Question

using guess law derive the r psrision for thr electric field due to an infinitly long straight uniformely charged wire​

Answer 1

SolutioN :

Consider a thin infinitely long straight wire having uniform line charge having Density as λ. The direction of electric field is radially outwards. We choose a Cylindrical Gussain Surface to determine electric field at a distance r from line charge.

$\dashrightarrow \displaystyle \tt{\phi \: = \: \oint_ s \vec E. \vec{ds} \: = \: \oint _{s_1} \vec E_1 \vec{ds_1} \: + \: \oint_{s_3} \vec E_3 \vec{ds_3}} \\ \\ \dashrightarrow \tt{E \oint _{s_1} ds_1 \cos 0^{\circ} \: + \: E \oint_{s_2} \cos 90^{\circ} \: + \: E \oint_{s_3} \cos 90^{\circ}} \\ \\ \dashrightarrow \tt{E \oint _{s_1} ds_1 \: = \: E \: \times \: 2 \pi rl \: ----(1)} \\ \\ \footnotesize {\underline{\sf{ \: \: \: \: \: \: \: \pink{Charge \: Enclosed} \: = \: \blue{q} \: = \: \green{\lambda l} \: \: \: \: \: \: \:}}} \\ \\ \dashrightarrow \tt{\phi \: = \: \dfrac{q}{\epsilon_o}} \\ \\ \dashrightarrow \tt{\phi \: = \: \dfrac{\lambda l}{\epsilon _o} \: = \: E \: \times \: 2 \pi r l} \\ \\ \dashrightarrow \boxed{\boxed{\tt{E \: = \: \dfrac{\lambda}{2 \pi \epsilon_o r}}}} \\ \\ \therefore \: \sf{E \: \propto \: \dfrac{1}{r}} \\ \\ {\blue{\boxed{\sf{Where, \: {\red{\boxed{\sf{\dfrac{\lambda}{2 \pi \epsilon _o}}}}} \: is \: constant}}}}$

Answer 2

Answer:

We shall be using Gauss' Law for deriving the expression of electric field due to an infinite wire.

Now , we shall consider a cylindrical Gaussian surface . And since the wire is infinite , the field lines will be radially outwards.

Another very important thing to notice is that :

The electric flux through tge plane surface of the cylinders will be zero because the the angle between Area vector and Flux vector is 90°

$\therefore \displaystyle \: \int d \phi = \dfrac{q}{\epsilon_{0}}$

$\implies \displaystyle \: \int E \:. ds = \dfrac{q}{\epsilon_{0}}$

$\implies \displaystyle \: E \int ds = \dfrac{q}{\epsilon_{0}}$

$\implies \displaystyle \: E \times (2\pi rh) = \dfrac{q}{\epsilon_{0}}$

Now the charge enclosed can be written in terms of Linear Charge Density ( λ) :

$\implies \displaystyle \: E \times (2\pi rh) = \dfrac{ (\lambda \times h)}{\epsilon_{0}}$

Cancelling h on both sides :

$\implies \displaystyle \: E \times (2\pi r) = \dfrac{ (\lambda )}{\epsilon_{0}}$

$\implies \displaystyle \: E = \dfrac{ (\lambda )}{2\pi r\epsilon_{0}}$

So final answer :

$\boxed{ \red{ \bold{ \huge{ \displaystyle \: E = \dfrac{ (\lambda )}{2\pi r\epsilon_{0}} }}}}$