Consider the Earth and a cloud layer 900 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 5.0 km2 = 5000000 m2, what is the capacitance?  F (b) If an electric field strength greater than 3.0 ✕ 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?  C

A parallel-plate capacitor has capacitance C0 = 8.00 pF when there is air between the plates. The separation between the plates is 1.20 mm. Express your answer with the appropriate units. Part A: What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00×104 V/m? Part B: A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00×104 V/m?

An air-filled parallel-plate capacitor has plates of area 2.80 cm separated by 1.80 mm. The capacitor is connected to a(n) 18.0 v battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? N/C

A charged air-filled parallel plate capacitor of 22 nF has a energy density of 30x10-3 J/m³. If the distance between the plates is 200 µm, calculate the magnitude of the charge, in µC, of the capacitor.