## 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}
`