Chapter 16, Problem 94E

(a)

To determine

The expression for the electric field for $r>b$.

Answer to Problem 94E

The expression for the electric field for $r>b$ is $\underset{\_}{\frac{Q}{4\pi r^2{\epsilon }_0}}$.

Explanation of Solution

The below Figure 1 shows the metallic sphere of outer radius $b$ and inner radius $a$.

Write the expression for the Gauss’s law.

  ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{S}=\frac{Q}{{\epsilon }_0}$        (I)

Here, ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{A}$ is the area integral of the electric field $\overrightarrow{E}$ over a closed surface $d\overrightarrow{S}$, $Q$ is the amount of charge encompassed by that surface, and ${\epsilon }_0$ is the permittivity of the free space.

Write the expression for the surface area of a hollow sphere.

  ${\displaystyle \oint d\overrightarrow{S}}=4\pi r^2$        (II)

Conclusion:

Substitute Equation (II) in (I) to find the electric field for $r>b$.

  $\begin{array}{c}E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}\\ E=\frac{Q}{4\pi r^2{\epsilon }_0}\end{array}$

Therefore, the expression for the electric field for $r>b$ is $\underset{\_}{\frac{Q}{4\pi r^2{\epsilon }_0}}$.

(b)

To determine

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

Answer to Problem 94E

The expression for the electric field for $a<r<b$ is $\underset{\_}{\frac{Q\left(r^3-a^3\right)}{4\pi r^2{\epsilon }_0\left(b^3-a^3\right)}}$.

Explanation of Solution

From part (a), the expression for the Gauss’s law is,

  ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{S}=\frac{Q}{{\epsilon }_0}$

And the expression for the surface area of a hollow sphere is,

  ${\displaystyle \oint d\overrightarrow{S}}=4\pi r^2$

Use Equation (II) in (I).

$\begin{array}{c}E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}\\ E=\frac{Q}{4\pi r^2{\epsilon }_0}\end{array}$        (III)

The expression for the volume charge density of a hollow sphere is,

  $\rho =\frac{Q}{\frac{4}{3}\pi \left(b^3-a^3\right)}$        (IV)

Here, $\rho$ is the volume charge density of a hollow sphere.

The volume of the insulating hollow sphere for a small element is,

  $dV=4\pi r^2\cdot dr$        (V)

Here, $dV$ is the volume of the smallest element, r is the radius of the sphere, and $dr$ is the radius of the smallest element.

The charge enclosed by the smallest element in the hollow sphere between $a<b$ is,

$Q={\displaystyle \underset{a}{\overset{r}{\int }}\rho \left(r\right)}dV$        (VI)

Use Equation (V) in (VI).

$Q={\displaystyle \underset{a}{\overset{r}{\int }}\rho \left(r\right)}\left(4\pi r^{2}dr\right)$        (VII)

Use Equation (VII) in Equation (III).

$\begin{array}{c}E=\frac{{\displaystyle \underset{a}{\overset{r}{\int }}\rho \left(r\right)}\left(4\pi r^{2}dr\right)}{4\pi r^2{\epsilon }_0}\\ =\frac{4\pi \rho }{4\pi r^2{\epsilon }_0}{\displaystyle \underset{a}{\overset{r}{\int }}{r}^2\left(dr\right)}\\ =\frac{\rho }{r^2{\epsilon }_0}{\left[\frac{r^3}{3}\right]}_a^r\\ =\frac{\rho }{3r^2{\epsilon }_0}\left(r^3-a^3\right)\end{array}$        (VIII)

Conclusion:

Substitute Equation (IV) in (VIII) to find the electric field for $a<r<b$.

  $\begin{array}{c}E=\frac{Q\left(r^3-a^3\right)}{\frac{4}{3}\pi \left(b^3-a^3\right)\left(3{\epsilon }_0{r}^2\right)}\\ =\frac{Q\left(r^3-a^3\right)}{4\pi {\epsilon }_0{r}^2\left(b^3-a^3\right)}\end{array}$

Therefore, the expression for the electric field for $a<r<b$ is $\underset{\_}{\frac{Q\left(r^3-a^3\right)}{4\pi r^2{\epsilon }_0\left(b^3-a^3\right)}}$.

(c)

To determine

The expression for the electric field of $r>b$.

Answer to Problem 94E

The expression for the electric field for $r>b$ is  zero.

Explanation of Solution

From part (a), the expression for the Gauss’s law is,

  ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{S}=\frac{Q}{{\epsilon }_0}$

The charge $q$ induces an equal and opposite charge on the inner surface of the spherical shell and will have equal charge on the outer surface of the sphere.

Any charge placed inside hollow spherical conductor attracts opposite charge from sphere.

Since sphere is neutral, an equal and opposite charge appears on inner surface of sphere. The charge inside a conducting shell is zero.

Conclusion:

Therefore, expression for the electric field for $r>b$ is zero.