Chapter 25, Problem 25.54AP

Review. In fair weather, the electric field in the air at a particular location immediately above the Earth's surface is 120 N/C directed downward, (a) What is the surface charge density on the ground? Is it positive or negative? (b) Imagine the surface charge density is uniform over the planet. What then is the charge of the whole surface of the Earth? (c) What is the Earth’s electric potential due to this charge? (d) What is the difference in potential between the head and the feet of a person 1.75 m tall? (Ignore any charges in the atmosphere.) (e) Imagine the Moon, with 27.3% of the radius of the Earth, had a charge 27.3% as large, with the same sign. Find the electric force the Earth would then exert on the Moon, (f) State how the answer to part (e) compares with the gravitational force the Earth exerts on the Moon.

To determine

The surface charge density on the ground.

Answer to Problem 25.54AP

The surface charge density on the ground is $-1.06\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward.

Write the expression for the surface charge density.

$\sigma =E{\epsilon }_0$

Here,

$\sigma$ is the surface charge density.

$E$ is the electric field.

${\epsilon }_0$ is the permittivity of free space.

The value of ${\epsilon }_0$ is $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$.

Substitute $120\text{N}/\text{C}$ for $E$ and $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ in above equation.

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

The direction of the electric field is downwards, so the direction of the surface charge density is negative.

Conclusion:

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

To determine

The charge of whole earth.

Answer to Problem 25.54AP

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

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward.

From part (a) the surface charge density on the ground is $-1.06\text{nC}/{\text{m}}^{\text{2}}$.

The radius of the earth is $6.37\times {10}^6\text{m}$.

The area of the earth is expressed as,

$A=4\pi r^2$

Here,

$r$ is the radius of the earth.

Substitute $6.37\times {10}^6\text{m}$ for $r$ in above equation.

$\begin{array}{c}A=4\pi {\left(6.37\times {10}^6\text{m}\right)}^2\\ =5.099\times {10}^{14}{\text{m}}^2\end{array}$

Thus, the area of the earth is $5.099\times {10}^{14}{\text{m}}^2$.

Write the expression for the charge of whole the earth.

$Q=\sigma A$

Here,

$Q$ is the charge.

$A$ is the area that encloses the charge.

Substitute $5.099\times {10}^{14}{\text{m}}^2$ for $E$ and $-1.06\text{nC}/{\text{m}}^{\text{2}}$ for $\sigma$ in above equation.

$\begin{array}{c}Q=-1.06\text{nC}/{\text{m}}^{\text{2}}\left(\frac{{10}^{-9}\text{nC}}{1\text{C}}\right)\times 5.099\times {10}^{14}{\text{m}}^2\\ =-540498.62\text{C}\left(\frac{1\text{kC}}{1000\text{C}}\right)\\ =-540.498\text{kC}\\ \approx -541\text{kC}\end{array}$

Conclusion:

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

To determine

The electric potential of earth.

Answer to Problem 25.54AP

The electric potential of earth is $-7.64\times {10}^2\text{MV}$.

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward.

From part (b) the charge of the earth is $-541\text{kC}$.

Write the expression for the electric potential of earth.

$V=\frac{k_{\text{e}}Q}{r}$

Here,

$V$ is the electric potential of earth.

$k_{\text{e}}$ is the coulomb’s constant.

The value of $k$ is $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^{\text{2}}$.

Substitute $-541\text{kC}$ for $Q$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$ and $6.37\times {10}^6\text{m}$ for $r$ in above equation.

$\begin{array}{c}V=\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^{\text{2}}\left(-541\text{kC}\left(\frac{1000\text{C}}{1\text{kC}}\right)\right)}{6.37\times {10}^6\text{m}}\\ =7.635\times {10}^8\text{V}\left(\frac{1\text{MV}}{{10}^6\text{V}}\right)\\ \approx -7.64\times {10}^2\text{MV}\end{array}$

Conclusion:

Therefore, the electric potential of earth is $-7.64\times {10}^2\text{MV}$.

To determine

The electric potential difference between the head and feet of a $1.75\text{m}$ tall person.

Answer to Problem 25.54AP

The electric potential difference between the head and feet of a $1.75\text{m}$ tall person is $210\text{V}$.

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward.

Write the expression for the electric potential difference between the head and feet of a person.

