A charge of uniform linear density 2.0nC/m is distributed along a long, thin, non-conducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius=5.0 cm , outer radius=10 cm ). The net charge on the shell is zero. (a) What is the magnitude of the electric field from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?

a) Linear charge density, $\lambda =2\times {10}^{-9}C/m$

b) The inner radius of the shell,$r_i=5\mathrm{cm}$

c) The outer radius of the shell,$r_0=10\mathrm{cm}$

d) Distance of the point from the field,d=15 cm

Using the concept of the electric field of a Gaussian cylindrical surface, we can get the value of the net electric field on the shell. Now, using the surface charge density formula of the cylinder, we can get the required charge value on the inner and the outer surface of the shell.

Formulae:

The magnitude of the electric field of a Gaussian cylindrical surface,$\left|E\right|=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ (1)

The surface density of a cylinder, $\sigma =\frac{\lambda }{2\mathrm{\pi r}}$ (2)

The field is being measured outside the system (the charged rod coaxial with the neutral cylinder) so that the net enclosed charge is only that which is on the rod. Thus, the net electric field on the cylindrical shell is given using equation (1) as follows:

$$\begin{gathered} \left|\mathrm{E}\right|=\frac{2\times \left(2.0\times {10}^{-9}\mathrm{C}/\mathrm{m}\right)}{4{\mathrm{\pi \epsilon }}_0\left(0.15\mathrm{m}\right)} \\ =2.4\times {10}^2\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the magnitude of the electric field is$2.4\times {10}^{-9}\mathrm{N}/\mathrm{C}$ .

Since the field is zero inside the conductor (in an electrostatic configuration), then there resides on the inner surface charge $-q$, and on the outer surface, charge $+q$(where q is the charge on the rod at the center).

Now, the charge on the inner surface of the shell is given using equation (2) as:

$$\begin{gathered} {\sigma }_{inner}=-\frac{\lambda }{2{\mathrm{\pi r}}_{\mathrm{i}}} \\ =-\frac{\left(2.0\times {10}^{-9}C/m\right)}{2\mathrm{\pi }\left(0.05\mathrm{m}\right)} \\ =-6.4\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the charge for the inner surface is$-6.4\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2$ .

With $r_0$= 0.10 m, the surface charge density of the outer surface is given using equation (2) as follows:

$$\begin{gathered} {\sigma }_{outer}=\frac{\left(2.0\times {10}^{-9}C/m\right)}{2\mathrm{\pi }\left(0.1\mathrm{m}\right)} \\ =3.2\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the surface charge density is$3.2\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2$ .