About \(5.0 \times 10^{4} \mathrm{m}\) above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. The Earth and the ionosphere form a giant spherical capacitor, with the lower atmosphere acting as a leaky dielectric. (a) Find the capacitance \(C\) of the Earth-ionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Com- pare Earth's radius to the distance between the "plates." (b) The fair-weather electric field is about \(150 \mathrm{V} / \mathrm{m},\) downward. How much energy is stored in this capacitor? (c) Due to radioactivity and cosmic rays, some air molecules are ionized even in fair weather. The resistivity of air is roughly \(3.0 \times 10^{14} \Omega \cdot \mathrm{m}\) Find the resistance of the lower atmosphere and the total current that flows between Earth's surface and the ionosphere. [Hint: since we treat the system as a parallel plate capacitor, treat the atmosphere as a dielectric of uniform thickness between the plates.] (d) If there were no lightning, the capacitor would discharge. In this model, how much time would elapse before Earth's charge were reduced to \(1 \%\) of its normal value? (Thunderstorms are the sources of emf that maintain the charge on this leaky capacitor.)

Step by step solution

01

(a) Find the capacitance of the Earth-ionosphere system

To find the capacitance, we will treat the Earth-ionosphere system as a parallel plate capacitor. We will consider that Earth's surface and ionosphere will be acting as two plates, separated by the atmosphere. Given, Earth's radius (R) = \(6.37 \times 10^{6} m\) atmosphere height (h) = \(5.0 \times 10^4 m\) Capacitance of a parallel plate capacitor is given by: C = \(\frac{\varepsilon_{0}A}{d}\), where \(\varepsilon_{0}\) is the vacuum permittivity, A is the area, and d is the separation between the plates. In our case, A = \(4\pi R^2\), and the distance d = h. Then, C = \(\frac{\varepsilon_{0}(4\pi R^2)}{h}\) Using the given values, we get: C = \(\frac{(8.85 \times 10^{-12} \,\mathrm{F/m})(4\pi (6.37 \times 10^6 \,\mathrm{m})^2)}{5.0 \times 10^4 \,\mathrm{m}}\). Calculating the value of C, we get the capacitance of the Earth-ionosphere system.

02

(b) Calculate the energy stored in the capacitor

The energy stored in a capacitor is given by: U = \(\frac{1}{2}CV^2\), where C is the capacitance and V is the potential difference across the capacitor. For the fair-weather electric field, the field (E) is given as 150 V/m downward. Hence, the potential difference V = E * h, where h is the distance between the two "plates" (Earth and ionosphere). V = 150 V/m * \(5.0 \times 10^4\) m Using the value of C from the previous step and V found now, we can calculate the energy stored in the capacitor U = \(\frac{1}{2}CV^2\).

03

(c) Calculate the resistance of the lower atmosphere and the total current flow

To determine the resistance of the lower atmosphere, we use a uniform resistivity value, \(\rho = 3.0 \times 10^{14} \Omega \cdot \mathrm{m}\). Resistance (R) is calculated using the relation: R = \(\frac{\rho l}{A}\), where l is the height of the lower atmosphere (5.0 × 10^4 m) and A is the cross-sectional area. In our case, A = 4πR², where R is Earth's radius. R = \(\frac{(3.0 \times 10^{14} \,\Omega \cdot \mathrm{m})(5.0 \times 10^4 \,\mathrm{m})}{4\pi (6.37 \times 10^6 \,\mathrm{m})^2}\) Now we have to calculate the total current that flows between Earth's surface and the ionosphere. The current (I) can be found using Ohm's law: I = \(\frac{V}{R}\), where V is the potential difference, and R is the resistance. Using the values of V and R, find the total current I.

04

(d) Calculate the time for Earth's charge to reduce to 1%

In the absence of lightning, the capacitor discharges through the resistor. The time constant (τ) for a capacitor's discharge is given by: τ = RC, where R is the resistance and C is the capacitance. Using the values of R and C, calculate τ. Now, when Earth's charge is reduced to 1% of its normal value, let t be the time elapsed. The equation for a discharging capacitor's voltage is given by: \(V(t) = V_{0} e^{-t/RC}\), where \(V_{0}\) is the initial voltage, and V(t) is the voltage at t. Here, we want \(V(t)=0.01 V_{0}\), where V₀ is the initial voltage across the capacitor. Solving for time t, we get: \(t = -RC \cdot ln(\frac{V(t)}{V_{0}})\) \(t = -RC \cdot ln(0.01)\) By substituting the values of R and C, calculate the elapsed time t.

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