using guess law derive the r psrision for thr electric field due to an

1.	using guess law derive the r psrision for thr electric field due to an infinitly long straight uniformely charged wire
Answer» 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. $\dashrightarrow \displaystyle \tt{\phi \: = \: \oint_ s \vec E. \vec{ds} \: = \: \oint _{s_1} \vec E_1 \vec{ds_1} \: + \: \oint_{s_3} \vec E_3 \vec{ds_3}} \\ \\ \dashrightarrow \tt{E \oint _{s_1} ds_1 \cos 0^{\circ} \: + \: E \oint_{s_2} \cos 90^{\circ} \: + \: E \oint_{s_3} \cos 90^{\circ}} \\ \\ \dashrightarrow \tt{E \oint _{s_1} ds_1 \: = \: E \: \times \: 2 \pi rl \: ----(1)} \\ \\ \footnotesize {\underline{\sf{ \: \: \: \: \: \: \: \pink{Charge \: Enclosed} \: = \: \blue{q} \: = \: \green{\lambda l} \: \: \: \: \: \: \:}}} \\ \\ \dashrightarrow \tt{\phi \: = \: \dfrac{q}{\epsilon_o}} \\ \\ \dashrightarrow \tt{\phi \: = \: \dfrac{\lambda l}{\epsilon _o} \: = \: E \: \times \: 2 \pi r l} \\ \\ \dashrightarrow \boxed{\boxed{\tt{E \: = \: \dfrac{\lambda}{2 \pi \epsilon_o r}}}} \\ \\ \therefore \: \sf{E \: \propto \: \dfrac{1}{r}} \\ \\ {\blue{\boxed{\sf{Where, \: {\red{\boxed{\sf{\dfrac{\lambda}{2 \pi \epsilon _o}}}}} \: is \: constant}}}}$

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

Answer»

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.

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