A thundercloud and the earth can be regarded as a parallel plate capacitor. Taking the area of the thundercloud to be $50 \;km^2$, its height above the earth as $1\; km$ and it's potential $100\; KV$, calculate the energy stored.

Given,
Area of the thundercloud $(A) = 50 km^2 = 50\;*\;10^{6}\;m^2$
Distance of cloud from the earth $(d) = 1\;km = 1 \; * \; 10^{3}\;m$
Potential $(V) = 100\;kV = 100 \; * \; 10^{3}\;V$
Energy Stored $(E) = ?$

Since the capacitance of parallel plates capacitor is $C = $ $\frac{1}{2}$ $\;C\;V^2 $

Energy stored $(E)= $ $\frac{1}{2}\;*\; \frac{\epsilon_0\;A}{d}$ $V^2 = $ $\frac{1}{2}\;*\; \frac{8.85\;*10^{-12}\;50\;*\;10^6\;*\;(10^5)^2}{10^3}$

∴ $E = 2.2 \; * \; 10^3\;J$

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