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Problem with solution Charge in a thundercloud

Thundercloud - lightning discharge
Level 2 (suitable for students)
Level 2 requires school mathematics. Suitable for pupils.

How much charge \(Q\) is in a thundercloud of area \(A = 1 \, \text{km}^2 \) and with an electric field \(E\) - of magnitude \( 3 \cdot 10^6 \, \text{V}/\text{m} \) - built up between the upper and lower parts of the cloud?

Solution tips

The cloud can be considered as a plate capacitor.

Solution

Solution

Assume that the upper and lower parts of the cloud behave like two capacitor plates. Then the following equation is valid for the voltage: 1 \[ U ~=~ E \, d \]

The capacitance \( C \) of a plate capacitor is given by the following formula: 2 \[ C ~=~ \varepsilon_0 \, \frac{A}{d} \]

In general, the charge is proportional to the voltage, where the constant of proportionality is the capacitance: 3 \[ Q ~=~ C \, U \]

Insert Eq. 1 and 2 into Eq. 3 and eliminate the distance \( d \). Then you get: 4 \[ Q ~=~ \varepsilon_0 \, A \, E \]

Inserting the concrete values from the exercise results in the approximate amount of charge in a thundercloud: 5\begin{align} Q &~=~ 8.854 \cdot 10^{-12} \, \frac{\text{As}}{\text{Vm}} ~\cdot~ 10^6 \, \text{m}^2 ~\cdot~ 3 \cdot 10^6 \, \frac{\text{V}}{\text{m}} \\\\ &~=~ 26.6 \, \text{C} \end{align}