|Oct16-05, 12:35 PM||#1|
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?
|Oct16-05, 01:09 PM||#2|
12.8kV ? That's the wrong units for a field.
PS : Post your textbook questions in the Coursework forum above.
|Oct16-05, 01:24 PM||#3|
I meant kVm-1
|Oct16-05, 04:44 PM||#4|
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 ?
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