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## Homework Statement

A charge is distributed evenly along the ##z## axis with a density ##\lambda = 3 \frac{\mu C}{m}##. There is also a cylinder with radius ##r = 2## and on his side there is a superficial density charge of ##\sigma = -\left( \frac{1.5}{4 \pi} \right) \frac{\mu C}{m^2}##. Both distributions are unlimited on the ##z## axis that coincide with the axis of the system.

## Homework Equations

Gauss Theorem:

##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##

##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##

## The Attempt at a Solution

So I started first by calulating the electric field inside the cylinder with a ##r_0 < r##. Since it is only the charge on the ##z## axis creating the field, we will use ##\lambda## and integrate on the line.

##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##

##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda dl##

##E_0 2 \pi r_0 l = \frac{1}{\epsilon_0} \int_l \lambda dl##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r_0 l} \int_l \lambda dl##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0 l} \int_l dl##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda l}{r_0 l}##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##

Then I calculated the electric field at ##r_0 = r##. Now there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.

For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r}##.

For the contribution of the superficial charge we have:

##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##

##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma dS##

##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{0}^{r} 2 \pi l dr##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} \int_{0}^{r} dr##

##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi r l}{r l}##

##E_0 = \frac{\sigma}{\epsilon_0}##

So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r} + \frac{\sigma}{\epsilon_0}##

In the end I calculated the electric field for ##r_0 > r##. Like before, there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.

For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##.

For the contribution of the superficial charge we have ##\frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##.

So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0} + \frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##

Is it all correct?