Irodov 3.352: The Moving Charge's Displacement Current

Problem Irodov 3.352

A point charge $q$ moves at a constant velocity $\vect v$ , slow compared with the speed of light. Find the displacement current density $\vect j_d$ at a point a distance $r$ from the charge that lies

on the line of the charge’s motion;
on the line through the charge perpendicular to its motion.

The charge q moving with v along the dashed path. The point A lies on the path and the point B on the perpendicular through the charge, each a distance r from it. — The charge $q$ moving with $\vect v$ along the dashed path. The point $A$ lies on the path and the point $B$ on the perpendicular through the charge, each a distance $r$ from it.

Given

$q$
$\vect v = \text{const}$ , $v \ll c$
distance $r$
(a) point on the path; (b) point on the perpendicular

Find

\vect j_d

at each point

Solution

Step 1The field at a fixed point as the charge passes

For $v \ll c$ the field at each instant is the Coulomb field of the charge where it is. Let $\vect\rho$ be the vector from the charge to a fixed point of space; it changes only because the charge moves, so

\vect E = \frac{q}{4\pi\varepsilon_0}\,\frac{\vect\rho}{\rho^3}, \qquad \frac{d\vect\rho}{dt} = -\vect v \tag{1}

Step 2The displacement current density in general

The displacement current density in vacuum is $\varepsilon_0\,\partial\vect E/\partial t$ at the fixed point. First the distance: differentiating $\rho^2 = \vect\rho\cdot\vect\rho$ and using (1),

\frac{d\rho}{dt} = \frac{\vect\rho}{\rho}\cdot\frac{d\vect\rho}{dt} = -\frac{\vect\rho\cdot\vect v}{\rho}

Then the field’s vector factor, by the product rule:

\frac{\partial}{\partial t}\,\frac{\vect\rho}{\rho^3} = \frac{1}{\rho^3}\,\frac{d\vect\rho}{dt} - \frac{3\vect\rho}{\rho^4}\,\frac{d\rho}{dt} = -\frac{\vect v}{\rho^3} + \frac{3(\vect\rho\cdot\vect v)\,\vect\rho}{\rho^5}

Multiplying by $\varepsilon_0 q/(4\pi\varepsilon_0)$ :

\vect j_d = \frac{q}{4\pi}\left[\frac{3(\vect\rho\cdot\vect v)\,\vect\rho}{\rho^5} - \frac{\vect v}{\rho^3}\right] \tag{2}

Step 3(a) A point on the path

On the path $\vect\rho$ is parallel or antiparallel to $\vect v$ , ahead of the charge or behind it, and in both cases $(\vect\rho\cdot\vect v)\,\vect\rho = r^2\vect v$ . With $\rho = r$ in (2):

\vect j_d = \frac{q}{4\pi}\left[\frac{3\vect v}{r^3} - \frac{\vect v}{r^3}\right]

\vect j_d = \frac{2q\,\vect v}{4\pi r^3}

(3)

For $q > 0$ it points along $\vect v$ , ahead of the charge and behind it alike.

Step 4(b) A point on the perpendicular

On the perpendicular through the charge $\vect\rho\cdot\vect v = 0$ , so only the second term of (2) survives, with $\rho = r$ :

\vect j_d = -\frac{q\,\vect v}{4\pi r^3}

(4)

For $q > 0$ it points against $\vect v$ , half as large as on the path.

The displacement current density for q > 0: at A on the path it points along v; at B on the perpendicular it points against v and is half as large. — The displacement current density for $q > 0$ : at $A$ on the path it points along $\vect v$ ; at $B$ on the perpendicular it points against $\vect v$ and is half as large.

Check

Units: $qv/r^3$ is $\text{C}\cdot(\text{m/s})/\text{m}^3 = \text{A/m}^2$ , a current density. Sense on the path, behind the charge at $A$ : the field there points away from the charge, back along the path, and weakens as the charge recedes, so its rate of change points along $\vect v$ , as (a) says. At $B$ the charge is momentarily closest and $\vect E$ is turning: its component along $\vect v$ goes from positive (charge behind) to negative (charge ahead), so $\partial\vect E/\partial t$ points against $\vect v$ , as (b) says.

Remark

Expression (2) has exactly the shape of the field of a dipole whose moment points along $\vect v$ . Together with the charge’s own convection current $q\vect v$ it forms closed lines of total current, forward along the path and through the charge, back round the sides, and those loops are what give the slow charge its Biot and Savart magnetic field.