_{\text{enc}} = \lambda L \] Step 4: Apply Gauss’s Law \[ E(2\pi rL) = \frac{\lambda L}{\varepsilon_0} \] Cancel \( L \): \[ E(2\pi r) = \frac{\lambda}{\varepsilon_0} \] \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Final Result: \[ \boxed{E = \frac{\lambda}{2\pi \varepsilon_0 r}} \]
Field decreases inversely with distance.
Field is radially outward for positive charge.