PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 108P

(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.

(a)

Expert Solution
Check Mark

Answer to Problem 108P

The capacitance of Earth ionosphere when it is treated as parallel plate capacitor is 0.090 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,

dR=5.0×104 m6.371×106 m102

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

Write the equation for capacitance.

C=ε0Ad=ε04πr2d (I)

Here, C is capacitance, A is the area, r is radius, ε0 permittivity of free space and d is distance between the plates.

Conclusion:

Substitute 8.854×1012 C2/N.m2 for ε0, 6.371×106 m for r and 5.0×104 m for d in equation I.

C=(8.854×1012 C2/N.m2)4π(6.371×106 m)25.0×104 m=0.090 F

Therefore, the capacitance of Earth ionosphere when it is treated as parallel plate capacitor is 0.090 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.

(b)

Expert Solution
Check Mark

Answer to Problem 108P

The energy stored in the capacitor is 2.5 TJ.

Explanation of Solution

Write the equation for energy.

U=12CV2=12C(Ed)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 F for C, 150 V/m for E and 5.0×104 m for d in equation II.

U=12(0.090 F)((150 V/m)(5.0×104 m))2=2.5 TJ

Therefore, the energy stored in the capacitor is 2.5 TJ.

(c)

To determine

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

(c)

Expert Solution
Check Mark

Answer to Problem 108P

The resistance of lower atmosphere is 29 kΩ and the total current flows from Earth’s surface to ionosphere is 260 A.

Explanation of Solution

Write the equation for resistance.

R=ρLA=ρL4πr2 (III)

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

Write the equation for current.

I=VR=EdR (IV)

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

Conclusion:

Substitute 3.0×1014 Ω.m for ρ, 6.371×106 m for r and 5.0×104 m for L in equation III.

R=(3.0×1014 Ω.m)5.0×104 m4π(6.371×106 m)2=29408 Ω29 kΩ

Substitute 29408 Ω for R, 150 V/m for E and 5.0×104 m for d in equation IV.

I=(150 V/m)(5.0×104 m)29408 Ω=260 A

Therefore, the resistance of lower atmosphere is 29 kΩ and the total current flows from Earth’s surface to ionosphere is 260 A.

(d)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 108P

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

Explanation of Solution

Write the equation for charge.

Q=Q0et/RC

Here, Q is the charge, Q0 is the initial charge, t is the time, R is resistance and C is the capacitance.

Rewrite the above equation to get time.

et/RC=Q0QtRC=lnQ0Q

t=RClnQ0Q (V)

Conclusion:

Substitute 0.090 F for C, 29408 Ω for R, 10.01 for Q0Q in equation V.

t=[(29408 Ω)(0.090 F)ln10.01](1 min60 s)=200 min

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

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Chapter 18 Solutions

PHYSICS

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