Calculating Electric Potential for an Earth-Cloud Capacitor

[Boozehound]

this question has three parts. i figured out the first and second one. i just don't understand electrical potential energy.

Consider the Earth and a cloud layer 820m above the Earth to be the plates of a parallel-plate capacitor. If the cloud layer has an area of 1.27 km2 = 1.27E+6 m2, what is the capacitance? 1.37×10-8 F

If an electric field strength greater than 4.91E+6 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? 55.15C

If the cloud is holding this maximum charge, what is the magnitude of the difference in electric potential between the cloud and the ground?

i took the V the i found in the second problem which was 4.026E9V. and i took q from the second problem and multiplied it by V. i ended up getting 2.22E11. which is wrong..what am i doing wrong? i think i might be using the wrong formulas.

 

[member 553083]

The correct answer to the third part of the question is 2.22E17 Joules (J). The formula used to calculate the electrical potential energy of a system is U = qV, where q is the charge and V is the voltage. In this case, q is the maximum charge the cloud can hold (55.15 C) and V is the difference in electric potential between the cloud and the ground (4.026E9 V). Therefore, U = qV = 55.15C * 4.026E9 V = 2.22E17 J.