Question

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350MV . The bottoms of the clouds are typically 1500m above the Earth, and may have an area of 120km2 . Modeling the Earth-cloud system as a huge capacitor, calculate

Answer 1

Complete Question

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350 MV (35,000,000 V). The bottoms of the thunderclouds are typically 1500 m above the earth, and can have an area of 120 km^2. Modeling the earth/cloud system as a huge capacitor, calculate

a. the capacitance of the earth-cloud system

b. the charge stored in the "capacitor"

c. the energy stored in the capacitor

Answer:

a

 $C = 7.08 *10^{-7} \ F$

b

  $Q = 24.78 \ C$

c

 $E =433650000 \ J$

Explanation:

From the question we are told that

The potential difference is  $V = 35000000 V$

The distance of the bottom of the thunderstorm from the earth is  d = 1500 m

The area is  $A = 120 \ km^2 = 120 *10^{6} \ m^2$

Generally the capacitance of the earth cloud system is mathematically represented as

         $C = \epsilon_o * \frac{A}{d}$

 Here $\epsilon_o$ is the permitivity of free space with as value $\epsilon_o = 8.85 *10^{-12} \ C/(V\cdot m)$

So

     $C = 8.85*10^{-12} * \frac{120*10^{6}}{1500}$

=>  $C = 7.08 *10^{-7} \ F$

Generally the charge stored in the capacitor (earth-cloud system) is mathematically represented as

       $Q = C * V$

=>    $Q = 7.08 *10^{-7} * 35000000$

=>    $Q = 24.78 \ C$

Generally the energy stored in the capacitor is mathematically represented as

       $E = \frac{1}{2} * Q * V$

=>    $E = \frac{1}{2} * 24.78 * 35000000$

=>    $E =433650000 \ J$