91Ó°ÊÓ

a) Radii of the concentric cylindrical shells,$r_i=3cm$and ${\mathrm{r}}_0=6\mathrm{cm}$.

b) Linear charge density on the inner shell,${\mathrm{λ}}_i=5×{10}^{-6}C/m$

c) Linear charge density on the outer shell, ${\mathrm{λ}}_0=-7×{10}^{-6}\mathrm{C}/\mathrm{m}$

Using the concept of the electric field of a solid cylinder, we can get the required magnitude and direction of the electric fields at different radial distances.

Formula:

The electric field due to a cylinder, $E\left(r\right)=\frac{\mathrm{λ}}{2{\mathrm{Ï€°ùÎ\mu }}_0}$ (1)

Since${\mathrm{r}}_{\mathrm{i}}<\mathrm{r}=4.0\mathrm{cm}<{\mathrm{r}}_0$the magnitude of the electric field using equation (1) and due to the contribution of the inner charge density is given as:

role="math" localid="1657346140910" $$\begin{gathered} \mathrm{E}\left(\mathrm{r}\right)=\frac{5.0×{10}^{-6}\mathrm{C}/\mathrm{m}}{2\mathrm{π}\left(0.04\mathrm{m}\right)\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)} \\ =2.3×{10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $2.3×{10}^6\mathrm{N}/\mathrm{C}$.

The electric field E(r) points radially outward as the radial vector of this field is pointing in an outward direction.

Since,$r=8cm>r_0$, the electric field is given using equation (1) and due to the contribution from both the inner and outer linear densities as follows:

$$\begin{gathered} \mathrm{E}\left(\mathrm{r}\right)=\frac{5.0×{10}^{-6}\mathrm{C}/\mathrm{m}-\left(7.0×{10}^{-6}\mathrm{C}/\mathrm{m}\right)}{2\mathrm{π}\left(0.08\mathrm{m}\right)\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)} \\ =-4.5×{10}^5\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the magnitude of the electric field is $-4.5×{10}^5\mathrm{N}/\mathrm{C}$.

As a positive value of electric field indicates the outward direction of the electric field, hence, here in the above case, the minus sign indicates that E (r) points radially inward.