Question

Two long, charged, thin-walled, concentric cylindrical shells have radii of3.0 cm and 6.0 cm . The charge per unit length is $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}\mathbf{/}\mathbf{m}$on the inner shell and $\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}\mathbf{/}\mathbf{m}$on the outer shell. What are the (a) magnitude Eand (b) direction (radially inward or outward) of the electric field at radial distance r=4.0 cm ? What are (c) Eand (d) the direction at r=8.0 cm?

Short answer

a) The magnitude of the electric field at a radial distance r=4 cm is $2.3\times {10}^6\mathrm{N}/\mathrm{C}$.

b) The direction of the electric field at a radial distance r=4 cm is radially outward.

c) The magnitude of the electric field at a radial distance r=8 cm is $4.5\times {10}^5\mathrm{N}/\mathrm{C}$.

d) The direction of the electric field at a radial distance r=8 cm is radially inward.

Step-by-step solution

The given data

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,${\lambda }_i=5\times {10}^{-6}C/m$

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

Understanding the concept of Gauss law-planar symmetry

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

a) Calculation of the electric field at r = 4cm

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

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

b) Calculation of the direction of the electric field at r = 4cm

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

c) Calculation of the electric field at r = 8cm

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:

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

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

d) Calculation of the direction of the electric field at r = 8cm

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.