Force on a dipole from a point charge?

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SUMMARY

The discussion focuses on calculating the vector force \( \mathbf{F}_p \) acting on a dipole \( \mathbf{p} \) placed in the electric field of a point charge \( q \) located at the origin. The electric field \( E \) of the point charge is given by \( E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \). The force on the dipole is derived using the equation \( \mathbf{F} = (\mathbf{p} \cdot \nabla) \mathbf{E} \), leading to the result \( \mathbf{F}_p = \frac{-q p_r}{2\pi \epsilon_0 r^3} \hat{r} \). The dipole experiences torque and aligns with the electric field, resulting in an attractive force towards the point charge.

PREREQUISITES
  • Understanding of electric fields, specifically the electric field of a point charge.
  • Knowledge of dipole moments and their behavior in electric fields.
  • Familiarity with vector calculus, particularly the gradient operator \( \nabla \).
  • Basic concepts of torque and its relation to dipoles in electric fields.
NEXT STEPS
  • Study the derivation of the electric field from point charges using Gauss's Law.
  • Learn about the torque experienced by dipoles in non-uniform electric fields.
  • Explore the concept of dipole alignment in external electric fields and its applications.
  • Investigate the effects of angle \( \theta \) on the force and torque experienced by dipoles.
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and researchers interested in the behavior of dipoles in electric fields.

warfreak131
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Homework Statement



A point charge q is situated at the origin. A dipole p is placed at r. The angle θ is defined by [itex]\hat{p}\cdot\hat{r}=cos(\theta)[/itex]

Calculate the vector force Fp acting on the dipole by the nonuniform E field of the point charge.

Homework Equations





The Attempt at a Solution



I know that the E field of a point charge is 1/4pi e0 * q/r2, and that [itex]F=(p\cdot\nabla)E=\nabla E\cdot p[/itex]

So since the E field of the point charge doesn't rely on phi or theta, i can just say that [itex]\nabla E=\frac{dE}{dr}\hat{r}=\frac{-q}{2\pi\epsilon_{0}r^{3}}\hat{r}[/itex]

Now if I dot that result with p, only the radial portions will remain since [itex]E_{\theta}=E_{\phi}=0[/itex]

All that remains is [itex]\frac{-qp_{r}}{2\pi\epsilon_{0}r^{3}}[/itex]
 
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Won't the dipole tend to do react to charge q in the following ways. It will align with the electric field of the charge q ( let q be positive for this example) such that the negative end of the dipole is closest to q, there is a torque on the dipole which depends on the angle between the vectors r and p.

Once the dipole aligns the negative end of the dipole is slightly closer to the positive charge then the positive end of the dipole so this will result in a slight attractive force.

Part of your answer should have a theta dependance?
 

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