Answer:
[tex]3.51\cdot 10^{10}J[/tex]
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
[tex]U=\frac{1}{2}CV^2[/tex] (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
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where in this case:
[tex]\epsilon_0 = 8.8542\cdot 10^{-12} C^2 /N m^2[/tex] is the vacuum permittivity
[tex]A=0.8 km^2 = 0.8\cdot 10^6 m^2[/tex] is the area of each plate (the area of the cloud)
[tex]d=910 m[/tex] is the distance between the plates (in this case, between the cloud and the ground)
Rewriting eq.(1),
[tex]U=\frac{\epsilon_0 A V^2}{2d}[/tex]
Moreover, the electric field in a parallel plate capacitor is uniform, so we can write
[tex]V=Ed[/tex]
where
[tex]E=3.3\cdot 10^6 N/C[/tex] 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:
[tex]U=\frac{\epsilon_0 AE^2 d}{2}=\frac{(8.85\cdot 10^{-12})(0.8\cdot 10^{6})(3.3\cdot 10^6)^2(910)}{2}=3.51\cdot 10^{10}J[/tex]
