# Problem with solution Charge in a thundercloud 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}