I'm stuck on a question involving dipoles and what I am guessing to be the method of images...here goes:(adsbygoogle = window.adsbygoogle || []).push({});

The electrical system of a thundercloud can be represented by a vertical dipole consisting of a charge +40C at a height of 10km and a charge of -40C vertically below it at a height of 6km. What is the electric field at the ground immediately below the cloud, treating the ground as a perfect conductor?

To tackle this question I tried using the method of images to create an image dipole on the other side of the ground. Then using the formula for the E field of a dipole:

E = 1/(4pie0r^5)(3p.rr- r^2p)

Summing the two dipole electric fields, and taking into account the coefficient in thermust go to 0 so that there is no horizontal electric field, I get:

E = -p/(2pie0r^3)

But using the values given, I do not get the answer which is supposed to be 12.8kV, what have I done wrong?

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# Dipole question

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