📂Electrodynamics

Alignment of Polar Molecules by a Non-uniform Electric Field

Explanation ¹

Polar molecules possess a dipole moment even in the absence of an external electric field. If there is a constant external electric field, the dipole moment aligns with the direction of the electric field.

However, if the external electric field is not constant, $\mathbf{F}_+$ and $\mathbf{F}_-$ are not the same, resulting in a net force as well as a torque. The net force can be calculated as follows. If the electric field at $\pm q$ is $\mathbf{E}_\pm$,

$$ \mathbf{F} = \mathbf{F}_+ + \mathbf{F}_- = q(\mathbf{E}_+ - \mathbf{E}_-) = q(\Delta \mathbf{E}) $$

$$ \begin{equation} \Delta \mathbf{E} = \Delta E_{x} \hat{\mathbf{x}} + \Delta E_{y} \hat {\mathbf{y}} + \Delta E_{z} \hat {\mathbf{z}} \label{1} \end{equation} $$

If the length of the dipole is very short, $\Delta E_{x}$ can be approximated to the total differential $dE_{x}$ as $\mathbf{d} = dx \hat{\mathbf{x}} + dy \hat {\mathbf{y}} + dz \hat {\mathbf{z}}$

$$ \begin{align*} \Delta E_{x} \approx dE_{x} =&\ \dfrac{\partial E_{x}}{\partial x} dx + \dfrac{\partial E_{x}}{\partial y}dy + \dfrac{\partial E_{z}}{\partial z}dz \\ =&\ \left( \dfrac{\partial E_{x}}{\partial x}\hat {\mathbf{x}} + \dfrac{\partial E_{x}}{\partial y} \hat {\mathbf{y}} + \dfrac{\partial E_{z}}{\partial z} \hat {\mathbf{z}} \right) \cdot (dx \hat {\mathbf{x}} + dy \hat {\mathbf{y}} + dz \hat {\mathbf{z}} ) \\ =&\ \nabla E_{x} \cdot \mathbf{d} \end{align*} $$

By calculating $E_{y}$ and $E_{z}$ in the same manner and substituting them into $\eqref{1}$,

$$ \begin{align*} \Delta \mathbf{E} =&\ (\nabla E_{x} \cdot \mathbf{d} ) \hat {\mathbf{x}} +( \nabla E_{y} \cdot \mathbf{d}) \hat {\mathbf{y}} +( \nabla E_{z} \cdot \mathbf{d} )\hat {\mathbf{z}} \\ =&\ \left( dx\dfrac{\partial E_{x}}{\partial x} + dy\dfrac{\partial E_{x}}{\partial y} + dz \dfrac{\partial E_{x} }{\partial z} \right) \hat {\mathbf{x}} + \left( dx\dfrac{\partial E_{y}}{\partial x} + dy\dfrac{\partial E_{y}}{\partial y} + dz \dfrac{\partial E_{y} }{\partial z} \right) \hat {\mathbf{y}} \\ &+ \left( dx\dfrac{\partial E_{z}}{\partial x} + dy\dfrac{\partial E_{z}}{\partial y} + dz \dfrac{\partial E_{z} }{\partial z} \right) \hat {\mathbf{z}} \\ =&\ \left( dx\dfrac{\partial }{\partial x} + dy\dfrac{\partial }{\partial y} + dz \dfrac{\partial }{\partial z} \right)E_{x} \hat {\mathbf{x}} + \left( dx\dfrac{\partial}{\partial x} + dy\dfrac{\partial }{\partial y} + dz \dfrac{\partial }{\partial z} \right) E_{y} \hat {\mathbf{y}} \\ &+ \left( dx\dfrac{\partial }{\partial x} + dy\dfrac{\partial }{\partial y} + dz \dfrac{\partial }{\partial z} \right) E_{z}\hat {\mathbf{z}} \\ =&\ \left( dx\dfrac{\partial }{\partial x} + dy\dfrac{\partial }{\partial y} + dz \dfrac{\partial }{\partial z} \right) \left( E_{x} \hat {\mathbf{x}} + E_{y} \hat {\mathbf{y}} +E_{z}\hat {\mathbf{z}} \right) \\ =&\ \left[ (dx \hat {\mathbf{x}} + dy \hat {\mathbf{y}} + dz \hat {\mathbf{z}} ) \cdot \left( \dfrac{\partial }{\partial x}\hat {\mathbf{x}} + \dfrac{\partial }{\partial y} \hat {\mathbf{y}}+ \dfrac{\partial }{\partial z} \hat {\mathbf{z}}\right)\right] \left( E_{x} \hat {\mathbf{x}} + E_{y} \hat {\mathbf{y}} +E_{z}\hat {\mathbf{z}} \right) \\ =&\ (\mathbf{d} \cdot \nabla ) \mathbf{E} \end{align*} $$

Therefore, the net force acting on polar molecules is $\mathbf{F}$

$$ \begin{align*} \mathbf{F} =&\ q(\Delta \mathbf{E} ) \\ =&\ (q\mathbf{d} \cdot \nabla ) \mathbf{E} \\ =&\ (\mathbf{p} \cdot \nabla ) \mathbf{E} \end{align*} $$

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