# How do I find the force on the charged sphere due to a dipole sitting inside?

I have a dipole sitting inside a charged sphere i.e. at the centre of the sphere. As far as I can see the force on the Dipole is zero because the external electric field on the dipole at $r=0$ is $0$. But what about the force due to the dipole on the sphere? We know that the electric field due to a dipole is given by $-\bar p\cdot \frac {\nabla \bar r}{r^3}$. I have come to the step (I don't know if I am right):

$$\frac{-3Q}{4\pi R^3} \int_{\partial V}\frac{\bar p\cdot\bar r}{r^3}.$$

Another case, If the sphere was a hollow sphere with the thickness $t$, what would be the force on the sphere due to this dipole?

-
I deleted my post because I started second guessing if was going about things the right way. But I did want mention that my electric field formula was correct and in standard form. In your formula you appear to be taking the gradient of the position vector, so I assume you started by taking the gradient of the dot product in the potential, which can be tricky, but you do get something like what you have above. If you finish taking the gradient you'll arrive at my formula. – David H Nov 21 '12 at 22:57
Third law of Newton states (actio = reactio). Try to think how this might solve your problem. You find more about that here. – Fabian Nov 21 '12 at 23:27