Chapter 18, Problem 112P

(a)

To determine

What is the capacitance of Earth-ionosphere when it is treated as parallel plate capacitor and why is it ok to consider like that.

Answer to Problem 112P

The capacitance of Earth ionosphere when it is treated as parallel plate capacitor is $0.090\text{ F}$. It is ok to consider the earth as parallel plate since the distance between the plates are high.

Explanation of Solution

The Earth-ionosphere system can be considered as parallel plate since,

$\frac{d}{R}=\frac{5.0\times {10}^4\text{ m}}{6.371\times {10}^6\text{ m}}\approx {10}^{-2}$

Thus, the Earth can be considered as flat when compared to the distance between the plates.

Write the equation for capacitance.

$C=\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_{0}4\pi r^2}{d}$ (I)

Here, $C$ is capacitance, $A$ is the area, $r$ is radius, ${\epsilon }_0$ permittivity of free space and $d$ is distance between the plates.

Conclusion:

Substitute $8.854\times {10}^{-12}{\text{ C}}^{\text{2}}\text{/N}{\text{.m}}^2$ for ${\epsilon }_0$, $6.371\times {10}^6\text{ m}$ for $r$ and $5.0\times {10}^4\text{ m}$ for $d$ in equation I.

$\begin{array}{c}C=\frac{\left(8.854\times {10}^{-12}{\text{ C}}^{\text{2}}\text{/N}{\text{.m}}^2\right)4\pi {\left(6.371\times {10}^6\text{ m}\right)}^2}{5.0\times {10}^4\text{ m}}\\ =0.090\text{ F}\end{array}$

Therefore, the capacitance of Earth ionosphere when it is treated as parallel plate capacitor is $0.090\text{ F}$. It is ok to consider the earth as parallel plate since the distance between the plates are high.

(b)

To determine

Find the energy stored in the capacitor.

Answer to Problem 112P

The energy stored in the capacitor is $2.5\text{ TJ}$.

Explanation of Solution

Write the equation for energy.

$U=\frac{1}{2}C{V}^2=\frac{1}{2}C{\left(Ed\right)}^2$ (II)

Here, $U$ is the energy, $C$ is the capacitance, $E$ is the electric field, $d$ is distance between plates and $V$ is the voltage.

Conclusion:

Substitute $0.090\text{ F}$ for $C$, $150\text{ V/m}$ for $E$ and $5.0\times {10}^4\text{ m}$ for $d$ in equation II.

$\begin{array}{c}U=\frac{1}{2}\left(0.090\text{ F}\right){\left(\left(150\text{ V/m}\right)\left(5.0\times {10}^4\text{ m}\right)\right)}^2\\ =2.5\text{ TJ}\end{array}$

Therefore, the energy stored in the capacitor is $2.5\text{ TJ}$.

(c)

To determine

Find the resistance of lower atmosphere and the total current flows from Earth’s surface to ionosphere.

Answer to Problem 112P

The resistance of lower atmosphere is $\text{29 k}\Omega$ and the total current flows from Earth’s surface to ionosphere is $\text{260 A}$.

Explanation of Solution

Write the equation for resistance.

$R=\rho \frac{L}{A}=\rho \frac{L}{4\pi r^2}$ (III)

Here, $R$ is resistance, $\rho$ is the resistivity, $L$ is distance between the plates and $r$ is radius of Earth.

Write the equation for current.

$I=\frac{V}{R}=\frac{Ed}{R}$ (IV)

Here, $I$ is current, $V$ is voltage and $R$ is resistance.

Conclusion:

Substitute $3.0\times {10}^{14}\Omega \text{.m}$ for $\rho$, $6.371\times {10}^6\text{ m}$ for $r$ and $5.0\times {10}^4\text{ m}$ for $L$ in equation III.

$\begin{array}{c}R=\left(3.0\times {10}^{14}\Omega \text{.m}\right)\frac{5.0\times {10}^4\text{ m}}{4\pi {\left(6.371\times {10}^6\text{ m}\right)}^2}\\ =29408\Omega \\ \simeq 29\text{ k}\Omega \end{array}$

Substitute $29408\Omega$ for $R$, $150\text{ V/m}$ for $E$ and $5.0\times {10}^4\text{ m}$ for $d$ in equation IV.

$\begin{array}{c}I=\frac{\left(150\text{ V/m}\right)\left(5.0\times {10}^4\text{ m}\right)}{29408\Omega }\\ =260\text{ A}\end{array}$

Therefore, the resistance of lower atmosphere is $\text{29 k}\Omega$ and the total current flows from Earth’s surface to ionosphere is $\text{260 A}$.

(d)

To determine

Find the time it would take Earth’s charge to reduce to $1\%$ from its normal value.

Answer to Problem 112P

The time it would take Earth’s charge to reduce to $1\%$ from its normal value is $\text{200 min}$.

Explanation of Solution

Write the equation for charge.

$Q=Q_0{e}^{-t/RC}$

Here, $Q$ is the charge, $Q_0$ is the initial charge, $t$ is the time, $R$ is resistance and $C$ is the capacitance.

Rewrite the above equation to get time.

$\begin{array}{l}{e}^{t/RC}=\frac{Q_0}{Q}\\ \frac{t}{RC}=\ln \frac{Q_0}{Q}\end{array}$

$t=RC\ln \frac{Q_0}{Q}$ (V)

Conclusion:

Substitute $0.090\text{ F}$ for $C$, $29408\Omega$ for $R$, $\frac{1}{0.01}$ for $\frac{Q_0}{Q}$ in equation V.

$\begin{array}{c}t=\left[\left(29408\Omega \right)\left(0.090\text{ F}\right)\ln \frac{1}{0.01}\right]\left(\frac{1\text{ min}}{60\text{ s}}\right)\\ =200\text{ min}\end{array}$

Therefore, the time it would take Earth’s charge to reduce to $1\%$ from its normal value is $\text{200 min}$.