Chapter 1

# lightning

Take 15 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

The electric field in a cloud is essentially vertical. We will suppose in this exercise that it is only vertical.

The following graph is an example of the algebric value of the field. It is counted as positive for a field oriented upward. Electric field in a cloud

The cloud is charged. We can distinguish two positively charged zones and one negatively charged.

Near the ground is developing a positively charged zone due to air ionization by point effect. Its height can reach .

The first phase of a thunderbolt is the formation of a faintly luminous pre-discharge called leader which progresses through the air with a relatively small velocity.

This pre-discharge is created by both the ground (ascending thunderbolts) and in the cloud (descending thunderbolts).

When the leaders meet, a conductive link is created between the cloud and the ground. It will allow the flow of a very high intensity current.

The following graph shows an example of intensity of a thunderbolt as a function of the discharge time . Intensity as a function of the discharge time

### To go further ...

Website "Culture Sciences - Physiques (ENS Lyon)" :

Where does thunder come from ?

## Question

Evaluate the total charge that has flowed and the average intensity of the lightning current.

### Hint

Use the definition of the intensity : ### Solution

The total charge that has flowed is equal to the area delimited by the curve and the abscises axis ( is the total length of the lightning) : If we consider as three half-lines, as seen on the graph, we can evaluate numerically : The average intensity of the lightning current is equal to : ## Question

By modeling the algebric value of the electric field, estimate the voltage difference between the ground and the bottom of the cloud. Its altitude is .

### Solution

Suppose the electric field fluctuates in a linear way between the ground (here it is equal to ) and its height (bottom of the cloud where it is estimated to be ), as we can see on the following picture.

Then the field can be modeled at any altitude by the following expression : We know that : We can deduce the potential : With in m and in kV.

Thus the voltage difference between the ground and the bottom of the cloud is equal to : The voltage difference can reach the incredible value of 82 millions of volt !

## Question

During the discharge, we admit the electrical energy dissipated is the same as the energy of a capacitor of charge under the voltage difference .

Evaluate the dissipated energy during this discharge.

Determine the capacitance of the equivalent capacitor.

Could we, in practice, retrieve this energy ?

### Solution

The electric energy dissipated during the discharge of the capacitor (whose capacitance is ) can be written under these equivalent forms : Numerically : And : It surely is appealing to recover this energy, because of how important it is !

Moreover, approximately 1 million lightning bolts strike France each year.

Thus the total power provided could reach 70 MW.

Yet how could we solve all the technical issues that would erupt : the a priori irregular and not repeatable location of the lightning impact, the immediate recovery of a large amount of energy and its storage problem, other issues linked to the safety of the power plant...

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Capacitors
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Yukawa Potential