
#1
Oct1605, 12:35 PM

P: 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
Oct1605, 01:09 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

12.8kV ? That's the wrong units for a field.
PS : Post your textbook questions in the Coursework forum above. 



#3
Oct1605, 01:24 PM

P: 23

I meant kVm1




#4
Oct1605, 04:44 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Dipole question
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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