Question

Earnshaw's theorem (Prob. 3.2) says that you cannot trap a charged

particle in an electrostatic field. Question:Could you trap a neutral (but polarizable) atom in an electrostatic field?

(a) Show that the force on the atom is $\overrightarrow{\mathrm{F}}=\frac{1}{2}\mathrm{\alpha }\overrightarrow{\nabla }{\mathrm{E}}^2$

(b) The question becomes, therefore: Is it possible for ${\mathbf{E}}^{\mathbf{2}}$ to have a local maximum (in a charge-free region)? In that case the force would push the atom back to its equilibrium position. Show that the answer is no. [Hint:Use Prob. 3.4(a).]

Short answer

(a) In the presence of an electrostatic field $\overrightarrow{\mathbf{E}}$, the force on an atom with polarizability $\alpha$is$\frac{1}{2}\alpha \overrightarrow{\nabla }{E}^2$.

(b) In a charge free region, ${\mathit{E}}^{\mathbf{2}}$ does not have local maxima.

Step-by-step solution

Given data

An atom of polarizability $\alpha$ is kept in an electrostatic field $\overrightarrow{\mathit{E}}$.

Electrostatic force on an atom

The force on an atom in an electrostatic field is

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{a}\mathbf{(}\overrightarrow{\mathbf{E}}\mathbf{.}\overrightarrow{\mathbf{\nabla }}\mathbf{)}\overrightarrow{\mathbf{E}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\alpha$is polarizability and $\overrightarrow{E}$is electrostatic field.

Derivation of force on an atom from an electrostatic field

The gradient of square of the electrostatic field is

$\overrightarrow{\nabla }{E}^2=2\overrightarrow{E}\times (\overrightarrow{\nabla }\times \overrightarrow{E})+(\overrightarrow{E}.\overrightarrow{\nabla })\overrightarrow{E}$

But the curl of an electrostatic field is zero. Hence

$\overrightarrow{\nabla }{E}^2=2\left(\overrightarrow{E}.\overrightarrow{\nabla }\right)\overrightarrow{E}$

Substitute this in equation (1) and get

$\overrightarrow{F}=\frac{\alpha \overrightarrow{\nabla }{E}^2}{2}$

Thus, the force on the atom is $\frac{1}{2}\alpha \overrightarrow{\nabla }{E}^2$.

Local maxima of electrostatic field

If $E^2$has a maxima, then there is a sphere about that maxima point $P$such that for all other points $P\text{'}\ne P$ on the sphere

$\left|\overrightarrow{E}(p\text{'})\right|<\left|\overrightarrow{E}\left(p\right)\right|........\left(2\right)$

In the absence of any charge inside the sphere, the average field over the surface is equal to the field on the maxima, that is,

$\frac{1}{4{\mathrm{\pi R}}^2}\int \overrightarrow{E}da=\overrightarrow{E}\left(P\right).......\left(3\right)$

Here, is the radius of the sphere and is the infinitesimal area element.

From equations (2) and (3) it follows

$$\begin{gathered} \frac{1}{4{\mathrm{\pi R}}^2}\int \overrightarrow{E}da<\frac{1}{4{\mathrm{\pi R}}^2}\int \overrightarrow{E}\left(p\right)da \\ \overrightarrow{E}\left(p\right)<\frac{1}{4{\mathrm{\pi R}}^2}\int \overrightarrow{E}\left(p\right)da \\ \overrightarrow{E}\left(p\right)<\overrightarrow{E}\left(p\right) \end{gathered}$$

This is a contradiction and hence the electrostatic field does not have a maxima.