# Force on a charge from an induced dipole

by Yitzach
Tags: dipole moment, electric field, electric force, induced dipole, polarization
 P: 61 1. The problem statement, all variables and given/known data A point charge q is situated a large distance r from a neutral atom of polarizability $$\alpha$$. Find the force of attraction between them. 2. Relevant equations $$\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}$$ $$\vec{E}_{dip}(r,\theta)=\frac{p}{4\pi\epsilon_0r^3}(2\cos\theta\hat{r}+ \sin\theta\hat{\theta})$$ $$\vec{p}=\alpha\vec{E}$$ 3. The attempt at a solution $$\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}$$ $$\vec{p}=\alpha\vec{E}$$ $$\vec{p}=\frac{\alpha q}{4\pi\epsilon_0r^2}\hat{r}$$ $$\vec{E}_{dip}(r,\theta)=\frac{p}{4\pi\epsilon_0r^3}(2\cos\theta\hat{r}+ \sin\theta\hat{\theta})$$ $$\vec{E}_{dip}(r,\pi)=\frac{\alpha q}{16\pi^2\epsilon^2_0r^5}(-2\hat{r})$$ $$\vec{E}_{dip}(r,\pi)=-\frac{\alpha q}{8\pi^2\epsilon^2_0r^5}\hat{r}$$ $$\vec{F}=q\vec{E}$$ $$\vec{F}=-\frac{\alpha q^2}{8\pi^2\epsilon^2_0r^5}\hat{r}$$ The result I got was unexpected because that is a repulsive force. Do I need to go about a longer way or did I mess it up somewhere?

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