Irodov 3.90: Pulled by Its Own Bound Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.90

A uniform isotropic dielectric of permittivity $\varepsilon$ fills a half-space with a plane face, and a point charge $q$ sits in the vacuum outside, at distance $l$ from the face. The face carries the bound surface charge $\sigma' = -\dfrac{\varepsilon - 1}{\varepsilon + 1}\,\dfrac{ql}{2\pi r^3}$ , where $r$ is the distance from $q$ to the point of the face. Find the magnitude of the force that this bound charge exerts on $q$ .

The point charge q in vacuum at height l above the plane face of the dielectric ε. — The point charge $q$ in vacuum at height $l$ above the plane face of the dielectric $\varepsilon$ .

Given

point charge $q$ at distance $l$ from the face
half-space of permittivity $\varepsilon$
$\sigma' = -\dfrac{\varepsilon - 1}{\varepsilon + 1}\,\dfrac{ql}{2\pi r^3}$

Find

the force

F

of the bound charge on

q

Solution

Step 1The force of one ring of bound charge

Divide the face into rings about the foot of the perpendicular from $q$ . A ring of radius $\rho$ and width $d\rho$ has area $2\pi\rho\,d\rho = 2\pi r\,dr$ , because $r^2 = l^2 + \rho^2$ , and every point of it is at the distance $r$ from $q$ . Its charge is

dq' = \sigma'\cdot 2\pi r\,dr = -\frac{\varepsilon - 1}{\varepsilon + 1}\,\frac{ql}{r^2}\,dr \tag{1}

One ring of the face, edge on. Its bound charge is opposite to q, so each piece of it pulls q toward itself; d F is the pull of the piece on the right. — One ring of the face, edge on. Its bound charge is opposite to $q$ , so each piece of it pulls $q$ toward itself; $d\vect F$ is the pull of the piece on the right.

Each piece of the ring pulls $q$ along the line joining them. By symmetry the components parallel to the face cancel around the ring, and the component normal to the face, toward it, survives with the factor $l/r$ :

dF = \frac{q\,|dq'|}{4\pi\varepsilon_0r^2}\cdot\frac{l}{r} = \frac{\varepsilon - 1}{\varepsilon + 1}\,\frac{q^2l^2}{4\pi\varepsilon_0}\,\frac{dr}{r^5}

where (1) was substituted.

Step 2Summing over the face

The rings run from $r = l$ under the charge out to infinity:

F = \frac{\varepsilon - 1}{\varepsilon + 1}\,\frac{q^2l^2}{4\pi\varepsilon_0}\int_l^{\infty}\frac{dr}{r^5} = \frac{\varepsilon - 1}{\varepsilon + 1}\,\frac{q^2l^2}{4\pi\varepsilon_0}\cdot\frac{1}{4l^4}

F = \frac{(\varepsilon - 1)\,q^2}{16\pi\varepsilon_0(\varepsilon + 1)\,l^2}

(2)

The force attracts $q$ toward the dielectric, whatever the sign of $q$ .

The field of q over the dielectric, drawn for ε = 3, with the image q' at the mirror point behind the face and the force F on q, toward the dielectric. — The field of $q$ over the dielectric, drawn for $\varepsilon = 3$ , with the image $q'$ at the mirror point behind the face and the force $\vect F$ on $q$ , toward the dielectric.

Check

The field of the bound charge in the vacuum is that of a point image charge $q' = -q(\varepsilon - 1)/(\varepsilon + 1)$ at the mirror point, a distance $2l$ from $q$ . Coulomb’s law for that pair gives

F = \frac{q\,|q'|}{4\pi\varepsilon_0(2l)^2} = \frac{(\varepsilon - 1)\,q^2}{16\pi\varepsilon_0(\varepsilon + 1)\,l^2},

the same as (2). With $\varepsilon = 1$ the force vanishes; as $\varepsilon \to \infty$ it tends to $q^2/(16\pi\varepsilon_0l^2)$ , the attraction of a charge to an earthed conducting plane.

Remark

Any dielectric attracts a charge brought near it, whatever the sign of the charge: the same effect lets a charged comb pick up scraps of paper. The attraction is the fraction $(\varepsilon - 1)/(\varepsilon + 1)$ of the pull of a conducting plane at the same distance: one third for $\varepsilon = 2$ , and already $0.98$ for water at $\varepsilon = 81$ . The force follows the same inverse square of the distance to the image, $2l$ , in both cases.