$\Delta V=Ed$

Here,

$\Delta V$ is the electric potential difference between the head and feet of a person.

$d$ is the height of the person.

Substitute $120\text{N}/\text{C}$ for $E$ and $\text{1}\text{.75}\text{m}$ for $d$ in above equation.

$\begin{array}{c}\Delta V=120\text{N}/\text{C}\times \text{1}\text{.75}\text{m}\\ =210\text{V}\end{array}$

Conclusion:

Therefore, the electric potential difference between the head and feet of a $1.75\text{m}$ tall person is $210\text{V}$.

To determine

The electric force between the earth and moon.

Answer to Problem 25.54AP

The electric force between the earth and moon is $4.87\times {10}^3\text{N}$.

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward, the radius of the moon is $27.3\%$ of radius of earth and the charge of the moon is $27.3\%$ of charge of earth.

The distance between moon and earth is $3.84\times {10}^8\text{m}$.

From part (b) the charge of earth is $-541\text{kC}$.

The charge of moon is given by,

$\begin{array}{c}{Q}_{\text{m}}=27.3\%\left(-541\text{kC}\right)\\ =-147.693\text{kC}\end{array}$

Thus, the value of $Q_{\text{m}}$ is $-147.693\text{kC}$.

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

$F=\frac{k_{\text{e}}Q{Q}_{\text{m}}}{r^2}$

Here,

$F$ is the electric force between moon and earth.

$Q_{\text{m}}$ is the charge of moon.

$r$ is the distance between moon and earth.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $-541\text{kC}$ for $Q$, $-147.693\text{kC}$ for $Q_{\text{m}}$ and $3.84\times {10}^8\text{m}$ for $r$ in above equation.

$\begin{array}{c}F=\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^{\text{2}}\times \left(-541\text{kC}\right)\times \left(-147.693\text{kC}\right)}{{\left(3.84\times {10}^8\text{m}\right)}^2}{\left(\frac{1000\text{C}}{1\text{kC}}\right)}^2\\ =4.87\times {10}^3\text{N}\end{array}$

Conclusion:

Therefore, the electric force between the earth and moon is $4.87\times {10}^3\text{N}$.

To determine

The comparison of the gravitation force and the electric force

Answer to Problem 25.54AP

The gravitation force is $4.08\times {10}^{16}$ times larger than the electric force and electric force is negligible in comparison to gravitation force.

Explanation of Solution

Given info: The electric field above the earth’s surface is $120\text{N}/\text{C}$ directed towards downward, the radius of the moon is $27.3\%$ of radius of earth and the charge of the moon is $27.3\%$ of charge of earth.

The mass of earth is $5.98\times {10}^{24}\text{kg}$ and the mass of the moon is $7.36\times {10}^{22}\text{kg}$.

From part (e) the electric force between the earth and moon is $4.87\times {10}^3\text{N}$.

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

$F_{\text{g}}=\frac{G{m}_{\text{e}}{m}_{\text{m}}}{r^2}$

Here,

$F_{\text{g}}$ is the gravitational force between moon and earth.

$m_{\text{m}}$ is the mass of moon.

$m_{\text{e}}$ is the mass of earth.

$G$ is the gravitational constant.

The value of $G$ is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $7.36\times {10}^{22}\text{kg}$ for $m_{\text{m}}$, $5.98\times {10}^{24}\text{kg}$ for $m_{\text{e}}$ and $3.84\times {10}^8\text{m}$ for $r$ in above equation.

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

Thus, the value of $F_{\text{g}}$ is $1.99\times {10}^{20}\text{N}$.

Divide $F_{\text{g}}$ with $F$ to compare.

$\frac{F_{\text{g}}}{F}$

Substitute $1.99\times {10}^{20}\text{N}$ for $F_{\text{g}}$ and $4.87\times {10}^3\text{N}$ for $F$ in above equation.

$\frac{1.99\times {10}^{20}\text{N}}{4.87\times {10}^3\text{N}}=4.08\times {10}^{16}$

The gravitation force is $4.08\times {10}^{16}$ times larger than the electric force and opposite in direction.

Conclusion:

Therefore, the gravitation force is $4.08\times {10}^{16}$ times larger than the electric force and electric force is negligible in comparison to gravitation force.