# Charged wire and Gauss theorem

Consider a long straight line of charge with 0.05 C/m of charge. Centered on this line of charge is a enclosing Gaussian surface that is a cylinder of radius 0.2 m and length of 0.3 m, as indicated in the adjacent figure.

1. What is the total net electric flux coming out of this enclosing Gaussian surface?

2. How much of the electric flux (prob, 7) comes out through the ends of the cylinder?

3. What is the electric field at a distance 0.2 m from the line of charge?

Answers

2) the electric field through the both ends of the cilider is opposite. thus the flux through the ends of the cylinder is zero.

1)

$\epsilon*E = Q (=\lambda*L)$

Gauss law is

$\phi = E*S = \lambda*L/\epsilon_0$

$Phi=0.05*0.3/8.856*10^{-12} =1.69*10^9 (V/m)*m^2$

3)

from above we have

$\Phi = E*S =E*2\pi r L$

$E*2\pi r L = \lambda L/\epsilon_0$

$E = \lambda / (2\pi \epsilon_0*r)$

$E = 0.05/(2\pi*8.856*10^{-12}*0.2) = 4.49*10^9 V/m =4.5*10^9 V/m$