Irodov 3.353: Peak Displacement Current at Arrival – Amit & Deepali Sinha Foundation

Problem Irodov 3.353

A thin wire ring of radius $a$ carries a charge $q$ . Its centre moves in a straight line at a constant speed $v$ toward an observation point $P$ , with the ring’s plane kept perpendicular to the direction of motion. How far from $P$ , $x_m$ , is the ring at the instant the displacement current density at $P$ is largest, and what is that largest value?

The ring of radius a and charge q, its plane perpendicular to its velocity v, approaching the point P on its axis, now a distance x away. — The ring of radius $a$ and charge $q$ , its plane perpendicular to its velocity $\vect v$ , approaching the point $P$ on its axis, now a distance $x$ away.

Given

$a$
$q$
$v = \text{const}$ , $v \ll c$ (assumed)
$P$ on the line of the centre’s motion

Find

x_m

and

j_{d,\max}

Solution

Step 1The field of the ring at a point on its axis

The centre moves straight toward $P$ , so $P$ lies on the ring’s axis. For $v \ll c$ (assumed) the field at each instant is the Coulomb field of the ring where it is. Every charge element $dq$ is the same distance $\sqrt{a^2+x^2}$ from $P$ , and by symmetry only the axial components survive, each a fraction $x/\sqrt{a^2+x^2}$ of the element’s field; adding them over the whole charge $q$ , along the axis away from the ring, which is the direction of $\vect v$ :

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

Step 2The displacement current density at P

At the fixed point $P$ the field changes only because $x$ does, and the ring approaches, so $dx/dt = -v$ . The displacement current density is along the axis, and its component along $\vect v$ is

j_d = \varepsilon_0\,\frac{\partial E}{\partial t} = \varepsilon_0\,\frac{dE}{dx}\,\frac{dx}{dt} = -\varepsilon_0 v\,\frac{dE}{dx}

Differentiating (1) with respect to $x$ :

\frac{d}{dx}\,\frac{x}{(a^2+x^2)^{3/2}} = \frac{(a^2+x^2) - 3x^2}{(a^2+x^2)^{5/2}} = \frac{a^2 - 2x^2}{(a^2+x^2)^{5/2}}

so that

j_d = -\frac{qv}{4\pi}\,\frac{a^2 - 2x^2}{(a^2+x^2)^{5/2}} \tag{2}

Step 3Where its magnitude is largest

Write $g(x) = (a^2-2x^2)/(a^2+x^2)^{5/2}$ , so that $|j_d|$ is proportional to $|g|$ . Its derivative, over a common denominator:

g'(x) = \frac{-4x(a^2+x^2) - 5x(a^2-2x^2)}{(a^2+x^2)^{7/2}} = \frac{3x\,(2x^2 - 3a^2)}{(a^2+x^2)^{7/2}}

It vanishes at $x = 0$ and at $x = a\sqrt{3/2}$ , and $g$ falls to zero far away. Comparing the two stationary values:

|g(0)| = \frac{1}{a^3}, \qquad \left|g\big(a\sqrt{3/2}\big)\right| = \frac{2a^2}{(5a^2/2)^{5/2}} \approx \frac{0.20}{a^3}

The larger is at $x = 0$ , the instant the ring’s centre passes through $P$ . Putting $x = 0$ into (2):

x_m = 0, \qquad j_{d,\max} = \frac{qv}{4\pi a^3}

(3)

For $q > 0$ it points against $\vect v$ .

The displacement current density at P (component along v, for q > 0) against the ring's distance x: largest in magnitude, -j_m, at x = 0, zero at x = a/2, a weaker opposite peak at 1.22a. — The displacement current density at $P$ (component along $\vect v$ , for $q > 0$ ) against the ring’s distance $x$ : largest in magnitude, $-j_{\mathrm{m}}$ , at $x = 0$ , zero at $x = a/\sqrt2$ , a weaker opposite peak at $1.22a$ .

Check

Units: $qv/a^3$ is $\text{C}\cdot(\text{m/s})/\text{m}^3 = \text{A/m}^2$ . Sense at $x = 0$ : just before arrival the field at $P$ points along $\vect v$ (the ring is behind $P$ ), just after it points against $\vect v$ , and at $x = 0$ it is zero; so it is decreasing along $\vect v$ there, and $j_d$ points against $\vect v$ , as (2) gives. Far away, $x \gg a$ , (2) becomes $2qv/(4\pi x^3)$ , the result for a point charge on its own path.

Remark

The field at $P$ is largest at $x = a/\sqrt2$ , yet the displacement current there is zero: $j_d$ follows the rate of change of $\vect E$ , not its size. The rate is greatest where the field swings through zero, as the ring’s centre crosses $P$ .