Chapter 19, Problem 62P

(a)

To determine

The surface charge density on the ground, and whether the surface charge density is positive or negative.

Answer to Problem 62P

The surface charge density on the ground is $\underset{\_}{1.06\text{nC}/{\text{m}}^2}$, and the surface charge density is negative.

Explanation of Solution

Consider the earth’s surface as a charged conducting plane.

Write the expression for electric field due to charged conducting plane.

    $E=\sigma /{\epsilon }_0$        (I)

Here, $E$ is the electric field, $\sigma$ is the surface charge density, and ${\epsilon }_0$ is the free space of permittivity.

Rearrange equation (I), to find $\sigma$.

    $\sigma =E{\epsilon }_0$        (II)

Conclusion:

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

    $\begin{array}{c}\sigma =\left(120\text{N}/\text{C}\right)\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\\ =1.06\times {10}^{-9}\frac{\text{C}}{{\text{m}}^2}\times \frac{1\text{nC}}{{10}^{-9}\text{C}}\\ =1.06\text{nC}/{\text{m}}^2\end{array}$

The electric field inside the conducting earth is zero, the electric field is in downward direction. thus the earth is negatively charged.

Therefore, the surface charge density on the ground is $\underset{\_}{1.06\text{nC}/{\text{m}}^2}$, and the surface charge density is negative.

(b)

To determine

The charge of total surface of earth.

Answer to Problem 62P

The charge of total surface of earth is $\underset{\_}{-542\text{kC}}$.

Explanation of Solution

Write the expression for charge in terms of surface charge density.

    $Q=\sigma A$        (III)

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

Due to spherical symmetry of the earth, the surface area is $A=4\pi r^2$.

Apply the condition in the above equation (III), to find $Q$.

    $Q=\sigma \left(4\pi r^2\right)$        (IV)

Conclusion:

Substitute $-1.06\times {10}^{-9}\text{C}/{\text{m}}^{\text{2}}$ for $\sigma$, and $6.37\times {10}^6\text{m}$ for $r$ in equation (IV), to find $Q$.

    $\begin{array}{c}Q=\left(-1.06\times {10}^{-9}\text{C}/{\text{m}}^{\text{2}}\right)\left(4\times 3.14\times {\left(6.37\times {10}^6\text{m}\right)}^2\right)\\ =-542\times {10}^3\text{C}\times \frac{\text{1}\text{kC}}{{10}^3\text{C}}\\ =-542\text{kC}\end{array}$

Therefore, the charge of total surface of earth is $\underset{\_}{-542\text{kC}}$.

(c)

To determine

The electric force on the moon exerted by the earth.

Answer to Problem 62P

The electric force on the moon exerted by the earth is $\underset{\_}{4.88\times {10}^3\text{N}}$.

Explanation of Solution

Given that the charge on the moon is $27.3\%$ greater than the charge on the earth.

Write the expression electrostatic force.

    $F_{\text{E}}=\frac{k_{\text{e}}{q}_1{q}_2}{r^2}$        (V)

Here, $F_{\text{E}}$ is the electric force on the moon exerted by the earth, $q_1$ is the charge of the moon, $q_2$ is the charge on the earth, and $r$ is the distance between the earth and the moon.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $0.273\left(5.42\times {10}^5\text{C}\right)$ for $q_1$, and $5.42\times {10}^5\text{C}$ for $q_2$, and $3.84\times {10}^8\text{m}$ for $r$ in equation (V), to find $F_{\text{E}}$.

    $\begin{array}{c}{F}_{\text{E}}=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(0.273\times 5.42\times {10}^5\text{C}\times 5.42\times {10}^5\text{C}\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}\\ =4.88\times {10}^3\text{N}\end{array}$        (VI)

Therefore, the electric force on the moon exerted by the earth is $\underset{\_}{4.88\times {10}^3\text{N}}$.

(d)

To determine

Compare the result in subpart (c) with gravitational force on moon due to the earth.

Answer to Problem 62P

The gravitational force is $\underset{\_}{4.08\times {10}^{16}}$ times larger than the electrical force.

Explanation of Solution

From subpart (c), the electric force on the moon exerted by the earth is $F_{\text{E}}=4.88\times {10}^3\text{N}$.

Write the expression for gravitational force.

    $F_{\text{G}}=\frac{G{M}_{\text{E}}{M}_{\text{M}}}{r^2}$        (VII)

Here, $F_{\text{G}}$ is the gravitational force, $G$ is the universal gravitational constant, $M_{\text{E}}$ is the mass of the earth, $M_{\text{M}}$ is the mass of the moon, $r$ is the distance between the earth and the moon.

Conclusion:

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

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

Divide equation (VII) by (VIII), to find $\frac{F_{\text{G}}}{F_{\text{E}}}$.

    $\begin{array}{c}\frac{F_{\text{G}}}{F_{\text{E}}}=\frac{1.99\times {10}^{20}\text{N}}{4.88\times {10}^3\text{N}}\\ =4.08\times {10}^{16}\end{array}$

Rearrange the above equation to find $F_{\text{G}}$.

    $F_{\text{G}}=4.08\times {10}^{16}{F}_{\text{E}}$        (IX)

Therefore, the gravitational force is $\underset{\_}{4.08\times {10}^{16}}$ times larger than the electrical force.