The earth can be considered as a single-conductor capacitor . It can also be considered in combination with a charged layer of the atmosphere, the ionosphere, as a spherical capacitor with two plates, the surface of the earth being the negative plate. The ionosphere is at a level of about 70 km, and the potential difference between earth and ionosphere is about 350,000 V. Calculate: (a) The capacitance of this system; (b) The total charge on the capacitor; (c) The energy stored in the system.

A)C= 0.066F

B)Q= 23100C

C)Uc= 4×10³*³j

Explanation:

A) capacitance of a spherical capacitor

C =4 πϵ. R1 R2 ÷ R2 - R1

Where:

C: Capacitance

R1: radius to the earth surface which is 6400× 10³ m

R2: radius to the ionosphere is 6400 + 70 = 6470 × 10³ m

ϵ. : permittivity of free space is given as 8.85 × 10¹/¹² C²/N.m²

Potential difference a cross the capacitor is 350,000V

C= 4 π×8.85 × 10¹/¹² (6400× 10³ ×6470 × 10³ / 70×10³) = 0.066F

C= 0.066F

B) Total charge of the capacitor

C= Q/V

C: is the capacitance

Q: is the charge

V: P.d a cross the capacitor

Since we are solving for the total charge Q= CV

Q= 0.066 × 350000 = 23100C

Q= 23100C

C) the energy stored in the system

Uc = 1/2 QV

Uc = 1/2 ×23100 × 350000 = 4×10³*³j

Uc= 4×10³*³j

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

365.25 earth days are in the year of a light year