Q.(a) Obtain the expression for electric field due to an infinitely long charged wire. OR

(a) Gauss's law with a cylindrical Gaussian surface gives the field of an infinite line charge as $E=\lambda/(2\pi\varepsilon_0r)$; (b) DeMorgan's two theorems, $\overline{A+B}=\bar A\bar B$ and $\overline{AB}=\bar A+\bar B$, are proved by truth table. Both alternatives answered below.

(a) Electric field due to an infinitely long charged wire

1. Setup. Consider an infinitely long straight wire with uniform linear charge density $\lambda$. By symmetry, the field at a perpendicular distance $r$ from the wire is radial (pointing directly away from the wire) and has the same magnitude at every point on a cylinder of radius $r$ coaxial with the wire.

2. Gaussian surface. Choose a cylindrical Gaussian surface of radius $r$ and length $l$, coaxial with the wire.

3. Flux calculation. The flux through the two flat circular end-caps is zero (since $\vec E$ is radial, perpendicular to the end-cap's outward normal along the axis). The flux through the curved lateral surface, where $\vec E$ is parallel to the outward normal everywhere:

$\Phi = E \times (2\pi r l)$

4. Applying Gauss's law. The charge enclosed is $q_{enc}=\lambda l$:

$E(2\pi rl) = \dfrac{\lambda l}{\varepsilon_0}$

$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$

The field falls off as $1/r$ (slower than the $1/r^2$ of a point charge), directed radially away from the wire (for $\lambda>0$).

(b) DeMorgan's theorems

First theorem: $\overline{A+B} = \bar A \cdot \bar B$ ("the complement of a sum equals the product of the complements").

Second theorem: $\overline{A\cdot B} = \bar A + \bar B$ ("the complement of a product equals the sum of the complements").

Proof by truth table (first theorem):