Irodov 3.73: A Molecule Seeks the Strongest Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.73

A thin ring of radius $R$ carries a uniformly distributed charge, and a non-polar molecule sits on the ring’s axis. At what distance $x$ from the center of the ring is the magnitude of the force $F$ on the molecule

zero;
largest? Sketch the graph of the axial force component $F_x(x)$ .

The uniformly charged ring of radius R (drawn for positive charge q), its axis, and the non-polar molecule of polarizability β on the axis at distance x from the center. — The uniformly charged ring of radius $R$ (drawn for positive charge $q$ ), its axis, and the non-polar molecule of polarizability $\beta$ on the axis at distance $x$ from the center.

Given

ring of radius $R$ , uniformly charged
a non-polar molecule on the axis at distance $x$

Find

(a) the

x

where

F = 0

; (b) the

x

where

F

is largest, and the graph of

F_x(x)

Solution

Step 1The field on the axis

Call the ring’s charge $q$ . By symmetry the field on the axis is axial: the transverse parts of opposite elements cancel, and every element is the same distance $(R^2 + x^2)^{1/2}$ from the point, its field making the same angle with the axis. The axial parts add:

E_x = \frac{q}{4\pi\varepsilon_0}\,\frac{x}{(R^2 + x^2)^{3/2}} \tag{1}

Step 2The force on the molecule

The field induces in the molecule the moment $\vect p = \beta\varepsilon_0\vect E$ , along the axis. The force on a dipole is $(\vect p\cdot\nabla)\vect E$ ; with $\vect p$ and $\vect E$ both axial on the axis, only the axial derivative of $E_x$ enters:

F_x = p\,\frac{dE_x}{dx} = \beta\varepsilon_0 E_x\,\frac{dE_x}{dx}

Differentiating (1) by the quotient rule:

\frac{dE_x}{dx} = \frac{q}{4\pi\varepsilon_0}\,\frac{(R^2 + x^2) - 3x^2}{(R^2 + x^2)^{5/2}} = \frac{q}{4\pi\varepsilon_0}\,\frac{R^2 - 2x^2}{(R^2 + x^2)^{5/2}}

Multiplying by $\beta\varepsilon_0 E_x$ :

F_x = \beta\varepsilon_0\left(\frac{q}{4\pi\varepsilon_0}\right)^2\frac{x\,(R^2 - 2x^2)}{(R^2 + x^2)^4}

(2)

Step 3(a) Where the force vanishes

The force (2) vanishes where either factor in its numerator does:

x = 0 \qquad\text{or}\qquad R^2 - 2x^2 = 0

x = 0 \quad\text{and}\quad x = \frac{R}{\sqrt 2}

(3)

At the center the field itself is zero; at $R/\sqrt 2$ the field is not zero but is at its largest, so its gradient vanishes there.

Step 4(b) Where the force is largest

In the dimensionless distance $u = x/R$ the shape of (2) is

f(u) = \frac{u\,(1 - 2u^2)}{(1 + u^2)^4}

Its derivative, over the common denominator $(1 + u^2)^5$ :

\begin{aligned} f'(u) &= \frac{(1 - 6u^2)(1 + u^2) - 8u^2(1 - 2u^2)}{(1 + u^2)^5}\\ &= \frac{1 - 13u^2 + 10u^4}{(1 + u^2)^5} \end{aligned}

Setting the numerator to zero gives a quadratic in $u^2$ :

u^2 = \frac{13 \pm \sqrt{169 - 40}}{20} = \frac{13 \pm 11.36}{20}

so $u^2 = 0.0821$ or $u^2 = 1.218$ . Taking square roots:

x_1 \approx 0.29R, \qquad x_2 \approx 1.1R

(4)

At $x_1$ the force is positive, pushing the molecule away from the ring’s plane; at $x_2$ it is negative, pulling it back toward the ring. The values $f(0.287) = 0.175$ and $f(1.104) = -0.065$ show the push at $x_1$ is the stronger of the two by a factor of about 2.7.

The axial force F_x against x: positive up to R/ 2 with its peak at x_1, negative beyond with its trough at x_2, and tending to zero far away. — The axial force $F_x$ against $x$ : positive up to $R/\sqrt 2$ with its peak at $x_1$ , negative beyond with its trough at $x_2$ , and tending to zero far away.

Check

Far from the ring, $x \gg R$ , (2) becomes $-2\beta\varepsilon_0(q/4\pi\varepsilon_0)^2/x^5$ . That is what a point charge $q$ gives: the induced moment falls as $1/x^2$ and the field gradient as $1/x^3$ , and the force points toward the charge. Near the center, $x \ll R$ , it becomes $\beta\varepsilon_0(q/4\pi\varepsilon_0)^2 x/R^6$ , growing linearly from zero.

Remark

A neutral polarizable body always drifts toward stronger field, so on the axis it is drawn to $x = R/\sqrt 2$ from either side: that zero of the force is a stable equilibrium along the axis, while the center, where the field vanishes, is unstable along it. The same rule, force along the gradient of $E^2$ , is what lets a charged comb pick up bits of paper and what traps particles in the focus of a laser beam.