Problem with solution: Charge in a thundercloud
Solution tips
The cloud can be considered as a plate capacitor.
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}
