Irodov 3.89: The Charge a Dielectric Gathers – Amit & Deepali Sinha Foundation

Problem Irodov 3.89

A uniform isotropic dielectric of permittivity $\varepsilon$ fills a half-space with a plane face. A point charge $q$ sits in the vacuum outside, at distance $l$ from the face. Find:

the surface density of the bound charge on the face as a function of the distance $r$ from $q$ to the point of the face, and what it becomes as $l \to 0$ ;
the total bound charge on the face.

The point charge q in vacuum at height l above the plane face of the dielectric ε; r is the distance from q to a point of the face. — The point charge $q$ in vacuum at height $l$ above the plane face of the dielectric $\varepsilon$ ; $r$ is the distance from $q$ to a point of the face.

Given

point charge $q$ in vacuum
distance $l$ to the face
half-space of permittivity $\varepsilon$

Find

(a)

\sigma'(r)

and its limit as

l \to 0

; (b) the total bound charge

q'

Solution

Step 1The normal field on each side of the face

Take a point of the face at distance $r$ from $q$ , and the normal pointing into the dielectric. The field of $q$ alone there has magnitude $q/(4\pi\varepsilon_0r^2)$ , and the fraction $l/r$ of it is normal to the face:

E_{q,n} = \frac{q}{4\pi\varepsilon_0r^2}\cdot\frac{l}{r} = \frac{ql}{4\pi\varepsilon_0r^3}

The bound charge $\sigma'$ on the face adds its own field: very close to the face it acts as an infinite plane, $\sigma'/(2\varepsilon_0)$ pointing away from it on each side, while the rest of the face, lying in the same plane, adds no normal component. So just outside (1) and just inside (2):

\begin{aligned} E_{n1} &= \frac{ql}{4\pi\varepsilon_0r^3} - \frac{\sigma'}{2\varepsilon_0}\\ E_{n2} &= \frac{ql}{4\pi\varepsilon_0r^3} + \frac{\sigma'}{2\varepsilon_0} \end{aligned}

Step 2(a) The bound surface density

The face carries no free charge, so the normal component of $\vect D$ is continuous across it, $\varepsilon_0E_{n1} = \varepsilon_0\varepsilon E_{n2}$ :

\begin{aligned} \frac{ql}{4\pi r^3} - \frac{\sigma'}{2} &= \varepsilon\left(\frac{ql}{4\pi r^3} + \frac{\sigma'}{2}\right)\\ \frac{\varepsilon + 1}{2}\,\sigma' &= -(\varepsilon - 1)\,\frac{ql}{4\pi r^3} \end{aligned}

Solving for $\sigma'$ :

\sigma' = -\frac{\varepsilon - 1}{\varepsilon + 1}\,\frac{ql}{2\pi r^3}

(1)

The bound charge has the sign opposite to $q$ and is densest right under it, at $r = l$ , falling off as $1/r^3$ .

The field of q over the dielectric, drawn for ε = 3 and q > 0. Of the lines reaching the face, some go on into the dielectric and the rest end on the negative bound charge of the face. — The field of $q$ over the dielectric, drawn for $\varepsilon = 3$ and $q > 0$ . Of the lines reaching the face, some go on into the dielectric and the rest end on the negative bound charge of the face.

As $l \to 0$ with $r$ held fixed, $\sigma'$ goes to zero at every point of the face except the foot of the perpendicular from $q$ , where $r = l$ and $\sigma' \propto 1/l^2$ grows without bound. The bound charge gathers into a point at the charge itself, and, as part (b) shows, its total does not depend on $l$ .

Step 3(b) The total bound charge

A ring of the face of radius ρ and width dρ, edge on; every point of it is at the distance r from q. — A ring of the face of radius $\rho$ and width $d\rho$ , edge on; every point of it is at the distance $r$ from $q$ .

Divide the face into rings of radius $\rho$ about the foot of the perpendicular, each of area $2\pi\rho\,d\rho$ , all of whose points are at the distance $r$ from $q$ . Since $r^2 = l^2 + \rho^2$ , the ring area is also $2\pi r\,dr$ , and $r$ runs from $l$ to infinity. Integrating (1):

\begin{aligned} q' &= \int_l^{\infty}\sigma'\,2\pi r\,dr = -\frac{\varepsilon - 1}{\varepsilon + 1}\,ql\int_l^{\infty}\frac{dr}{r^2}\\ &= -\frac{\varepsilon - 1}{\varepsilon + 1}\,ql\cdot\frac{1}{l} \end{aligned}

q' = -\frac{\varepsilon - 1}{\varepsilon + 1}\,q

(2)

Check

With $\varepsilon = 1$ there is no dielectric and both results vanish. As $\varepsilon \to \infty$ the dielectric behaves like a conductor: $\sigma' \to -ql/(2\pi r^3)$ and $q' \to -q$ , which are the induced density and the total induced charge on an earthed conducting plane facing $q$ .

Remark

The total bound charge (2) is the image charge of this geometry. In the vacuum the field is exactly that of $q$ together with a point charge $q'$ at the mirror image of $q$ in the face, and inside the dielectric it is the field of the single point charge $q + q' = 2q/(\varepsilon + 1)$ sitting where $q$ is. Seen from the vacuum side, the bound charge spread over the face is indistinguishable from that one point charge behind it.