Irodov 3.41: A Sphere Where Two Fields Balance – Amit & Deepali Sinha Foundation

Problem Irodov 3.41

A point electric dipole of moment $\vect{p}$ sits in an external uniform electric field $\vect{E}_0$ , with $\vect{p}$ pointing in the same direction as $\vect{E}_0$ . One of the equipotential surfaces of the combined field encloses the dipole and is a sphere. Find its radius.

The point dipole p in the uniform external field E_0, both pointing the same way. — The point dipole $\vect{p}$ in the uniform external field $\vect{E}_0$ , both pointing the same way.

Given

point dipole $\vect{p}$
uniform field $\vect{E}_0$
$\vect{p}$ parallel to $\vect{E}_0$

Find

the radius

R

of the spherical equipotential

Solution

Step 1The two potentials

Put the dipole at the origin and measure the polar angle $\theta$ from the common direction of $\vect{p}$ and $\vect{E}_0$ . The uniform field’s potential falls along the field; taking it to be zero in the plane through the dipole, it is $-\vect{E}_0\cdot\vect{r}$ . The dipole adds its own:

\varphi = -E_0 r\cos\theta + \frac{p\cos\theta}{4\pi\varepsilon_0 r^2}

Both terms carry the same angular factor, which comes out as a common factor:

\varphi = \cos\theta\left(\frac{p}{4\pi\varepsilon_0 r^2} - E_0 r\right) \tag{1}

Step 2The zero surface

The potential (1) vanishes at every $\theta$ wherever the bracket vanishes, and that fixes $r$ alone:

\begin{aligned} \frac{p}{4\pi\varepsilon_0 r^2} &= E_0 r\\ r^3 &= \frac{p}{4\pi\varepsilon_0 E_0} \end{aligned}

So the surface $\varphi = 0$ contains a sphere centered on the dipole, which encloses it:

R = \sqrt[3]{\frac{p}{4\pi\varepsilon_0 E_0}}

(2)

The same level $\varphi = 0$ also contains the plane $\theta = 90^\circ$ through the dipole, where $\cos\theta = 0$ .

Step 3Where the two pieces meet

An equipotential can cross itself only where the field vanishes. On the equator the dipole field points against $\vect{p}$ with strength $p/(4\pi\varepsilon_0 r^3)$ , and it cancels $E_0$ exactly where the sphere meets the plane:

E_z = E_0 - \frac{p}{4\pi\varepsilon_0 R^3} = 0

by (2). In the map every field line that meets the sphere crosses it at right angles, and the circle of zero field, seen edge-on, is the pair of dots.

Field lines (crimson) of the dipole at the center in the uniform field pointing up, and equipotentials (teal): the level φ = 0, made of the sphere of radius R and the equatorial plane crossing on the circle where E = 0 (dots), and two further levels above and below, close round the dipole. — Field lines (crimson) of the dipole at the center in the uniform field pointing up, and equipotentials (teal): the level $\varphi = 0$ , made of the sphere of radius $R$ and the equatorial plane crossing on the circle where $\vect{E} = 0$ (dots), and two further levels above and below, close round the dipole.

Check

Units: $p/(\varepsilon_0 E_0)$ is (C $\cdot$ m)/(C/(V $\cdot$ m) $\cdot$ V/m) $=$ m $^3$ , so $R$ is a length. On the sphere the field must be normal to it. The tangential component, along growing $\theta$ , follows from (1):

E_\theta = -\frac{1}{r}\frac{\partial\varphi}{\partial\theta} = \sin\theta\left(\frac{p}{4\pi\varepsilon_0 r^3} - E_0\right)

which is zero at $r = R$ for every $\theta$ , as it must be. A stronger dipole or a weaker field makes the sphere larger, as it should: the dipole’s field then dominates further out.

Remark

On the sphere the radial field is

E_r = -\frac{\partial\varphi}{\partial r} = \cos\theta\left(\frac{2p}{4\pi\varepsilon_0 R^3} + E_0\right) = 3E_0\cos\theta

which is exactly the field at the surface of an uncharged conducting ball of radius $R$ placed in $\vect{E}_0$ . Read backwards: the charges induced on such a ball act outside it as a point dipole $p = 4\pi\varepsilon_0 R^3 E_0$ at its center, which is how the polarizability of a conducting sphere is found.