1
$\begingroup$

Vqµ = − qµ / (4π ε0εr r2)

Say you had a negative charge, interacting with the positive end of a dipole then the charge would be -1.602 x 10-19 C and the dipole moment would always be a positive value. This means Vqµ would always be positive.

However a negative charge interacting with the positive end of a dipole is an attractive interaction so Vqµ should be negative. What am I missing here?

|improve this question
$\endgroup$
  • $\begingroup$ You're missing the fact that $\mu$, your quantity that measures the dipole only measures the magnitude but not the direction or alignment of the dipole. Your equation is flawed, I have no idea what it should properly be because as far as I know the interaction of a point charge with a dipole is more complex than merely a modified Coulomb law $\endgroup$ – orthocresol♦ Apr 13 '16 at 12:53
  • $\begingroup$ So if you are talking about the positive end of the dipole, then μ becomes negative? My equation is from a 3rd year biophysical chem course, so I assume there is much more to charge-dipole interactions than this. $\endgroup$ – draksi Apr 13 '16 at 13:01
  • $\begingroup$ The electric dipole moment is a vector quantity: $\mathbf{p} = q\mathbf{d}$ where $\mathbf{d}$ is a vector pointing from the negative end to the positive end. You have the dipole moment as a scalar quantity: $\mu = qd$ where $d$ is the norm (aka length) of the vector $\mathbf{d}$. In simplifying from $\mathbf{p}$ to $\mu$ there is a loss of information about the orientation of the dipole - analogous to velocity/speed. The eqn doesn't capture this and you could add negative signs to make it work but 1) it is only a quick fix 2) the choice of sign is only conventional ie there is no right/wrong. $\endgroup$ – orthocresol♦ Apr 13 '16 at 13:35
  • $\begingroup$ (cont) for example if you define the positive end of the dipole as having $\mu < 0$ then you need to retain the negative sign at the front of your formula. Whereas if you define the positive end of the dipole as having $\mu > 0$ and the negative end as having $\mu < 0$ then you need to get rid of the negative sign in the formula. I cannot say which approach is correct, because both give you the correct answer. Strictly speaking I would say neither are correct because $q$ is (by definition of the dipole moment) positive and $d$ being a length has to be non-negative, so $\mu$ cannot be negative. $\endgroup$ – orthocresol♦ Apr 13 '16 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.