Chapter 25, Problem 54AP

(a)

To determine

The surface charge density of the ground.

Answer to Problem 54AP

The surface charge density of the ground is $1.062\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Write the expression for the electric field due to charged earth immediately above its surface.

    $E=\frac{Q}{4\pi {\epsilon }_0{R}^2}$                            (I)

Here, $E$ is the electric field just above the surface of the earth, $Q$ is the charge, $R$ is the radius of the sphere and ${\epsilon }_0$ is the permittivity of the free space.

Write the expression for the surface charge density.

    $\sigma =\frac{Q}{4\pi R^2}$                   (II)

Here, $\sigma$ is the surface charge density.

Substitute $\sigma$ for $\frac{Q}{4\pi R^2}$ in equation (I).

    $E=\frac{\sigma }{{\epsilon }_0}$                                         (III)

Conclusion:

Substitute $120\text{N}/\text{C}$ for $E$ and $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/{\text{N-m}}^{\text{2}}$ for ${\epsilon }_0$ in equation (III) to solve for $\sigma$.

    $\begin{array}{c}120\text{N}/\text{C}=\frac{\sigma }{8.85\times {10}^{-12}{\text{C}}^{\text{2}}/{\text{N-m}}^{\text{2}}}\\ \sigma =1.062\times {10}^{-9}\text{C}/{\text{m}}^{\text{2}}\times \frac{{10}^9\text{nC}}{\text{C}}\\ =1.062\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Since the electric field is directed downward. Hence, the surface charge density will be negative.

Therefore, the surface charge density of the ground is $1.062\text{nC}/{\text{m}}^{\text{2}}$.

(b)

To determine

The charge of the whole surface of the earth.

Answer to Problem 54AP

The charge of the whole surface of the earth is $-\text{541}\text{.51}\text{kC}$.

Explanation of Solution

Conclusion:

Substitute $6.37\times {10}^6\text{m}$ for $R$ and $-1.062\text{nC}/{\text{m}}^{\text{2}}$ for $\sigma$ in equation (II) to solve for $Q$.

    $\begin{array}{c}\left(-1.062\text{nC}/{\text{m}}^{\text{2}}\times \frac{{10}^{-9}\text{C}}{1\text{nC}}\right)=\frac{Q}{4\pi {\left(6.37\times {10}^6\text{m}\right)}^2}\\ Q=-\left(541518.4343\text{C}\times \frac{{10}^{-3}\text{kC}}{\text{C}}\right)\\ =-\text{541}\text{.51}\text{kC}\end{array}$

Therefore, the charge of the whole surface of the earth is $-\text{541}\text{.51}\text{kC}$.

(c)

To determine

The earth’s electric potential.

Answer to Problem 54AP

The earth’s electric potential is $-763\text{MV}$.

Explanation of Solution

Write the expression for the electric potential.

    $V=\frac{k_{\text{e}}Q}{R}$                              (IV)

Here, $V$ is the electric potential and $k_{\text{e}}$ is the Coulomb’s constant.

Conclusion:

Substitute $6.37\times {10}^6\text{m}$ for $R$, $8.9768\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$ and $-\text{541}\text{.51}\text{kC}$ for $Q$ in equation (IV) to solve for $V$.

    $\begin{array}{c}V=\frac{\left(8.9768\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-\text{541}\text{.51}\text{kC}\times \frac{{10}^3\text{C}}{1\text{kC}}\right)}{6.37\times {10}^6\text{m}}\\ =-763112553.8\text{V}\times \frac{{10}^{-6}\text{MV}}{1\text{V}}\\ =-763\text{MV}\end{array}$

Therefore, The earth’s electric potential is $-763\text{MV}$.

(d)

To determine

The difference in potential between the head and foot of a person.

Answer to Problem 54AP

The difference in potential between the head and foot of a person is $210\text{V}$.

Explanation of Solution

Write the relationship between the electric potential and the electric field.

    $V_{\text{H}}-V_{\text{F}}=Ed$                       (V)

Here, $V_{\text{H}}$ is the potential at head of a person, $V_{\text{F}}$ is the potential at the foot of a person and $d$ is the height of the person.

