Chapter 4: Q31P (page 206)

A point charge Qis "nailed down" on a table. Around it, at radius R,
is a frictionless circular track on which a dipole $\vec{p}$ rides, constrained always to point tangent to the circle. Use Eq. 4.5 to show that the electric force on the dipole is
$\vec{F} = \frac{Q}{4 {ττε}_{0}} \frac{\vec{p}}{R^{3}}$
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

Expert verified

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

around a point charge Qis

$\vec{F} = \frac{Q}{4 τ τ ε_{0}} \frac{\vec{p}}{R^{3}}$

Step by step solution

Given data

A point charge Qis fixed on a table.

With Q at the center, a dipole $\vec{p}$ 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 $\vec{p}$ in the presence of an electric field $\vec{E}$ is

$\vec{F} = (\vec{p} . \vec{\nabla}) \vec{E} . . . . . (1)$

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

$\vec{E} = \frac{Q}{4 π ε_{0} s^{2}} \hat{s}$

Here, $ε_{0}$ is the permittivity of free space.

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

$\vec{F} = \frac{p}{s} \frac{\partial}{\partial θ} \frac{Q}{4 π ε_{0} s^{2}} \hat{s} = \frac{p}{s} \frac{Q}{4 π ε_{0} s^{2}} \frac{\partial \hat{s}}{\partial θ} = \frac{p}{s} \frac{Q}{4 π ε_{0} s^{2}} \hat{θ} = \frac{Q}{4 π ε_{0} s^{3}} \vec{p}$

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

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Short Answer

Step by step solution

Given data

Force on a dipole

Derivation of force on a dipole rotating around a point charge

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