tripiverded9

Answered

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 $\pi$.

Answer & Explanation

Mary Herrera

Expert

2022-01-16Added 37 answers

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

Given:

Radius of the cylinder

Linear Charge density of the cylinder

Step 2

Calculating the electric field outside the cylinder:

Let’s assume a Cylindrical Gaussian surface of radius ‘r’

Charge enclosed by the Guassian surface,

Using Gausss

Piosellisf

Expert

2022-01-17Added 40 answers

Given Data:

- Radius of cylinder is

- 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,

$\int}_{S}\overrightarrow{E}.d\overrightarrow{S}=\frac{\lambda L}{{\in}_{0}$

Where$\lambda L$ is the charge enclosed by the surface.

Thus,

${\int}_{S}\overrightarrow{E}.d\overrightarrow{S}=E\int dS$

$=E.2\pi rL$

$=\frac{\lambda L}{{\in}_{0}}$

So,$E=\frac{1}{4\pi {\in}_{0}}\cdot \frac{2\lambda}{r}\cdot \hat{r}$

Where$\hat{r}$ is the direction of electric field and it is normal to the curved portion.

The cylinder's electric field strength outside the cylinder is$E=\frac{1}{4\pi {\in}_{0}}\cdot \frac{2\lambda}{r}\cdot \hat{r}$

Part b

Now, we have to find out the electric field strength inside the cylinder$r\le R$

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,

$\int}_{S}\hat{E}.d\hat{S}=\frac{{Q}_{enclosed}}{{\in}_{0}$

So,${\int}_{S}\hat{E}.d\hat{S}=E\int dS$

$=E\cdot 2\pi r\text{}L$

$=\frac{\pi {r}^{2}L\cdot \rho}{{\in}_{0}}$

Now,$E\cdot 2\pi r\text{}L=\frac{\pi {r}^{2}L}{{\in}_{0}}\cdot \frac{\lambda}{\pi {R}^{2}}$

So,$E=\frac{1}{4\pi {\in}_{0}}\cdot \frac{2\lambda r}{{R}^{2}}\cdot \hat{r}$

Where$\hat{r}$ is the direction of electric field and it is normal to the curved portion.

The cylinder's electric field strength inside the cylinder is

$E=\frac{1}{4\pi {\in}_{0}}\cdot \frac{2\lambda r}{{R}^{2}}\cdot \hat{r}$

- Radius of cylinder is

- Linear charge density is

Part a

We have to find out the electric field strength outside the cylinder

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,

Where

Thus,

So,

Where

The cylinder's electric field strength outside the cylinder is

Part b

Now, we have to find out the electric field strength inside the cylinder

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,

Where

Now, volume charge density is related to linear charge density as,

From Gauss law,

So,

Now,

So,

Where

The cylinder's electric field strength inside the cylinder is

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1
a) Apply Gausss

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