# Problem with solution Charge in a thundercloud

**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} `