91Ó°ÊÓ

1. A narrow charged solid cylinder is coaxial with a larger charged cylindrical shell.
2. The vertical scale of magnitude of the electric field,${\mathrm{E}}_{\mathrm{s}}=3.0×{10}^3\mathrm{N}/\mathrm{C}$

Using the concept of the electric field of a cylinder, we can get the values of the internal and external fields. Subtracting these values will give us the net electric field, now solving it using the given data; we will have the linear charged density of the shell.

Formula:

The electric field of a Gaussian cylindrical surface, $\mathrm{E}=\frac{\mathrm{λ}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}}$ (1)

As we approach r=3.5 cm from the inside, we have the internal field value from the graph and the equation (1) such that,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{internal}}=\frac{\mathrm{λ}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}} \\ =1000\mathrm{N}/\mathrm{C} \end{gathered}$$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(2)

And as we approachr=3.5cm from the outside, we have the external field value from the graph and the equation (i) as:

${\mathrm{E}}_{\mathrm{external}}=\frac{\mathrm{λ}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}}+\frac{\mathrm{λ}\text{'}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(3)

Now, subtracting the equations (2) and (3), we can get the value of the linear density of the cylindrical shell as given:

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{λ}\text{'}}{2{\mathrm{Ï€Î\mu }}_0\mathrm{r}} \\ -4000\mathrm{N}/\mathrm{C}=\frac{\mathrm{λ}}{2\mathrm{Ï€}\left(8.85×{10}^{-12}\mathrm{F}/\mathrm{m}\right)\left(0.035\mathrm{m}\right)} \\ \mathrm{λ}\text{'}=-5.8×{10}^{-9}\mathrm{C}/\mathrm{m} \end{gathered}$$

Hence, the value of the linear density is $-5.8×{10}^{-9}\mathrm{C}/\mathrm{m}$.