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Homework Statement
A solid sphere, radius ##R##,is centered at the origin. The "northern" hemisphere carries a uniform charge density ##\rho_0## and the "southern" hemisphere a uniform charge density ##-\rho_0##. Find the approximate field ##\mathbf E(r,\theta)## for points far from the sphere (##r >> R##).
Homework Equations
For a dipole where ##\mathbf p## is located at the origen and points in ##z## direction the electric field is given by
##\mathbf E_{\text{dip}}(r,\theta) = \frac{p}{4\pi \epsilon r^3}(2\cos \theta \hat r + \sin \theta \hat \theta)##.
Dipole moment
##\mathbf p = \int \mathbf r' \rho(\mathbf r') d\tau'##.
The Attempt at a Solution
I think I solved this one but I'm not entirely sure so I'm hoping someone could take a look and see if I'm doing this roughly right.
The dipole has to point in the ##\hat z## direction since symmetry. Our ##\mathbf r'## vector should then be ##\mathbf r' = R\cos \theta \hat z##. So the dipole moment is
##\mathbf p = \hat z \int_S R\cos \cdot R^2\sin \theta d\theta d\phi = \pi R^3\hat z\left[ \int_0^{\pi/2} \sin 2\theta d\theta - \int_{\pi/2}^\pi \sin 2\theta d\theta \right] = 2\pi R^3\hat z##
using that the area element of an ##r=R## surface is ##R^2\sin \theta d\theta d\phi##.
Our electric field is then (only taking account of the dipole term in the multipole expansion)
##\mathbf E(r,\theta) =\frac{R^3}{2\epsilon_0 r^3}(2\cos \theta \hat r +\sin \theta \hat \theta)##.
Does this look correct? It's certainly the field of a dipole and it's stronger when ##\theta=0## than when ##\theta=\pi/2## which makes sense but I don't know how to check it really.