40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
 Coating Ring-down Measurement Lab elog Not logged in
Message ID: 366     Entry time: Thu Jul 6 12:48:54 2017
 Author: Zach Type: Electronics Category: Modeling Subject: Resolving the factor of two

# 2017-07-06

I resolved the factor of two from Griffiths' discussion of dipoles in non-uniform electric fields. The force on a dipole in a non-uniform field is $\textbf{F}=\textbf{F}_+ + \textbf{F}_-=q(\Delta \textbf{E})$ where $\Delta \textbf{E}$ is the difference in the field between the plus end and the minus end. Component wise, $\Delta E_x = (\nabla E_x) \cdot \textbf{d}$ where d is a unit vector. This holds for y and z, the whole thing can also be written as $\Delta \textbf{E} = (\textbf{d} \cdot \nabla) \textbf{E}$. Since p=qd, we can write $\textbf{F} = (\textbf{p} \cdot \nabla) \textbf{E}$

Jackson derives it differently by deriving the electrostatic energy of a dielectric from the energy of a collection of charges in free space. He then derives the change in energy of a dielectric placed in a fixed source electric field to derive that the energy density w is given by $w = -\frac{1}{2} \textbf{P} \cdot \textbf{E}_0$. This explicity explains the factor of two and is an interesting alternative explanation.

ELOG V3.1.3-