Chapter 19, Problem 75P

A solid, insulating sphere of radius a has a uniform charge density throughout its volume and a total charge Q. Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are b and c as shown in Figure P19.75. We wish to understand completely the charges and electric fields at all locations. (a) Find the charge contained within a sphere of radius r < a. (b) From this value, find the magnitude of the electric field for r < a. (c) What charge is contained within a sphere of radius r when a < r < b? (d) From this value, find the magnitude of the electric field for r when a < r < b. (e) Now consider r when b < r < c. What is the magnitude of the electric field for this range of values of r? (f) From this value, what must be the charge on the inner surface of the hollow sphere? (g) From part (f), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical surfaces of radii a, b, and c. Which of these surfaces has the largest magnitude of surface charge density?

To determine

The charge enclosed by the Gaussian surface in the region $r<a$.

Answer to Problem 75P

The charge enclosed by the Gaussian surface in the region $r<a$ is $\underset{\_}{Q{\left(\frac{r}{a}\right)}^3}$.

Explanation of Solution

Figure 1 represents an insulating sphere of radius $a$ concentric with a conducting hollow sphere of inner radii $b$ and outer radii $c$.

Consider a Gaussian surface of radius $r$.

Write the expression for charge enclosed by the Gaussian surface in the region $r<a$.

    $q_{\text{in}}=\rho V_{\text{encl}}$        (I)

Here, $q_{\text{in}}$ is the charge enclosed, $\rho$ is the volume charge density, and $V_{\text{encl}}$ is the volume enclosed.

Write the expression for volume enclosed by the Gaussian surface.

    $V_{\text{encl}}=\frac{4}{3}\pi r^3$        (II)

Here, $r$ is the radius of the Gaussian surface.

Use equation (II) in (I).

    $q_{\text{in}}=\rho \left(\frac{4}{3}\pi r^3\right)$        (III)

Write the expression for volume charge density for the insulating sphere of radius $a$.

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

Use equation (IV) in (III),to find $q_{\text{in}}$.

    $\begin{array}{c}{q}_{\text{in}}=\frac{Q}{\left(4/3\pi a^3\right)}\left(\frac{4}{3}\pi r^3\right)\\ =Q{\left(\frac{r}{a}\right)}^3\end{array}$        (V)

Conclusion:

Therefore, the charge enclosed by the Gaussian surface in the region $r<a$ is $\underset{\_}{Q{\left(\frac{r}{a}\right)}^3}$.

To determine

The electric field in the region $r<a$.

Answer to Problem 75P

The electric field in the region $r<a$ is $\underset{\_}{k_{\text{e}}\frac{Qr}{a^3}}$.

Explanation of Solution

Write the expression for Gauss law.

    ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{q_{\text{in}}}{{\epsilon }_0}$        (VI)

Here, $q_{\text{in}}$ is the charge enclosed by the Gaussian surface, ${\epsilon }_0$ is the free space of permittivity, $\overrightarrow{E}$ is the electric field, and $d\overrightarrow{A}$ is the elemental area.

Due to spherical symmetry, the element of area $d\overrightarrow{\text{A}}$ is $4\pi r^2$. The above equation is written as

    $E\left(4\pi r^2\right)=\frac{q_{\text{in}}}{{\epsilon }_0}$        (VII)

From subpart (a), the charge enclosed in the region $r<a$ is given by $Q{\left(\frac{r}{a}\right)}^3$.

Use equation (V) in (VII), and rearrange.

    $\begin{array}{c}E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}{\left(\frac{r}{a}\right)}^3\\ E=\frac{1}{4\pi {\epsilon }_0}\frac{Qr}{a^3}\end{array}$        (VIII)

The above is written as

    $E=k_{\text{e}}\frac{Qr}{a^3}$        (IX)

Here, $k_e$ is the coulomb’s constant.

Conclusion:

Therefore, the electric field in the region $r<a$ is $\underset{\_}{k_{\text{e}}\frac{Qr}{a^3}}$.

To determine

The charge enclosed by the Gaussian surface in the region $a<r<b$.

Answer to Problem 75P

The charge enclosed by the Gaussian surface in the region $a<r<b$ is $\underset{\_}{Q}$.

Explanation of Solution

Write the expression for charge enclosed by the Gaussian surface in the region $a<r<b$.

    $q_{\text{in}}=Q$        (X)

Here, $q_{\text{in}}$ is the charge enclosed.

Conclusion:

Therefore, charge enclosed by the Gaussian surface in the region $a<r<b$ is $\underset{\_}{Q}$

To determine

The magnitude of electric field in the region $a<r<b$.

