Consider the parallel-plate capacitor formed by Earth and a cloud layer 910 m above Earth, with an area of 0.8 km2 . Assume this capacitor will discharge (that is, lighting occurs) when the electric field strength between the plates reaches 3.3 × 106 N/C. What is the energy released if the capacitor discharges completely during a lightning strike? The permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m 2 . Answer in units of J.

Explanation:

The energy released when the capacitor is discharged is equal to its initial energy stored. The energy stored in a capacitor is given by

(1)

where

C is the capacitance

V is the potential difference across the capacitor

The capacitance of a parallel-plate capacitor can be written as

where in this case:

is the vacuum permittivity

is the area of each plate (the area of the cloud)

is the distance between the plates (in this case, between the cloud and the ground)

Rewriting eq.(1),

Moreover, the electric field in a parallel plate capacitor is uniform, so we can write

where

is the strength of the electric field

d = 910 m is the distance between the plates

Substituting again into the equation and solving, we find:

the answer is a 3.8cm

explanation:

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