Answer:
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
Explanation:
Let the linear charge density of the charged wire is given as
[tex]\frac{q}{L} = \lambda[/tex]
here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have
[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]
here we have
[tex]E \int dA = \frac{\lambda L}{\epsilon_0}[/tex]
[tex]E. 2\pi r L = \frac{\lambda L}{\epsilon_0}[/tex]
so we have
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]