Register to reply

Force on a charge from an induced dipole

Share this thread:
Yitzach
#1
Dec6-09, 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?
Phys.Org News Partner Science news on Phys.org
New type of solar concentrator desn't block the view
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Asian inventions dominate energy storage systems

Register to reply

Related Discussions
Surface charge induced in conductor by point charge Introductory Physics Homework 17
Force induced on a point charge by two conducting planes Introductory Physics Homework 6
Induced transitions, Dipole Approx. Advanced Physics Homework 2
Induced charge density by non-uniform dipole density in dielectric?! Classical Physics 1
Net charge VS dipole moment in E field by infinite line charge General Physics 5