Assume Earth and a cloud layer 0.900 Km above the Earth can be treated as plates of a parallel-plate capacitor. a. If the cloud layer has an area of 2.00 × 104 m2, what is the capacitance?
b. If an electric field strength of 2.0 × 105 N/C causes the air to conduct charge (lightning), what charge can the cloud hold?
c. Describe what must happen to its molecules for air to conduct electricity.
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(A)-1.96e-10 nanoFarads (B)- 0.03528 C (C)- The flow of electricity is the flow of electrons through a medium. The air has to partially ionize in order to allow for the flow of electrons.Explanation:

(A) Capacitance can be calculated using or . This results in a value of 1.96e-12 nanofarads

(B) The total charge can be calculated using the equation C*V=Q

We have to find the voltage across the capacitor. To do this use the equation E*D=V. This results in a value of 1.8e8 Volts. Plug this value and the capacitance value into the equation C*V=Q. This will result in a value of 0.03528 coulombs of charge.

(C) The flow of electricity is the flow of electrons through a medium. The air has to partially ionize in order to allow for the flow of electrons.