Irodov 3.46: Pushed Sideways by a Radial Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.46

A point dipole of moment $\vect{p}$ is at a distance $r$ from a long straight thread carrying a uniform charge of linear density $\lambda$ ; $\vect{r}$ is the radius vector from the thread to the dipole, perpendicular to the thread. Find the force $\vect{F}$ on the dipole when $\vect{p}$ points

along the thread;
along $\vect{r}$ ;
perpendicular to both the thread and $\vect{r}$ .

The thread with charge density λ, the radius vector r from it to the dipole, and the three orientations of p considered one at a time: (a) along the thread, (b) along r, (c) across both. — The thread with charge density $\lambda$ , the radius vector $\vect{r}$ from it to the dipole, and the three orientations of $\vect{p}$ considered one at a time: (a) along the thread, (b) along $\vect{r}$ , (c) across both.

Given

thread of density $\lambda$
dipole $\vect{p}$ at distance $r$

Find

\vect{F}

for

\vect{p}

(a) along the thread, (b) along

\vect{r}

, (c) across both

Solution

Step 1The force on a dipole

Treat the dipole as $-q$ at $\vect{s}$ and $+q$ at $\vect{s} + \vect{l}$ , with $\vect{p} = q\vect{l}$ . The two Coulomb forces differ only because the field changes over the short distance $\vect{l}$ , by $(\vect{l}\cdot\nabla)\vect{E}$ to first order:

\vect{F} = q\left[\vect{E}(\vect{s} + \vect{l}) - \vect{E}(\vect{s})\right] = (\vect{p}\cdot\nabla)\,\vect{E} \tag{1}

Step 2The field of the thread

By symmetry the field of a long uniform thread is radial and the same all round it, so Gauss’s law on a coaxial cylinder of radius $\rho$ and length $\ell$ gives

E\cdot 2\pi\rho\ell = \frac{\lambda\ell}{\varepsilon_0}

Take the $z$ axis along the thread and the $x$ axis along $\vect{r}$ , so the dipole sits at $(r, 0, 0)$ . The field at a point $(x, y, z)$ points straight away from the thread:

\vect{E} = \frac{\lambda}{2\pi\varepsilon_0}\,\frac{x\,\vect{i} + y\,\vect{j}}{x^2 + y^2} \tag{2}

Step 3(a) p along the thread

Now $\vect{p} = p\,\vect{k}$ , so (1) needs $\partial\vect{E}/\partial z$ . The field (2) does not depend on $z$ at all:

\vect{F} = p\,\frac{\partial\vect{E}}{\partial z} = 0

(3)

Step 4(b) p along r

Now $\vect{p} = p\,\vect{i}$ and (1) needs $\partial\vect{E}/\partial x$ , taken at $(r, 0)$ . Differentiating the two components of (2):

\begin{aligned} \frac{\partial E_x}{\partial x} &= \frac{\lambda}{2\pi\varepsilon_0}\,\frac{y^2 - x^2}{(x^2 + y^2)^2} = -\frac{\lambda}{2\pi\varepsilon_0 r^2}\\ \frac{\partial E_y}{\partial x} &= \frac{\lambda}{2\pi\varepsilon_0}\,\frac{-2xy}{(x^2 + y^2)^2} = 0 \end{aligned}

So the force is $-p\lambda/(2\pi\varepsilon_0 r^2)$ along $\vect{i}$ , which is along $\vect{p}$ with the opposite sense:

\vect{F} = -\frac{\lambda\vect{p}}{2\pi\varepsilon_0 r^2}

(4)

For $\lambda\vect{p}$ pointing away from the thread the dipole is pulled toward it, into the stronger field.

Case (b), seen along the thread: p along the field E, and the force F toward the thread (drawn for λ > 0). — Case (b), seen along the thread: $\vect{p}$ along the field $\vect{E}$ , and the force $\vect{F}$ toward the thread (drawn for $\lambda > 0$ ).

Step 5(c) p across the thread and r

Now $\vect{p} = p\,\vect{j}$ and (1) needs $\partial\vect{E}/\partial y$ at $(r, 0)$ :

\begin{aligned} \frac{\partial E_x}{\partial y} &= \frac{\lambda}{2\pi\varepsilon_0}\,\frac{-2xy}{(x^2 + y^2)^2} = 0\\ \frac{\partial E_y}{\partial y} &= \frac{\lambda}{2\pi\varepsilon_0}\,\frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{\lambda}{2\pi\varepsilon_0 r^2} \end{aligned}

So the force is $p\lambda/(2\pi\varepsilon_0 r^2)$ along $\vect{j}$ , along $\vect{p}$ itself:

\vect{F} = \frac{\lambda\vect{p}}{2\pi\varepsilon_0 r^2}

(5)

Case (c), seen along the thread: p across the field E, and the force F along p, at right angles to E (drawn for λ > 0). — Case (c), seen along the thread: $\vect{p}$ across the field $\vect{E}$ , and the force $\vect{F}$ along $\vect{p}$ , at right angles to $\vect{E}$ (drawn for $\lambda > 0$ ).

Check

Since the field has no curl, (1) is also the gradient of $\vect{p}\cdot\vect{E}$ at fixed $\vect{p}$ , so each case can be checked from that scalar. In (b), $\vect{p}\cdot\vect{E} = p\lambda/(2\pi\varepsilon_0 x)$ near the axis $y = 0$ , whose derivative in $x$ at $x = r$ is $-p\lambda/(2\pi\varepsilon_0 r^2)$ , as in (4). In (c), $\vect{p}\cdot\vect{E} = p\lambda y/(2\pi\varepsilon_0(x^2 + y^2))$ , whose derivative in $y$ at $(r, 0)$ is $p\lambda/(2\pi\varepsilon_0 r^2)$ , as in (5). In each case the other derivatives of $\vect{p}\cdot\vect{E}$ vanish at $(r, 0)$ , so the check covers the whole vector. Units: $\lambda p/(\varepsilon_0 r^2)$ is a charge times a field, a force.

Remark

Case (c) is the surprising one: the field at the dipole is radial, yet the force is purely sideways, along $\vect{p}$ , with the same strength as the radial pull of case (b). A dipole is not pushed along $\vect{E}$ but along the gradient of $\vect{p}\cdot\vect{E}$ . In (c) the dipole also feels a torque $\vect{p}\times\vect{E}$ that swings it toward the radial direction, and once it lines up with $\vect{E}$ it is in case (b) and is drawn in toward the thread.