Question

Assume that clouds are distributed around the entire earth at a height of 3000m above the ground. The atmosphere can be modeled as a spherical capacitor with the earth as one plate and the cloud as other. When the electric field between the earth and the cloud becomes large, the air begins to conduct and the phenomena is called lightning. On a typical day $4 \times 10^5 C$ of positive charge is spread over the surface of the earth and equal amount of negative charge is there on the cloud. Resistivity of the air is $\rho = 3 \times 10^{13} \Omega m$ and radius of the earth = 6000km. (a) Find the resistance of the air gap between the earth's surface and cloud. (b) Estimate the potential difference between the surface of the earth and the cloud (c) In how much time the capacitor formed between the earth and the cloud will lose 63% of the charge?

Answer 1

(a) 198.9 Ω (b) 3 × 10^5 V (c) 265.5 s

Answer 2

The problem describes the Earth-cloud system as a spherical capacitor, with the Earth as the inner plate and the cloud as the outer plate. The air between them acts as a dielectric and a resistive medium.

Given values:

• Height of cloud, $h = 3000 \text{ m} = 3 \times 10^3 \text{ m}$
• Charge, $Q = 4 \times 10^5 \text{ C}$
• Resistivity of air, $\rho = 3 \times 10^{13} \Omega \text{ m}$
• Radius of Earth, $R_E = 6000 \text{ km} = 6 \times 10^6 \text{ m}$
• Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

Let $R_1 = R_E$ be the inner radius and $R_2 = R_E + h$ be the outer radius. $R_1 = 6 \times 10^6 \text{ m}$ $R_2 = 6 \times 10^6 + 3 \times 10^3 = 6.003 \times 10^6 \text{ m}$

(a) Find the resistance of the air gap between the earth's surface and cloud.

For a spherical shell of thickness $dr$ at radius $r$, the differential resistance $dR$ is given by: $dR = \rho \frac{dr}{4\pi r^2}$ To find the total resistance $R$, we integrate $dR$ from $R_1$ to $R_2$: $R = \int_{R_1}^{R_2} \frac{\rho}{4\pi r^2} dr = \frac{\rho}{4\pi} \left[ -\frac{1}{r} \right]_{R_1}^{R_2}$ $R = \frac{\rho}{4\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ Substitute the values: $R = \frac{3 \times 10^{13} \Omega \text{ m}}{4\pi} \left( \frac{1}{6 \times 10^6 \text{ m}} - \frac{1}{6.003 \times 10^6 \text{ m}} \right)$ $R = \frac{3 \times 10^{13}}{4\pi} \left( \frac{6.003 \times 10^6 - 6 \times 10^6}{(6 \times 10^6)(6.003 \times 10^6)} \right)$ $R = \frac{3 \times 10^{13}}{4\pi} \left( \frac{0.003 \times 10^6}{36.018 \times 10^{12}} \right)$ $R = \frac{3 \times 10^{13}}{4\pi} \left( \frac{3 \times 10^3}{3.6018 \times 10^{13}} \right)$ $R = \frac{9 \times 10^{16}}{4\pi \times 3.6018 \times 10^{13}} = \frac{9 \times 10^3}{4\pi \times 3.6018}$ $R \approx \frac{9000}{12.566 \times 3.6018} \approx \frac{9000}{45.25} \approx 198.89 \Omega$

Since $h \ll R_E$, we can also use the parallel plate approximation for resistance: $R \approx \rho \frac{h}{A}$, where $A = 4\pi R_E^2$. $R \approx \frac{3 \times 10^{13} \Omega \text{ m} \times 3 \times 10^3 \text{ m}}{4\pi (6 \times 10^6 \text{ m})^2} = \frac{9 \times 10^{16}}{4\pi \times 36 \times 10^{12}} = \frac{9 \times 10^4}{144\pi} \approx \frac{90000}{452.39} \approx 198.9 \Omega$. The results are consistent.

(b) Estimate the potential difference between the surface of the earth and the cloud.

First, calculate the capacitance $C$ of the spherical capacitor: $C = 4\pi \epsilon_0 \frac{R_1 R_2}{R_2 - R_1} = 4\pi \epsilon_0 \frac{R_E (R_E+h)}{h}$ Substitute the values: $C = 4\pi (8.85 \times 10^{-12} \text{ F/m}) \frac{(6 \times 10^6 \text{ m})(6.003 \times 10^6 \text{ m})}{3 \times 10^3 \text{ m}}$ $C = 4\pi (8.85 \times 10^{-12}) \frac{36.018 \times 10^{12}}{3 \times 10^3}$ $C = 4\pi (8.85 \times 10^{-12}) (12.006 \times 10^9)$ $C = 4\pi \times 8.85 \times 12.006 \times 10^{-3} \text{ F}$ $C \approx 1.3357 \text{ F}$

Now, calculate the potential difference $V = Q/C$: $V = \frac{4 \times 10^5 \text{ C}}{1.3357 \text{ F}}$ $V \approx 2.994 \times 10^5 \text{ V} \approx 3 \times 10^5 \text{ V}$

(c) In how much time the capacitor formed between the earth and the cloud will lose 63% of the charge?

The discharge of a capacitor through a resistor is governed by the equation $Q(t) = Q_0 e^{-t/\tau}$, where $\tau = RC$ is the time constant. Losing 63% of the charge means the remaining charge is $Q(t) = Q_0 - 0.63 Q_0 = 0.37 Q_0$. So, $0.37 Q_0 = Q_0 e^{-t/\tau}$ $0.37 = e^{-t/\tau}$ Taking natural logarithm on both sides: $\ln(0.37) = -t/\tau$ $-0.994 = -t/\tau$ $t \approx 0.994 \tau \approx \tau$

This means that the time taken to lose 63% of the charge is approximately equal to the time constant $\tau$. $\tau = RC$ $\tau = (198.89 \Omega) \times (1.3357 \text{ F})$ $\tau \approx 265.5 \text{ s}$

The time required is approximately 265.5 seconds.