Chapter 22, Problem 63P

(a)

To determine

The total and individual charge on the inner and outer surface of the spherical shell.

Explanation of Solution

Given:

A positive point charge of $2.5\text{μC}$ is placed at the center of a conducting sphere. The radius of the inner surface is $60\text{cm}$ and the radius of the outer surface is $90\text{cm}$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

  ${\sigma }_{\text{surface}}=\frac{Q_{\text{enclosed}}}{A}$   ........ (1)

Here, ${\sigma }_{\text{surface}}$ is the surface charge density, $Q_{\text{enclosed}}$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

  $A=4\pi R^2$

Here $R$ is the radius of the spherical shell.

Substitute $4\pi R^2$ for $A$ in expression (1).

  ${\sigma }_{\text{surface}}=\frac{Q_{\text{enclosed}}}{4\pi R^2}$   ........ (2)

Forinner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge.

Calculation:

Substitute $-2.5\text{μC}$ for $Q_{\text{enclosed}}$ and $60\text{cm}$ for $R$ in the expression (2).

  $\begin{array}{c}{\sigma }_{\text{surface}}=\frac{-2.5\text{μC}}{4\pi {\left(60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =-0.55\text{μC}/{\text{m}}^{\text{2}}\end{array}$

For outer surface of the spherical shell

The induced charge on the outer surface has same magnitude of the point charge .The polarity of the charge is same to the point charge

Substitute $2.5\text{μC}$ for $Q_{\text{enclosed}}$ and $90\text{cm}$ for $R$ in the expression (2).

  $\begin{array}{c}{\sigma }_{\text{surface}}=\frac{2.5\text{μC}}{4\pi {\left(90\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =0.25\text{μC}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the total charge on the inner surface is $-2.5\text{μC}$ and the surface charge density is $-0.55\text{μC}/{\text{m}}^{\text{2}}$ .The total charge on the outer surface is $2.5\text{μC}$ and the surface charge density is $0.25\text{μC}/{\text{m}}^{\text{2}}$ .

(b)

To determine

The electric field due to a positive point charge located at the spherical shell.

Explanation of Solution

Given:

A positive point charge of $2.5\text{μC}$ is placed at the center of a conducting sphere. The radius of the inner surface is $60\text{cm}$ and the radius of the outer surface is $90\text{cm}$ .

Formula used:

Write the expression for Gauss’s law.

  ${\displaystyle \int \overrightarrow{E}}.d\overrightarrow{s}=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is the electric field, $d\overrightarrow{s}$ is the elementary surface area, $Q_{\text{enclosed}}$ is the total charge enclosed by the spherical shell and ${\epsilon }_0$ is the permittivity of free space.

Substitute $4\pi r^2$ for $d\overrightarrow{s}$ in the above expression.

  ${\displaystyle \int E\cdot 4\pi r^2}=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

  $E=\frac{Q_{\text{enclosed}}}{4\pi r^2\cdot {\epsilon }_0}$

Substitute $K$ for $1/4\pi {\epsilon }_0$ in the above equation.

  $E=\frac{K{Q}_{\text{enclosed}}}{r^2}$   ........ (3)

Calculation:

For $r<60\text{cm}$

Substitute $2.5\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(2.5\text{μC}\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)}{r^2}\\ =\left(2.25\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}\end{array}$

For $60\text{cm<r<90}\text{cm}$

Substitute $0\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(0\text{μC}\right)}{r^2}\\ =0\end{array}$

For $\text{r>90}\text{cm}$

Substitute $2.5\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(2.5\text{μC}\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)}{r^2}\\ =\left(2.25\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}\end{array}$

Conclusion:

Thus, the electric field of the spherical shell for $r<60\text{cm}$ is $\left(2.5\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}$ , For $60\text{cm<r<90}\text{cm}$ is $0$ and For $\text{r}\text{>}\text{90}\text{cm}$ is $\left(2.5\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}$ .

(c)

To determine

The electric field and surface charge density of the spherical shell.

Explanation of Solution

Given:

A positive point charge of $3.5\text{μC}$ is placed at the center of a conducting sphere. The radius of the inner surface is $60\text{cm}$ and the radius of the outer surface is $90\text{cm}$ .

