Question

Q. Obtain an expression for electric feild intensity due to an infinity long straight charged wire or charged conducting Cylinder ?

Need Content Quality answer (^^)​

Answer 1

$\Huge\mathcal{\underline{\blue{Answer\::-}}}$

Gauss law states that the total electric flux through any closed surface is equal to 1/epsilonnot times of total charge enclosed by that closed surface placed in free surface for the straight uniformly charged infinite line we have to draw gaussian surface and it is cylinder.

$\small\mathcal{\green{❖\:Diagram\:And\:Derivation\:Is\:In\:The\:Attachment.}}$

$\Huge\mathcal{\underline{\red{Hope\:It\: Helps\:U}}}$

Answer 2

$\underline{\underline{\pink{\huge\sf Answer}}}$

Considered an infinity long wire having a uniform linear charge of density λ C/m and let l be the length of the charged body

Considered a point P at a distance r from the centre. Also consider an imaginary Gaussian Surface in the form of a cylinder around the wire.

Part of an Answer has been attached!!

From 1 & 2

➝ $\sf E2\pi r l = \dfrac{q}{\epsilon_0}$

$\sf Where \ q = \lambda l$

➝ $\sf E2\pi r l = \dfrac{\lambda l}{\epsilon_0}$

Cancel out l on LHS & RHS

➝ $\sf \orange{E = \dfrac{\lambda}{2\pi\epsilon_0}}$

━━━━━━━━━━━━━━━━