About 5.0 ×10^4 m above Earth's surface, the atmosphere is...

About $5.0 \times 10^{4} \mathrm{m}$ above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. 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 Earthionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Compare 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.)

About $5.0 \times 10^{4} \mathrm{m}$ above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. 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 Earthionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Compare 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.)

Key Concepts

RC Circuit Discharge and Time Constant

An RC circuit, which includes a resistor and a capacitor, exhibits an exponential decay of voltage and charge over time. The time constant ? (tau) of the circuit, defined as ? = R × C, characterizes the rate at which the capacitor discharges. This concept is crucial for predicting how quickly a system loses stored charge in the presence of leakage or conductive pathways.

Capacitance

Capacitance is the property of a system that enables it to store electric charge. It is defined as the ratio of charge stored to the voltage applied between the conductors. In various geometries, including spherical and parallel plate configurations, the capacitance depends on the shape, size, and separation distance between the conductors, as well as the properties of the intervening medium.

Parallel Plate Capacitor Approximation

When the separation between two conductors is much smaller than their size or curvature radius, the electric field between them can be approximated as uniform. This approximation simplifies calculations by allowing the use of the parallel plate capacitor formulas, even when dealing with curved surfaces like a spherical Earth with an ionosphere. The key justification is that the curvature effects are negligible relative to the small gap between the conductors.

Stored Energy in Capacitors

The energy stored in a capacitor is given by the formula U = (1/2)CV², where C is the capacitance and V is the voltage across the capacitor. This concept is important for understanding how energy is accumulated in electric fields and for analyzing the energy balance in various circuit components and systems.

Resistivity and Resistance

Resistivity is a material property that quantifies how strongly a given medium opposes the flow of electric current. The resistance of a medium can be determined using its resistivity by taking into account the geometry of the conductor; for a slab of material, resistance is calculated as R = (resistivity × distance) / area. This relationship is used to model the atmospheric leakage current between Earth and the ionosphere.

Transcript

00:01 So given in the question about 5 .0 into 10 to the power 4 meters above the surface of the earth the atmosphere is sufficiently ionized that it behaves as a capacitor.

00:20 Above the height of about 5 .0 into 10 to the power 4 meters is the surface called ionosphere and thus earth and the ionosphere forms a giant capacitor.

00:37 The lower atmosphere in between the earth and the ionosphere it acts as a leaky dielectric.

00:48 The fair weather electric field when the weather is fine electric field e is given to be 150 volts per meter acting downwards from the ionosphere to the earth and resistivity of air is given to be 3 .0 into 10 to the power 14 ohm.

01:10 So let this be the ionosphere and this one be the earth.

01:14 So this be the earth and this is the ionosphere.

01:18 Ionosphere starts at a height 5 .0 into 10 to the power 4 meters and the electric fields are pointing in this direction acting downwards from the ionosphere to the earth and has a value 150 volts per meter.

01:37 The resistivity of the air in between the earth and the ionosphere is 3 .0 into 10 to the power 14 ohm.

01:47 It has been asked to find the capacitance of the earth ionized fair system treating it as a parallel plate capacitor.

02:00 Also the reason why the combination can be approximated as a parallel plate capacitor.

02:12 So although they are spherical structures yet their combination can be approximated as a parallel plate capacitor.

02:22 What is the reason behind it? the second part is to find the energy stored in this capacitor.

02:30 Part c is to find the resistance of the lower atmosphere and the total current that flows between the earth's surface and the ionosphere and part d is the time elapsed before the earth's charge the time elapsed before the earth's charge was reduced to one percent of its normal values ignoring lightning that means assuming the weather is fine there is no lightning or thunderstorm.

03:13 So focusing on the first part the capacitance c of a parallel plate capacitor is given by the formula c is epsilon naught a by d where epsilon naught is the permittivity of free space a is the area of the plate and d is the separation between it.

03:46 In this case the two plates are the earth's surface and the ionosphere surface and the separation d is the distance is the separation between the earth and the ionosphere surface.

04:01 The radius of the earth we know is 6378 kilometers which is 6 .378 into 10 to the power 6 meters.

04:12 Therefore surface area of the earth let it be represented by a e would be 4 pi r square and it comes out to be 5 .11 into 10 to the power 14 meters square.

04:26 Now the radius of the ionosphere would be equal to the radius of the earth plus this d that is the separation between the earth's surface and the ionosphere.

04:40 So if we take the sum it comes out to be 6 .428 into 10 to the power 6 meters.

04:48 Now the surface area of the ionosphere a let it be represented by a i it comes out to be 5 .19 into 10 to the power 14 meters.

05:07 Now since the difference in the surface area of the earth and the ionosphere is of the order is almost two orders of magnitude less than its value less than the surface area of the earth than that of the earth...

Please give Ace some feedback