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Force on a charge from an induced dipole 
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#1
Dec609, 11:38 PM

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 [tex]\alpha[/tex]. Find the force of attraction between them. 2. Relevant equations [tex]\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}[/tex] [tex]\vec{E}_{dip}(r,\theta)=\frac{p}{4\pi\epsilon_0r^3}(2\cos\theta\hat{r}+ \sin\theta\hat{\theta})[/tex] [tex]\vec{p}=\alpha\vec{E}[/tex] 3. The attempt at a solution [tex]\vec{E}_{mono}(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{r}[/tex] [tex]\vec{p}=\alpha\vec{E}[/tex] [tex]\vec{p}=\frac{\alpha q}{4\pi\epsilon_0r^2}\hat{r}[/tex] [tex]\vec{E}_{dip}(r,\theta)=\frac{p}{4\pi\epsilon_0r^3}(2\cos\theta\hat{r}+ \sin\theta\hat{\theta})[/tex] [tex]\vec{E}_{dip}(r,\pi)=\frac{\alpha q}{16\pi^2\epsilon^2_0r^5}(2\hat{r})[/tex] [tex]\vec{E}_{dip}(r,\pi)=\frac{\alpha q}{8\pi^2\epsilon^2_0r^5}\hat{r}[/tex] [tex]\vec{F}=q\vec{E}[/tex] [tex]\vec{F}=\frac{\alpha q^2}{8\pi^2\epsilon^2_0r^5}\hat{r}[/tex] 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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