A very long, uniformly charged cylinder has radius R and linear charge densityλ.a. Find the...

Step 1
Given:
Radius of the cylinder ${\displaystyle =R}$
Linear Charge density of the cylinder ${\displaystyle =\lambda }$
Step 2
Calculating the electric field outside the cylinder:
Let’s assume a Cylindrical Gaussian surface of radius ‘r’ ${\displaystyle \left(r>R\right)}$ and length ‘L’ around the given charged cylinder.
Charge enclosed by the Guassian surface, ${\displaystyle q_{enclosed}=\lambda L}$
Using Gausss

Given Data:
- Radius of cylinder is
- Linear charge density is ${\displaystyle \lambda }$
Part a
We have to find out the electric field strength outside the cylinder ${\displaystyle r\ge R}$
Let P be the external point, where we have to find the electric field.
So, we consider a gaussian surface of length L, at a distance r from the axis of the cylinder, which is also a coaxial cylinder.
From Gauss law,
${\displaystyle {\int }_S\overrightarrow{E}.d\overrightarrow{S}=\frac{\lambda L}{{\in }_0}}$
Where ${\displaystyle \lambda L}$ is the charge enclosed by the surface.
Thus,
${\displaystyle {\int }_S\overrightarrow{E}.d\overrightarrow{S}=E\int dS}$
${\displaystyle =E.2\pi rL}$
${\displaystyle =\frac{\lambda L}{{\in }_0}}$
So, ${\displaystyle E=\frac{1}{4\pi {\in }_0}\cdot \frac{2\lambda }{r}\cdot \hat{r}}$
Where ${\displaystyle \hat{r}}$ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength outside the cylinder is ${\displaystyle E=\frac{1}{4\pi {\in }_0}\cdot \frac{2\lambda }{r}\cdot \hat{r}}$
Part b
Now, we have to find out the electric field strength inside the cylinder ${\displaystyle r\le R}$
Let P be any internal point, where we have to find the electric field.
Similarly, we consider a gaussian surface of length L at a distance from the axis of the cylinder, which is also a coaxial cylinder.
Charge enclosed by it is,
${\displaystyle Q_{enclosed}=\pi r^{2}L\cdot \rho }$
Where ${\displaystyle \rho }$ is the volume charge density.
Now, volume charge density is related to linear charge density as,
${\displaystyle \pi R^2\cdot 1\cdot \rho =\lambda \rho =\frac{\lambda }{\pi R^2}}$
From Gauss law,
${\displaystyle {\int }_S\hat{E}.d\hat{S}=\frac{Q_{enclosed}}{{\in }_0}}$
So, ${\displaystyle {\int }_S\hat{E}.d\hat{S}=E\int dS}$
${\displaystyle =E\cdot 2\pi rL}$
${\displaystyle =\frac{\pi r^{2}L\cdot \rho }{{\in }_0}}$
Now, ${\displaystyle E\cdot 2\pi rL=\frac{\pi r^{2}L}{{\in }_0}\cdot \frac{\lambda }{\pi R^2}}$
So, ${\displaystyle E=\frac{1}{4\pi {\in }_0}\cdot \frac{2\lambda r}{R^2}\cdot \hat{r}}$
Where ${\displaystyle \hat{r}}$ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength inside the cylinder is
${\displaystyle E=\frac{1}{4\pi {\in }_0}\cdot \frac{2\lambda r}{R^2}\cdot \hat{r}}$

Step 1 a) Apply Gausss