A.) Consider the Earth and a cloud layer 845 m above the Earth to be the plates

Question

A.) Consider the Earth and a cloud layer 845 m above the Earth to be the plates of a parallel-plate capacitor. If the cloud layer has an area of 1.18 km2 = 1.18E+6 m2, what is the capacitance?
1.24×10-8 F
B.) If an electric field strength greater than 4.03E+6 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?
C.) If the cloud is holding this maximum charge, what is the magnitude of the difference in electric potential between the cloud and the ground?

AND
A.) A uniform electric field of magnitude 241 V/m is directed in the negative x direction. A +10.0 C charge moves from the origin to the point (x, y) = (23.5 cm, 48.6 cm). What was the change in the potential energy of this charge?
B.) Through what potential difference did the charge move?

Explanation / Answer

(A) We know that capacitance C = eoA/d
C =( 8.85*10-12*(1.186*106))/845 = 1.24*10-8 F
(B) We know that
Eo = Q /Aeo
Q = EoAe0 = 4.03*106*1.186*106*8.85*10-12 = 42.299 C
(c) We know that
C = Q/V
V = Q/C = 42.299/1.24*10-8 = 34.11*108 volt

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.