91Ó°ÊÓ

a) The magnitude of linear density,$\left|\mathrm{λ}\right|=3.6\mathrm{nC}/\mathrm{m}$

b) The wire has fixed negative charge as -q .

c) The radius of the cylindrical shell, r=1.5 cm

d) The net external field is zero.

Using the concept of the electric field of a Gaussian cylindrical surface, we can get the value of the linear density of the cylinder for the net external field to be zero. Now, using this linear density, we can get the surface density of the cylinder.

Formulae:

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

The surface density of a cylinder, $\mathrm{σ}=\frac{\mathrm{λ}}{2\mathrm{Ï€°ù}}$

The net electric field for r > R is given by using equation (1) as follows:

$$\begin{gathered} E_{net}=E_{wire}+E_{cyclinder} \\ =\frac{-\mathrm{λ}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}+\frac{{\mathrm{λ}}^{\text{'}}}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r} \end{gathered}$$

But, for the given net external field value to be zero, we get the linear charge density of the cylinder from the above equation as:${\mathrm{λ}}^{\text{'}}=3.6\mathrm{nC}/\mathrm{m}$

Now, substituting this value in equation (2), we can get the value of the surface charge density as follows:

$$\begin{gathered} \mathrm{σ}=\frac{3.6×{10}^{-6}\mathrm{C}/\mathrm{m}}{\left(2\mathrm{π}\right)\left(0.015\mathrm{m}\right)} \\ =3.8×{10}^{-8}\mathrm{C}/\mathrm{m} \end{gathered}$$

Hence, the value of the surface charge density is $3.8×{10}^{-8}\mathrm{C}/{\mathrm{m}}^2$.