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## Homework Statement

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.

## Homework 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]

## 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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