Formula used:

The positive charge at the center of the spherical sphere induces a charge of same magnitude but different in polarity on the inner surface. The negative charge on the inner surface induces a positive charge of same magnitude on the outer surface.

Write the expression for the surface charge density

  ${\sigma }_{\text{surface}}=\frac{Q_{\text{enclosed}}}{A}$

Here, ${\sigma }_{\text{surface}}$ is the surface charge density, $Q_{\text{enclosed}}$ is the total charge enclosed by the surface of the spherical shell and $A$ is the surface area of the spherical shell.

Write the expression for the surface area of the square sheet.

  $A=4\pi R^2$

Here $R$ is the radius of the spherical shell.

Substitute $4\pi R^2$ for $A$ in expression (1).

  ${\sigma }_{\text{surface}}=\frac{Q_{\text{enclosed}}}{4\pi R^2}$

For inner surface of the spherical shell.

The induced charge on the inner surface has same magnitude of the point charge .The polarity of the charge is opposite to the point charge

Write the expression for Gauss’s law.

  ${\displaystyle \int \overrightarrow{E}}.d\overrightarrow{s}=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is the electric field, $d\overrightarrow{s}$ is the elementary surface area, $Q_{\text{enclosed}}$ is the total charge enclosed by the spherical shell and ${\epsilon }_0$ is the permittivity of free space.

Substitute $4\pi r^2$ for $d\overrightarrow{s}$ in the above expression.

  ${\displaystyle \int E\cdot 4\pi r^2}=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$

Here $r$ is the distance from the center.

Rearrange the above equation for $E$ .

  $E=\frac{Q_{\text{enclosed}}}{4\pi r^2\cdot {\epsilon }_0}$

Substitute $K$ for $1/4\pi {\epsilon }_0$ in the above equation.

  $E=\frac{K{Q}_{\text{enclosed}}}{r^2}$

Calculation:

For inner surface $E=0$ inside a conductor.

Substitute for $-2.5\text{μC}$

  $Q_{\text{enclosed}}$ and $60\text{cm}$ for $R$ in the expression (2).

  $\begin{array}{c}{\sigma }_{\text{surface}}=\frac{-2.5\text{μC}}{4\pi {\left(60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =-0.55\text{μC}/{\text{m}}^{\text{2}}\end{array}$

For outer surface of the spherical shell

Write the expression for surface charge.

  $q_{\text{outer}}=3.50\text{μC}{\text{-q}}_{\text{inner}}$

Substitute $-2.5\text{μC}$ for ${\text{q}}_{\text{inner}}$

  $q_{\text{outer}}=6.0\text{μC}$

For outer surface.

Substitute for $\text{6}\text{.0}\text{μC}$

  $Q_{\text{enclosed}}$ and $90\text{cm}$ for $R$ in the expression (2).

  $\begin{array}{c}{\sigma }_{\text{surface}}=\frac{6.0\text{μC}}{4\pi {\left(90\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =0.59\text{μC}/{\text{m}}^{\text{2}}\end{array}$

For $r<60\text{cm}$

Substitute $2.5\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(2.5\text{μC}\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)}{r^2}\\ =\left(2.25\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}\end{array}$

For $60\text{cm<r<90}\text{cm}$

Substitute $0\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(0\text{μC}\right)}{r^2}\\ =0\end{array}$

For $\text{r>90}\text{cm}$

Substitute $6.0\text{μC}$ for $Q_{\text{enclosed}}$ and $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $K$ in the expression (3).

  $\begin{array}{c}E=\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(6.0\text{μC}\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)}{r^2}\\ =\left(5.4\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}\end{array}$

Conclusion:

Thus, the total charge on the inner surface is $-2.5\text{μC}$ and the surface charge density is $-0.55\text{μC}/{\text{m}}^{\text{2}}$ .The total charge on the outer surface is $\text{6}\text{.0μC}$ and the surface charge density is $0.59\text{μC}/{\text{m}}^{\text{2}}$ . Thus, the electric field of the spherical shell for $r<60\text{cm}$ is $\left(2.25\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}$ , for $60\text{cm<r<90}\text{cm}$ is $0$ and for $\text{r>90}\text{cm}$ is $\left(5.4\times {10}^4{\text{Nm}}^{\text{2}}/\text{C}\right)\frac{1}{r^2}$ .