Question

Figure 23-40 shows a section of a long, thin-walled metal tube of radius$\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{\text{cm}}$, with a charge per unit length of $\mathit{\lambda }\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\text{C/m}}$.

What is the magnitude Eof the electric field at radial distance

(a)$\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathit{R}\mathbf{/}\mathbf{2}\mathbf{.00}$ and

(b) $\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathit{R}$?

(c) Graph Eversus rfor the range$\mathit{r}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{2}\mathbf{.00}\mathit{R}$.

Short answer

1. The magnitude of the electric field at radial distance$r=R/2$ is$0\text{\hspace{0.17em}N/C}$ .
2. The magnitude of the electric field at radial distance$r=2R$ is$5.99\times {10}^3\text{\hspace{0.17em}N/C}$.
3. The graph of electric field versus radial distance is plotted for the range of $\mathit{r}\mathbf{=}\mathbf{0}\mathit{t}\mathit{o}\mathbf{2}\mathit{R}$.

Step-by-step solution

The given data

1. Linear charge density,$\lambda =2.00\times {10}^{-8}\text{C/m}$
2. The radius of the metal tube,$R=3.00\text{\hspace{0.17em}cm}$

Understanding the concept of Gauss law-planar symmetry

Using the concept of the electric field of a cylindrical Gaussian surface, we can get the electric field value at the given radial distances for the enclosed charge.

Formula:

The electric field of a cylindrical Gaussian surface,

$\left|E\right|=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ (1)

a) Calculation of the electric at r = R/2

We imagine a cylindrical Gaussian surface A of radius r and unit length concentric with the metal tube.

For $r<R$, $q_{enc}=0C$

Thus, from the enclosed charge value in this case and equation (1), we get, $E=0\text{\hspace{0.17em}N/C}$

Hence, the value of the electric field is$0\text{\hspace{0.17em}N/C}$.

b) Calculation of the electric at r = 2R

For ,the electric field, in this case, is given using $r=0.06\text{\hspace{0.17em}m}$the given data in equation (i) as follows:

$\begin{array}{c}E=\frac{(2.0\times {10}^{-8}\text{C/m})}{2\pi (0.06\text{m})(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}})}\\ =5.99\times {10}^3\text{N/C}\end{array}$

Hence, the value of the electric field is$5.99\times {10}^3\text{N/C}$

c) Calculation of the graph of the electric field with radial distances

The plot of E vs. r is shown.

Here, the maximum value of the electric field using the given data in equation (i) is given as:

$\begin{array}{c}{E}_{max}=\frac{(2.0\times {10}^{-8}\text{C/m})}{2\pi (0.03\text{m})(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}})}\\ =1.2\times {10}^4\text{N/C}\end{array}$

Here, the maximum value of the electric field is .$1.2\times {10}^4\text{N/C}$