91影视

1. Linear charge density,$位=2.00脳{10}^{鈭�8}\text{C/m}$
2. The radius of the metal tube,$R=3.00\text{鈥塩尘}$

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,

$\left|E\right|=\frac{位}{2蟺{蔚}_{0}r}$ (1)

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

For $r<R$, $q_{enc}=0\text{鈥�}C$

Thus, from the enclosed charge value in this case and equation (1), we get, $E=0\text{鈥塏/颁}$

Hence, the value of the electric field is$0\text{鈥塏/颁}$.

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

$\begin{array}{c}E=\frac{(2.0脳{10}^{鈭�8}\text{C/m})}{2蟺(0.06\text{m})(8.85脳{10}^{鈭�12}{\text{C}}^{\text{2}}\text{/N}鈰�{\text{m}}^{\text{2}})}\\ =5.99脳{10}^3\text{N/C}\end{array}$

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

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:

$\begin{array}{c}{E}_{max}=\frac{(2.0脳{10}^{鈭�8}\text{C/m})}{2蟺(0.03\text{m})(8.85脳{10}^{鈭�12}{\text{C}}^{\text{2}}\text{/N}鈰�{\text{m}}^{\text{2}})}\\ =1.2脳{10}^4\text{鈥�}\text{N/C}\end{array}$

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