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

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation: