Question

Figure 23-41ashows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are non-conducting and thin and have uniform surface charge densities on their outer surfaces. Figure 23-41bgives the radial component Eof the electric field versus radial distance rfrom the common axis, and. What is the shell’s linear charge density?

Short answer

The linear charge density of the shell is $-5.8\times {10}^{-9}\mathrm{C}/\mathrm{m}$.

Step-by-step solution

The given data

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

Understanding the concept of Gauss law-planar symmetry

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, $\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ (1)

Calculation of the linear charged density of the shell

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,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{internal}}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ =1000\mathrm{N}/\mathrm{C} \end{gathered}$$………………(2)

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

${\mathrm{E}}_{\mathrm{external}}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}+\frac{\mathrm{\lambda }\text{'}}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$……………….(3)

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

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{\lambda }\text{'}}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ -4000\mathrm{N}/\mathrm{C}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)\left(0.035\mathrm{m}\right)} \\ \mathrm{\lambda }\text{'}=-5.8\times {10}^{-9}\mathrm{C}/\mathrm{m} \end{gathered}$$

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