Voltage and current of positive lightning

Question

For a physics issues investigation I chose to investigate what effects lightning could have on an aeroplane while in flight if it was struck and then go on to discuss some possible implications of engineers not taking into account the power of positive lightning.

Just in-case you don't know what positive lightning is, my understanding of it at least is that when charges accumulate in clouds (I won't go into how) in most cases the underside of the cloud is negatively charged and the top of the cloud is positively charged. Basically positive lightning is a lot more powerful than negative lightning as it has a higher voltage and current.

Q1. How would you determine the potential difference between the underside of the cloud (given an overall charge) and the ground (given the overall charge) and hence the electric field strength. $E = V/d$ ? But how would I calculate the voltage?

Q2. I understand that $V = IR$. And this is why the voltage of a positive lightning strike is higher than a negative strike as the resistance for the positive strike is higher (it has to go out to the side of the cloud and THEN down). But why is the current higher? If $I = V/R$ and the resistance is higher, wouldn't the current be lower?

(This question probably isn't as high a level as many of the other questions on this site so you should find it quite easy to answer.)

Answer 1

(1) To address your first question: you have to treat the cloud and the earth below it as forming as a capacitor. There's a good popular description of this at http://micro.magnet.fsu.edu/electromag/java/lightning/. A capacitor is described by it's capacitance, and this is related to the voltage and charge by:

$$C = \frac{Q}{V}$$

where $Q$ is the electric charge and $V$ is the voltage difference across the capacitor. You can approximate the cloud and earth as a parallel plate capacitor, and the capacitance is given by:

$$C = \frac{\epsilon A}{d}$$

where $A$ is the area of the cloud base, $d$ is the spacing between the cloud base and the earth, and $\epsilon$ is the permittivity of air ($8.854 \times 10^{-12}C^2N^{-1}m^{-2}$). Combining the two equations and a quick rearrangement gives:

$$V = \frac{Qd}{\epsilon A}$$

This is obviously a gross simplification, but should give you a rough idea of the potential difference.

(2) As to your second question: as you say, positive lightning requires a higher voltage to get it started. Looking at the equation for the voltage, assuming the cloud stays the same the only way the voltage can be higher is if the charge is higher. Current is defined as charge per unit time, and if the duration of the lightening strike is roughly constant a positive lightning bolt has to transfer more charge in the same time and therefore has a higher current.

Answer 2

Ultimately, the one thing that matters for the airplane is the current, from which you can find the voltage across the airplane itself (knowing resistance of airplane). The length of pulse also may matter if the heating of materials is an issue. There's also the magnetic field, which also depends on the current.

The initial 'voltage' between ground and cloud is not very relevant, other than as perhaps in it's effect on length of the strike.

If one is to go into very short-time details one may need to consider inductance, but I don't think it is very relevant for the airplane itself.

And the airplane can be tested with relatively low voltages that induce same current through the airplane.