A) Consider the Earth and a cloud layer 760 m above the Earth to be the plates o

Question

A) Consider the Earth and a cloud layer 760 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?

B) If an electric field strength greater than 4.78E+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?

Explanation / Answer

a) C = epsilon *A/d
A = area of plates
d = distance between plates
C = epsilon* 1.18E+6 / 760 = 1.37 x 10^-8 = 13.7 nanoF
b) E = 4.78E+6 N/C
E = V/d
V = Ed = 4.78E+6 N/C * 760 = 3.633 x 10^9 volts
Q = CV = 13.7 x 10^-9 * 3.633 x 10^9 = 49.772 C
c) electric potential = V = 3.633 x 10^9 volts (calculated in previous step)

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