Force on a charge from an induced dipole

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SUMMARY

The discussion focuses on calculating the force of attraction between a point charge \( q \) and a neutral atom with polarizability \( \alpha \). The derived expression for the force is \( \vec{F} = -\frac{\alpha q^2}{8\pi^2\epsilon^2_0r^5}\hat{r} \), indicating a repulsive force rather than an attractive one. The confusion arises from the direction of the force vector, which points towards the origin, suggesting attraction, but the negative sign indicates repulsion. This highlights the importance of correctly interpreting vector directions in electrostatics.

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  • Understanding of electrostatics, specifically Coulomb's law
  • Familiarity with the concept of electric dipoles and polarizability
  • Knowledge of vector calculus in the context of electric fields
  • Basic grasp of Maxwell's equations and their implications in electrostatics
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  • Study the derivation of electric field equations for monopoles and dipoles
  • Learn about the implications of polarizability in different materials
  • Explore the concept of induced dipoles in various physical contexts
  • Investigate the relationship between force direction and vector signs in electrostatics
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Students and professionals in physics, particularly those specializing in electromagnetism, as well as educators looking to clarify concepts related to electric forces and dipoles.

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Homework Statement


A point charge q is situated a large distance r from a neutral atom of polarizability \alpha. Find the force of attraction between them.

Homework 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}

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?
 
Last edited:
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Why do you say that the force is not attractive? It points in the negative ##\hat r## direction, i.e. towards the origin where presumably you put the monopole.
 

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