Chapter 24, Problem 60AP

(a)

To determine

The electric field for $r<a$.

Answer to Problem 60AP

The electric field for $r<a$ is $\left(\frac{2\lambda k_{\text{e}}}{r}\right)$ directed radial outwards.

Explanation of Solution

The diagram for the cylindrical shell with inner radius $a$ and the outer radius $b$ is shown below.

Figure (1)

Here, $a$ is the inner radius, $b$ is the outer radius and $\lambda$ is the linear charge density.

Write the expression for the linear charge density.

    $\lambda =\frac{q_{\text{en}}}{L}$                                                                                                                      (I)

Here, $L$ is the length of the cylindrical shell and $q_{\text{en}}$ is the enclosed charge.

Rearrange equation (I) to get the expression for $q_{\text{en}}$.

    $q_{\text{en}}=\lambda L$

Write the expression to calculate the electric field for the region $r<a$.

    $EA=\frac{q_{\text{en}}}{{\epsilon }_0}$                                                                                                                  (II)

Here, $E$ is the electric field for the region $r<a$, ${\epsilon }_0$ is the permittivity of free space and $A$ is the area of cylinder.

Substitute $\lambda L$ for $q_{\text{en}}$ and $2\pi rL$ for $A$ in equation (II) to solve for $E$.

    $\begin{array}{c}E\left(2\pi rL\right)=\frac{\lambda L}{{\epsilon }_0}\\ E=\frac{\lambda }{2\pi r{\epsilon }_0}\times \frac{1}{2}\times 2\\ E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2\lambda }{r}\right)\end{array}$                                                                                       (III)

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in equation (III).

    $E=\left(\frac{2\lambda k_{\text{e}}}{r}\right)$

Positive sign indicate the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for $r<a$ is $\left(\frac{2\lambda k_{\text{e}}}{r}\right)$ directed radial outward.

(b)

To determine

The electric field for $a<r<b$.

Answer to Problem 60AP

The electric field for $a<r<b$ is $k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(r^2-a^2\right)\right)\right)}{r}\right)$ directed radially outward.

Explanation of Solution

Write the expression for the volume charge density.

    $\rho =\frac{q_{\text{v}}}{V}$                                                                                                                    (IV)

Here, $\rho$ is the volume charge density and $V$ is the volume of the enclosed charge.

Rearrange equation (IV) to obtain the expression for $q_{\text{en}}$.

    $q_{\text{v}}=\rho V$                                                                                                                   (V)

Here, $q_{\text{v}}$ is the volumetric charge.

Write the expression to calculate the volume of the enclosed charge.

    $V=V_{\text{r}}-V_{\text{a}}$                                                                                                               (VI)

Here, $V_{\text{r}}$ is the volume of cylinder with radius $r$ and $V_{\text{a}}$ is the volume of cylinder with radius $a$.

Substitute $\pi r^{2}L$ for $V_{\text{r}}$ and $\pi a^{2}L$ for $V_{\text{a}}$ in equation (VI) to solve for $V$.

    $\begin{array}{c}V=\pi r^{2}L-\pi a^{2}L\\ =\pi L\left(r^2-a^2\right)\end{array}$

Write the expression for total charge inside the cylinder.

    $q_{\text{en}}=q_{\text{L}}+q_{\text{V}}$                                                                                                          (VII)

Here, $q_{\text{L}}$ is the linear charge.

Substitute $\lambda L$ for $q_{\text{L}}$ and $\rho V$ for $q_{\text{v}}$ in equation (VII) to solve for $q_{\text{en}}$.

    $q_{\text{en}}=\lambda L+\rho V$

Write the expression to calculate the electric field for the region $a<r<b$.

    $EA=\frac{q_{\text{en}}}{{\epsilon }_0}$.                                                                                                             (VIII)

Here, $E$ is the electric field for the region $a<r<b$.

Substitute $\lambda L+\rho V$ for $q_{\text{en}}$, $2\pi rL$ for $A$ and $\pi L\left(r^2-a^2\right)$ for $V$ in equation (V) to solve for $E$.

    $\begin{array}{c}E\left(2\pi rL\right)=\frac{\lambda L+\rho \left(\pi L\left(r^2-a^2\right)\right)}{{\epsilon }_0}\\ E=\frac{\lambda +\rho \left(\pi \left(r^2-a^2\right)\right)}{2\pi r{\epsilon }_0}\times \frac{2}{2}\\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{2\left(\lambda +\rho \left(\pi \left(r^2-a^2\right)\right)\right)}{r}\right)\end{array}$                                                             (IX)

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in equation (IX).

    $E=k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(r^2-a^2\right)\right)\right)}{r}\right)$

Positive sign indicates the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for $a<r<b$ is $k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(r^2-a^2\right)\right)\right)}{r}\right)$ directed radially outward.

(c)

To determine

The electric field for $r>b$.

Answer to Problem 60AP

The electric field for $r>b$ is $k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(b^2-a^2\right)\right)\right)}{r}\right)$ directed radially outward.

Explanation of Solution

Write the expression to calculate the volume of the enclosed charge.

    $V=V_{\text{b}}-V_{\text{a}}$                                                                                                               (X)

Here, $V_{\text{b}}$ is the volume of cylinder with radius $b$ and $V_{\text{a}}$ is the volume of cylinder with radius $a$.

Substitute $\pi b^{2}L$ for $V_{\text{b}}$ and $\pi a^{2}L$ for $V_{\text{a}}$ in equation (X) to solve for $V$.

    $\begin{array}{c}V=\pi b^{2}L-\pi a^{2}L\\ =\pi L\left(b^2-a^2\right)\end{array}$

Write the expression to calculate the electric field for the region $r>b$.

    $EA=\frac{q_{\text{en}}}{{\epsilon }_0}$.                                                                                                                (XI)

Here, $E$ is the electric field for the region $r>b$.

Substitute $\lambda L+\rho V$ for $q_{\text{en}}$, $2\pi rL$ for $A$ and $\pi L\left(b^2-a^2\right)$ for $V$ in equation (XI) to solve for $E$.

    $\begin{array}{c}E\left(2\pi rL\right)=\frac{\lambda L+\rho \left(\pi L\left(b^2-a^2\right)\right)}{{\epsilon }_0}\\ E=\frac{\lambda +\rho \left(\pi \left(b^2-a^2\right)\right)}{2\pi r{\epsilon }_0}\times \frac{2}{2}\\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{2\left(\lambda +\rho \left(\pi \left(b^2-a^2\right)\right)\right)}{r}\right)\end{array}$                                                          (XII)

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in equation (XII).

    $E=k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(b^2-a^2\right)\right)\right)}{r}\right)$

Positive sign indicate the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for $r>b$ is $k_{\text{e}}\left(\frac{2\left(\lambda +\rho \left(\pi \left(b^2-a^2\right)\right)\right)}{r}\right)$ directed radially outward.