Answer to Problem 75P

The magnitude of electric field in the region $a<r<b$ is $\underset{\_}{E=k_{\text{e}}\frac{Q}{r^2}}$.

Explanation of Solution

Write the expression for charge enclosed by the Gaussian surface in the region $a<r<b$.

    $q_{\text{in}}=Q$        (XI)

Here, $q_{\text{in}}$ is the charge enclosed

Use equation (XI) in (VII).

    $\begin{array}{c}E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}\\ E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\end{array}$        (XII)

The above equation is written as

    $E=k_{\text{e}}\frac{Q}{r^2}$

Conclusion:

Therefore, the magnitude of electric field in the region $a<r<b$ is $\underset{\_}{E=k_{\text{e}}\frac{Q}{r^2}}$.

To determine

The magnitude of electric field in the region $b\le r\le c$.

Answer to Problem 75P

The magnitude of electric field in the region $b\le r\le c$ is $\underset{\_}{E=0}$.

Explanation of Solution

Write the expression for Gauss law.

    ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{q_{\text{in}}}{{\epsilon }_0}$

The Gaussian surface encloses zero charge in the region $b\le r\le c$.

Write the expression for charge enclosed by the Gaussian surface in the region $b\le r\le c$.

    $q_{\text{in}}=0$        (XIII)

Here, $q_{\text{in}}$ is the charge enclosed.

Use equation (XIII) in (VI).

    $E=0$

Conclusion:

Therefore, the magnitude of electric field in the region $b\le r\le c$ is $\underset{\_}{E=0}$.

To determine

The charge on the inner surface of the hollow sphere.

Answer to Problem 75P

The charge on the inner surface of the hollow sphere is $\underset{\_}{Q_{\text{outer}}=-Q}$.

Explanation of Solution

Write the expression for Gauss law.

    ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{q_{\text{in}}}{{\epsilon }_0}$

Write the expression for charge enclosed by the Gaussian surface.

    $q_{\text{in}}=Q+Q_{\text{inner}}$        (XIV)

Here, $q_{\text{in}}$ is the charge enclosed.

Use equation (XIV) in (VI).

    ${\displaystyle \oint \overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{Q+Q_{\text{inner}}}{{\epsilon }_0}$        (XV)

The electric field inside the conductor is zero, the charge enclosed by the Gaussian surface is zero. The above equation is reduced to

    $\begin{array}{c}0=\frac{Q+Q_{\text{inner}}}{{\epsilon }_0}\\ Q_{\text{inner}}=-Q\end{array}$        (XVI)

Conclusion:

Therefore, the charge on the inner surface of the hollow sphere is $\underset{\_}{Q_{\text{outer}}=-Q}$.

To determine

The charge on the outer surface of the hollow sphere.

Answer to Problem 75P

The charge on the outer surface of the hollow sphere is $\underset{\_}{+Q}$.

Explanation of Solution

Write the expression for total charge inside the hollow sphere.

    $q_{\text{tot}}=Q+Q_{\text{inner}}$        (XVII)

Here, $q_{\text{in}}$ is the charge enclosed.

The total charge inside the conductor is zero, the above equation is reduced to

    $0=Q_{\text{outer}}+Q_{\text{inner}}$

Rearrange the above equation, to find $Q_{\text{outer}}$.

    $Q_{\text{outer}}=-Q_{\text{inner}}$        (XVIII)

Conclusion:

Substitute $-Q$ for $Q_{\text{inner}}$ in equation (XVIII), to find $Q_{\text{outer}}$.

    $Q_{\text{outer}}=+Q$

Therefore, The charge on the outer surface of the hollow sphere is $\underset{\_}{+Q}$.

To determine

Among the three spherical surfaces having radii $a$, $b$, and $c$ has the largest magnitude of surface charge density.

Answer to Problem 75P

The inner surface of radius $b$ has largest surface charge density.

Explanation of Solution

Write the expression for surface charge density.

    $\sigma =\frac{q}{A}$        (XIX)

Here, $\sigma$ is the surface charge density, $q$ is the charge, and $A$ is the area.

The solid insulating sphere has small surface charge density since its total charge is uniformly distributed throughout its volume.

The inner surface of the hollow cylinder with radius $b$ has a smaller surface area compared to the outer surface of the hollow cylinder with radius $c$, thus from equation (XIX) it clear that inner surface of the hollow cylinder has greater surface charge density due to its less surface area.

Conclusion:

Therefore, the inner surface of radius $b$ has largest surface charge density.