# Electric field and Potential: Parallel plate capacitor.

Part 1

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 1.03 km2 = 1.03E+6 m2, what is the capacitance?

Part 2

If an electric field strength greater than 4.29E+6 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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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 1.03 km2 = 1.03E+6 m2, what is the capacitance?

The capacitance of a parallel plate capacitor is given by

Here A is the area of the capacitor plates equal to 1.03*106 m2

And d is the distance between the plates equal to 760 m.

is the permittivity constant 8.85*10-12 F/m.

Substituting the values in the equation above the capacitance of the capacitor formed by the Earth and the cloud layer is given by

F

Thus the capacitance is 1.20*10-8 F or 12.0 nF.

If an electric field strength greater than 4.29E+6 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?

The magnitude of the charge Q stored on the plates of a capacitor is given by

Q = C*V

Here V is the potential difference across the capacitor.

Now as the electric field can be defined as the potential gradient (potential difference per unit distance), the magnitude of the electric field between the plates of a parallel plate capacitor is given by

Thus the maximum possible voltage across the cloud and the earth will be given by

V = E*d = 4.29*106*760 = 3.26*109 V

Hence maximum charge the cloud can hold will be

Q = CV = 1.20*10-8*3.26*109 = 39.1 C

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