Irodov 3.92: Image Charges Across a Dielectric Face – Amit & Deepali Sinha Foundation

Problem Irodov 3.92

A uniform isotropic dielectric of permittivity $\varepsilon$ fills the half-space on one side of an infinite plane, with vacuum on the other side. A small conducting ball with charge $q$ sits inside the dielectric at a distance $l$ from the plane. Find the surface density of bound charge on the plane as a function of the distance $r$ from the ball, and discuss the result as $l \to 0$ .

The small ball with charge q inside the dielectric, a distance l from its plane face; vacuum lies below. A point of the plane is a distance r from the ball. Drawn for q > 0. — The small ball with charge $q$ inside the dielectric, a distance $l$ from its plane face; vacuum lies below. A point of the plane is a distance $r$ from the ball. Drawn for $q > 0$ .

Given

ball charge $q$
permittivity $\varepsilon$
distance $l$ to the boundary with vacuum

Find

\sigma'(r)

on the boundary, and its behavior for

l \to 0

Solution

Step 1Image charges for the two media

Seek the field in the dielectric as that of $q$ and an image $q_1$ at the mirror point of the ball, both screened by $\varepsilon$ , and the field in the vacuum as that of a single charge $q_2$ at the ball. With $r_1$ the distance from the ball and $r_2$ from the mirror point:

\begin{aligned} \vect E_{\mathrm{diel}} &= \frac{1}{4\pi\varepsilon_0\varepsilon}\left(\frac{q\,\vect r_1}{r_1^3} + \frac{q_1\vect r_2}{r_2^3}\right)\\ \vect E_{\mathrm{vac}} &= \frac{1}{4\pi\varepsilon_0}\,\frac{q_2\,\vect r_1}{r_1^3} \end{aligned}

Each form has the right sources in its own region: $\nabla\cdot\vect D$ is the free charge $q$ in the dielectric and zero in the vacuum. It remains to choose $q_1$ and $q_2$ so that the boundary conditions hold.

Step 2The boundary conditions fix the images

At a point of the plane $r_1 = r_2 = r$ . Let $\rho$ be its distance from the foot of the perpendicular from the ball, and take the normal $\vect n$ from the dielectric into the vacuum. The tangential components of the three fields are all along $\rho$ and proportional to $\rho/r^3$ ; continuity of tangential $E$ requires

\frac{q + q_1}{\varepsilon} = q_2 \tag{1}

Along $\vect n$ the field of $q$ has the component $q\,l/r^3$ (up to the common factor), and the image’s has $-q_1 l/r^3$ . Continuity of the normal component of $\vect D$ , with $D = \varepsilon_0\varepsilon E$ in the dielectric and $D = \varepsilon_0 E$ in the vacuum:

q - q_1 = q_2 \tag{2}

Eliminating $q_2$ between (1) and (2):

q + q_1 = \varepsilon(q - q_1)

\begin{gathered} q_1 = \frac{\varepsilon - 1}{\varepsilon + 1}\,q\\ q_2 = \frac{2q}{\varepsilon + 1} \end{gathered}

(3)

The field for ε = 3. In the dielectric the lines are those of q and its image q_1 (dashed); in the vacuum they run straight out from the ball's position. Lines that begin at the face start on positive bound charge. Drawn for q > 0. — The field for $\varepsilon = 3$ . In the dielectric the lines are those of $q$ and its image $q_1$ (dashed); in the vacuum they run straight out from the ball’s position. Lines that begin at the face start on positive bound charge. Drawn for $q > 0$ .

Step 3The normal fields on either side of the plane

On the vacuum side the normal component comes from $q_2$ alone. On the dielectric side it comes from $q$ and $q_1$ , and (2) turns $q - q_1$ into $q_2$ .

\begin{aligned} E_{n,\mathrm{vac}} &= \frac{q_2\,l}{4\pi\varepsilon_0 r^3}\\ E_{n,\mathrm{diel}} &= \frac{(q - q_1)\,l}{4\pi\varepsilon_0\varepsilon r^3} = \frac{q_2\,l}{4\pi\varepsilon_0\varepsilon r^3} \end{aligned}

Step 4The bound charge on the plane

Gauss’s theorem for $\vect E$ over a flat pillbox on the plane counts all the charge there, which is bound charge only, so $\sigma'$ is $\varepsilon_0$ times the jump of the normal field.

\sigma' = \varepsilon_0\left(E_{n,\mathrm{vac}} - E_{n,\mathrm{diel}}\right) = \frac{q_2\,l}{4\pi r^3}\left(1 - \frac{1}{\varepsilon}\right)

Substituting $q_2$ from (3):

\sigma' = \frac{(\varepsilon - 1)\,q\,l}{2\pi\varepsilon(\varepsilon + 1)\,r^3}

(4)

For $q > 0$ it is positive everywhere: the normal field is stronger on the vacuum side, so lines begin on the face.

Step 5The limit of a ball at the face

At any fixed point of the plane, $r$ tends to its distance $\rho$ from the ball’s foot as $l \to 0$ , while the numerator carries a factor $l$ .

\lim_{l\to 0}\sigma'(r) = 0 \quad \text{for every } r > 0

(5)

The total bound charge on the plane does not vanish, however. Sum over rings of area $2\pi\rho\,d\rho$ , with $r^2 = \rho^2 + l^2$ ; the radial integral comes first.

\begin{aligned} \int_0^\infty \frac{\rho\,d\rho}{(\rho^2 + l^2)^{3/2}} &= \left[-\frac{1}{(\rho^2 + l^2)^{1/2}}\right]_0^\infty = \frac{1}{l}\\ q'_{\mathrm{plane}} &= \int_0^\infty \frac{(\varepsilon - 1)\,q\,l}{2\pi\varepsilon(\varepsilon + 1)}\,\frac{2\pi\rho\,d\rho}{(\rho^2 + l^2)^{3/2}}\\ &= \frac{(\varepsilon - 1)\,q\,l}{\varepsilon(\varepsilon + 1)}\cdot\frac{1}{l} = \frac{(\varepsilon - 1)\,q}{\varepsilon(\varepsilon + 1)} \end{aligned}

It is the same for every $l$ , so as $l \to 0$ it is all squeezed into the point where the ball touches the plane.

Check

With $\varepsilon = 1$ the boxed $\sigma'$ is zero, as it must be with no dielectric. For the limit $l \to 0$ the ball lies on the plane, which is Problem 3.91: there the field is that of a charge $2q/(1+\varepsilon)$ in vacuum. Here the charges at the ball are $q$ , the bound charge $-(\varepsilon-1)q/\varepsilon$ that the dielectric places on the ball’s surface (Gauss’s theorem for $\vect D$ and $\vect E$ about the ball), and the plane’s $q'_{\mathrm{plane}}$ :

q - \frac{\varepsilon - 1}{\varepsilon}\,q + \frac{\varepsilon - 1}{\varepsilon(\varepsilon + 1)}\,q = \frac{(\varepsilon + 1) + (\varepsilon - 1)}{\varepsilon(\varepsilon + 1)}\,q = \frac{2q}{\varepsilon + 1}

which reproduces it.

Remark

The image $q_1$ has the same sign as $q$ , so the ball is pushed away from the face into the dielectric, with the force $q q_1/4\pi\varepsilon_0\varepsilon(2l)^2$ . A charge in a dielectric is repelled by a boundary with vacuum, while a charge in vacuum is attracted to a dielectric: in both cases it moves toward the medium of higher permittivity, where its field energy is lower.