Question

The electric field due to a uniformly charge infinite wire at a distance r is :

Answer 1

at zero potential

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Answer 2

The electric field due to uniformly chagred infinitewire is E = λ/(2πε₀r)

 

Step by step explanation:

According to gauss law:

The total flux of electric field E is $\frac{1}{\epsilon_0}$  times the net charge enclosed (q) in the closed surface.

so    

$\phi = \frac{q}{\epsilon_0 }$                             (equation 1)

$\textbf{Electric flux through curved surface =} \oint Eds\cos \theta\\\\\text{For infinite wire }\theta = 0 \\So,\cos \theta =1\\\\So, \text{Electric flux through curved surface =} \oint Eds\\\\\text{As E does not depends on s and integration of s become }2\pi rl \text{(Surface area of curved part)}$

$\phi = E(2\pi rl)$                     (equation 2)

As the total charge inside a wire = q = λl

putting th value of q in equation (1)

$\phi = \frac{\lambda \times l}{\epsilon_0 }$                         (equation 3)

On comparing equation 2 and 3.

$E(2\pi rl) = \frac{\lambda \times l}{\epsilon_0 }$

$E = \frac{\lambda }{(2\pi r)\epsilon_0 }$

$\textbf{\Large The electric field due to uniform infinite wire =}\\\\E = \frac{\lambda }{(2\pi r)\epsilon_0 }$