Question

A point charge Qis "nailed down" on a table. Around it, at radius R,

is a frictionless circular track on which a dipole$\overrightarrow{\mathrm{p}}$ rides, constrained always to point tangent to the circle. Use Eq. 4.5 to show that the electric force on the dipole is

$\overrightarrow{\mathrm{F}}=\frac{\mathrm{Q}}{4{\mathrm{\tau \tau \epsilon }}_0}\frac{\overrightarrow{\mathrm{p}}}{{\mathrm{R}}^3}$

Notice that this force is always in the "forward" direction (you can easily confirm

this by drawing a diagram showing the forces on the two ends of the dipole). Why

isn't this a perpetual motion machine?

Short answer

It is proved that the electric force on a dipole that moves in a circle ofradius R

around a point charge Qis

$\overrightarrow{\mathit{F}}\mathit{=}\frac{\mathit{Q}}{\mathit{4}\mathit{\tau }\mathit{\tau }{\mathit{\epsilon }}_{\mathit{0}}}\frac{\overrightarrow{\mathit{p}}}{{\mathit{R}}^{\mathit{3}}}$

Step-by-step solution

Given data

A point charge Qis fixed on a table.

With Q at the center, a dipole $\overrightarrow{p}$moves on a frictionless circular track of radius R,

and constrained always to point tangent to the circle.

Force on a dipole

The force on a dipole having moment $\overrightarrow{p}$in the presence of an electric field $\overrightarrow{E}$is

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

Derivation of force on a dipole rotating around a point charge

Consider cylindrical coordinates.

The expression for the electric field from a point charge Q is

$\overrightarrow{E}=\frac{Q}{4\pi {\epsilon }_0{s}^2}\hat{s}$

Here, ${\epsilon }_0$is the permittivity of free space.

Using equation (1), Force on the dipole $\overrightarrow{p}$ moving on a circular track with Q at its center is

$$\begin{gathered} \overrightarrow{F}=\frac{p}{s}\frac{\partial }{\partial \theta }\frac{Q}{4\pi {\epsilon }_0{s}^2}\hat{s} \\ =\frac{p}{s}\frac{Q}{4\pi {\epsilon }_0{s}^2}\frac{\partial \hat{s}}{\partial \theta } \\ =\frac{p}{s}\frac{Q}{4\pi {\epsilon }_0{s}^2}\hat{\theta } \\ =\frac{Q}{4\pi {\epsilon }_0{s}^3}\overrightarrow{p} \end{gathered}$$

The direction of the force is different for the positive and negative ends of the dipole. The net force acts tangential.