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) = ?
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}
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
Return to Main Menu
Comments
Post a Comment