Conclusion:

Substitute $120\text{N}/\text{C}$ for $E$ and $1.75\text{m}$ for $d$ in equation (V) to get the value of $V_{\text{H}}-V_{\text{F}}$.

    $\begin{array}{c}{V}_{\text{H}}-V_{\text{F}}=120\text{N}/\text{C}\left(1.75\text{m}\right)\\ =210\text{V}\end{array}$

Therefore, the difference in potential between the head and foot of a person is $210\text{V}$.

(e)

To determine

The force that the earth would exert on the moon.

Answer to Problem 54AP

The force that the earth would exert on the moon is $4.874\text{kN}$.

Explanation of Solution

Write the expression for the magnitude of electrostatic force exerted by the earth on the moon.

    $F_{\text{E}}=k_{\text{e}}\frac{Q{Q}^{\prime }}{r^2}$                         (VI)

Here, $F_{\text{E}}$ is the magnitude of electrostatic force exerted by the earth on the moon, $Q^{\prime }$ is the charge on the moon and $r$ is the distance between the earth and moon.

Conclusion:

Substitute $8.9768\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $-\text{541}\text{.51}\text{kC}$ for $Q$, $\left(27.3\%\times \left(-\text{541}\text{.51}\text{kC}\right)\right)$ for $Q^{\prime }$ and for $r$ in equation (VI) to solve for $F_{\text{E}}$.

    $\begin{array}{c}{F}_{\text{E}}=8.9768\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(-\text{541}\text{.51}\text{kC}\times \frac{{10}^3\text{C}}{1\text{kC}}\right)\left(\left(27.3\%\times \left(-\text{541}\text{.51}\text{kC}\times \times \frac{{10}^3\text{C}}{1\text{kC}}\right)\right)\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}\\ =8.9768\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 5.43\times {10}^{-7}{\text{C}}^{\text{2}}/{\text{m}}^{\text{2}}\\ =4874\text{N}\times \frac{{10}^{-3}\text{kN}}{1\text{N}}\\ =4.874\text{kN}\end{array}$

Therefore, the force that the earth would exert on the moon is $4.874\text{kN}$.

(f)

To determine

The comparison between the gravitational force and the electrostatic force exerted by the earth on the moon.

Answer to Problem 54AP

The gravitational force is $4.08\times {10}^{16}$ times greater than the electrostatic force exerted by the earth on the moon.

Explanation of Solution

Write the expression for the gravitational force between the earth and moon.

    $F_{\text{G}}=\frac{GMm}{r^2}$                                     (VII)

Here, $F_{\text{G}}$ is the gravitational force between the earth and moon, $G$ is the gravitational constant and $m$ is the moon.

Write the expression to calculate the ratio of gravitational force and the electrostatic force between the earth and the moon.

    $\frac{F_{\text{G}}}{F_{\text{E}}}$                                            (VIII)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}/\text{kg}$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M$, $7.36\times {10}^{22}\text{kg}$ for $m$ and $3.84\times {10}^8\text{m}$ for $r$ in equation (VII) to solve for $F_{\text{G}}$.

    $\begin{array}{c}{F}_{\text{G}}=\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}/\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(7.36\times {10}^{22}\text{kg}\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}\\ =1.99\times {10}^{20}\text{N}\end{array}$

Substitute $1.99\times {10}^{20}\text{N}$ for $F_{\text{G}}$ and $4.874\text{kN}$ for $F_{\text{E}}$ in equation (VIII) to solve for $\frac{F_{\text{G}}}{F_{\text{E}}}$.

    $\begin{array}{c}\frac{F_{\text{G}}}{F_{\text{E}}}=\frac{1.99\times {10}^{20}\text{N}}{4.874\text{kN}\times \frac{{10}^3\text{N}}{1\text{kN}}}\\ \frac{F_{\text{G}}}{F_{\text{E}}}=4.08\times {10}^{16}\\ F_{\text{G}}=4.08\times {10}^{16}{F}_{\text{E}}\end{array}$

Therefore, the gravitational force is $4.08\times {10}^{16}$ times greater than the electrostatic force exerted by the earth on the moon.