# Dipole question

1. ### Willa

23
I'm stuck on a question involving dipoles and what I am guessing to be the method of images...here goes:

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 the r must 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?

2. ### Gokul43201

11,044
Staff Emeritus
12.8kV ? That's the wrong units for a field.

PS : Post your textbook questions in the Coursework forum above.

3. ### Willa

23
I meant kVm-1

4. ### Gokul43201

11,044
Staff Emeritus
I get 12.8 kV/m, if I just find the field due to each charge (and it's image) and add the numbers.

I get 11.25 kV/m using the dipole equation you posted. I think you forgot a factor of 2 (when you add the field from the image). However, isn't this formula accurate only in the limit r >> a ?