tripiverded9

2022-01-15

A very long, uniformly charged cylinder has radius R and linear charge density $\lambda$.
a. Find the cylinder's electric field strength outside the cylinder, $r\ge R$. Give your answer as a multiple of $\frac{\lambda }{{\xi }_{0}}$.
Express your answer in terms of some or all of the variables R, r, and the constant $\pi$.
b. Find the cylinder's electric field strength inside the cylinder, $r\le R$. Give your answer as a multiple of $\frac{\lambda }{{\xi }_{0}}$.
Express your answer in terms of some or all of the variables R, r, and the constant $\pi$.

Mary Herrera

Expert

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

Piosellisf

Expert

Given Data:
- Linear charge density is $\lambda$
Part a
We have to find out the electric field strength outside the cylinder $r\ge R$
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,
${\int }_{S}\stackrel{\to }{E}.d\stackrel{\to }{S}=\frac{\lambda L}{{\in }_{0}}$
Where $\lambda L$ is the charge enclosed by the surface.
Thus,
${\int }_{S}\stackrel{\to }{E}.d\stackrel{\to }{S}=E\int dS$
$=E.2\pi rL$
$=\frac{\lambda L}{{\in }_{0}}$
So, $E=\frac{1}{4\pi {\in }_{0}}\cdot \frac{2\lambda }{r}\cdot \stackrel{^}{r}$
Where $\stackrel{^}{r}$ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength outside the cylinder is $E=\frac{1}{4\pi {\in }_{0}}\cdot \frac{2\lambda }{r}\cdot \stackrel{^}{r}$
Part b
Now, we have to find out the electric field strength inside the cylinder $r\le R$
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,
${Q}_{enclosed}=\pi {r}^{2}L\cdot \rho$
Where $\rho$ is the volume charge density.
Now, volume charge density is related to linear charge density as,
$\pi {R}^{2}\cdot 1\cdot \rho =\lambda \rho =\frac{\lambda }{\pi {R}^{2}}$
From Gauss law,
${\int }_{S}\stackrel{^}{E}.d\stackrel{^}{S}=\frac{{Q}_{enclosed}}{{\in }_{0}}$
So, ${\int }_{S}\stackrel{^}{E}.d\stackrel{^}{S}=E\int dS$

$=\frac{\pi {r}^{2}L\cdot \rho }{{\in }_{0}}$
Now,
So, $E=\frac{1}{4\pi {\in }_{0}}\cdot \frac{2\lambda r}{{R}^{2}}\cdot \stackrel{^}{r}$
Where $\stackrel{^}{r}$ is the direction of electric field and it is normal to the curved portion.
The cylinder's electric field strength inside the cylinder is
$E=\frac{1}{4\pi {\in }_{0}}\cdot \frac{2\lambda r}{{R}^{2}}\cdot \stackrel{^}{r}$

alenahelenash

Expert