Question

A thunder cloud and earth can be regarded as a parallel plate capacitor. Taking the area of thunder cloud to be 50 km2, its height above the earth is 1 km at potential 100 KV. The energy stored is
1) 2 × 103 J
2) 2.2 × 103 J
3)2.6 × 103 J
4) 3.2 × 103 J

Answer 1

Answer:

The energy stored is $2.2 * 10^{3} J$.

Explanation:

According to question , A thunder cloud and earth can be regarded as a parallel plate capacitor. Taking the area of thunder cloud to be 50 km2, its height above the earth is 1 km at potential 100 KV.

We need to find out the energy storage.

Area(A) = $50 km^{2}$ = $50 * 10^{6} m^{2}$

Cloud height above the earth = distance between cloud and earth (d)= $1 km = 1*10^{3} m$

Potential (V)= $100kV =100 * 10^{3} V$

This can be consider as a parallel plate capacitor.

So the energy stored = $\frac{1}{2} CV^{2}$ $\frac{1}{2} * \frac{e * A}{d} * V^{2}$ = $\frac{1}{2} * \frac{8.85 *10^{-12}*50*10^{6}*(10^{5}) ^{2} }{10^{3} }$ = $2.2 * 10^{3} J$

The energy stored is $2.2 * 10^{3} J$

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