Originally Posted by **bnay** Hello everybody. I've been having an issue with this question:
A point charge q is situated a large distance r from a neutral atom of
polarizability alpha. Find the force of attraction between them.
When I worked it out I found the force to be [alpha*q^2]/[8*pi^2*epsilon^2*r^5], but I know a dipole should fall off with 1/r^3, although I'm not sure how to eliminate a factor of 1/r^2.
I arrived at this answer by finding the electric field of the dipole of the atom, where the non-dipole field, E = qd/alpha=p/alpha (where d is the induced shift of from the center of the electron cloud of the nucleus), so the electric field of the dipole (where the angle between the center of the electron cloud and the nucleus is pi) should be
Edip = [p/(4*pi*epsilon*r^3)]*(2cos(pi))E = -2[p/(4*pi*epsilon*r^3)]*[alpha*q/(4*pi*epsilon*r^2)]
But this results in a force over r^5. Any ideas where I went wrong?
Thank you |

Yes, the electric field, and thus the force on a point charge, falls off as 1/r^2. That's IF the dipole moment has a fixed value.

The dipole moment here is induced - it's due to the electric field of the point charge. The electric field due to the point charge, falls off as 1/r^2, so the resulting 1/r^5 seems reasonable.