On the first problem set, one problem (1.5) asks us to show that along the axis of the dipole (the same line that contains +Q and -Q), the E field is E = 1/4pi eo (2p/r^3) for rl, where r is the distance from the point to the center of the dipole.

I ran through the derivations, but I get only 1/4pi eo (p/r^3). Where does the factor of 2 come from?

I start by adding up E+ and E-, which is (1/4pi eo) (Q/(r^2) - Q/(r + l/2)^2). I put the terms over a common denominator, perform the subtraction, and take the case where r l. I don't have a 2p in the numerator if I do that.

Thank

Question

On the first problem set, one problem (1.5) asks us to show that along the axis of the dipole (the same line that contains +Q and -Q), the E field is E = 1/4pi eo (2p/r^3) for r>>l, where r is the distance from the point to the center of the dipole.  

I ran through the derivations, but I get only 1/4pi eo (p/r^3).  Where does the factor of 2 come from?

I start by adding up E+ and E-, which is (1/4pi eo) (Q/(r^2) - Q/(r + l/2)^2).  I put the terms over a common denominator, perform the subtraction, and take the case where r >>l.    I don't have a 2p in the numerator if I do that.

Thank

anonymous · Answer

Your solution is OK. Only the way you calculate the field has a mistake. The right one is:

E= (1/4pi eo) (Q/(r - I/2)^2- Q/(r + l/2)^2), but developing this one I get the same solution you get