Integral of electric field of dipole moment over a sphere at r = 0

[rafaelpol]

Homework Statement

Can't get to the final equality ( integral = - 4*Pi/3).

Homework Equations

<br /> \int_V \mathbf{E }dV = - \int_F \frac{_{\mathbf{p}.\mathbf{e_{r}}}}{r^2}\mathbf{e_{r}}r^{2}d\Omega = \mathbf{p}\frac{-4\pi }{3} <br />

The Attempt at a Solution

Can't find how to get -4Pi/3. In the way I am doing the r^2 would cancel out, the unit radial vectors would multiply each other to give one, and since p is the dipole moment vector and is constant in the problem (two point charges), it would get out of the integral. The integral over the solid angle d(omega) would then be equal to 4*Pi, and the answer would p*4Pi. Something is wrong here and I don't know what. Can anyone help?

Thanks

 

[tiny-tim]

hi rafaelpol!

(have a pi: π and an omega: Ω and an integral: ∫ and try using the X² icon just above the Reply box )

… the unit radial vectors would multiply each other to give one

nooo … you can't do that …

it's not e_r.e_r ​

 

[rafaelpol]

Thanks. I've got the correct result now. However, when this result is added to the Electric field of the dipole computed by the grad of the potential, instead of adding -4*Pi/3 *delta(r) Greiner adds +4*Pi/3 *delta(r). Why, is that? It seems to be a contradiction since if you integrate over a sphere with small radius, you should get -4*Pi/3 (as given by the integral in the above post) and not 4*Pi/3.

Thanks