Irodov 3.91: One Field Shared by Two Media – Amit & Deepali Sinha Foundation

Problem Irodov 3.91

An infinite plane separates vacuum from a uniform isotropic dielectric of permittivity $\varepsilon$ that fills the other half of space, and a point charge $q$ sits on this plane. Find the magnitudes of $\vect D$ and $\vect E$ , and the potential $\varphi$ , as functions of the distance $r$ from the charge.

The charge q on the plane boundary, with vacuum above and the dielectric below; r is the distance from q to a field point. Drawn for q > 0. — The charge $q$ on the plane boundary, with vacuum above and the dielectric below; $r$ is the distance from $q$ to a field point. Drawn for $q > 0$ .

Given

$q$ on the boundary plane
vacuum on one side
dielectric of permittivity $\varepsilon$ on the other

Find

D(r)

E(r)

and

\varphi(r)

Solution

Step 1A radial field fits both media

Try a field that is radial from $q$ and has the same magnitude $E(r)$ in both halves. On the boundary such a field runs along the plane, so it is purely tangential there, and tangential $E$ is continuous across the boundary exactly when the magnitudes on the two sides agree.

E_1(r) = E_2(r) = E(r) \tag{1}

The normal component of $\vect D$ vanishes on both sides, so its continuity holds trivially. In each half a radial $1/r^2$ field has zero curl, and $\vect D = \varepsilon_0\varepsilon\vect E$ has zero divergence away from $q$ , so the trial field meets every condition and, by uniqueness, it is the field.

Step 2Gauss’s theorem for

\vect D

fixes the magnitude

Take a sphere of radius $r$ about $q$ . The plane cuts it into two hemispheres, one in each medium, each of area $2\pi r^2$ , and only the free charge $q$ counts.

\oint \vect D\cdot d\vect S = \varepsilon_0 E\cdot 2\pi r^2 + \varepsilon_0\varepsilon E\cdot 2\pi r^2 = q

Collecting the two hemispheres:

2\pi\varepsilon_0(1+\varepsilon)r^2 E = q

E = \frac{q}{2\pi\varepsilon_0(1+\varepsilon)r^2}

(2)

By (1) this holds in the vacuum and in the dielectric alike.

Step 3The displacement differs between the media

In vacuum $D = \varepsilon_0 E$ , in the dielectric $D = \varepsilon_0\varepsilon E$ . Substituting (2):

\begin{gathered} D_1 = \frac{q}{2\pi(1+\varepsilon)r^2} \ \ \text{(vacuum)}\\ D_2 = \frac{\varepsilon q}{2\pi(1+\varepsilon)r^2} \ \ \text{(dielectric)} \end{gathered}

(3)

The field lines leave q at equal angles in both media, so E is the same on both sides; D_2 in the dielectric is ε times D_1 in vacuum. Drawn for q > 0 and ε = 2. — The field lines leave $q$ at equal angles in both media, so $E$ is the same on both sides; $\vect D_2$ in the dielectric is $\varepsilon$ times $\vect D_1$ in vacuum. Drawn for $q > 0$ and $\varepsilon = 2$ .

Step 4The potential along a radial line

A radial line from $q$ to infinity stays inside one medium, and the field along it is (2). With $\varphi = 0$ at infinity:

\varphi(r) = \int_r^\infty E\,dr' = \frac{q}{2\pi\varepsilon_0(1+\varepsilon)}\int_r^\infty \frac{dr'}{r'^2}

\varphi = \frac{q}{2\pi\varepsilon_0(1+\varepsilon)r}

(4)

The potential is the same in both media, as it must be with the field the same.

Check

With $\varepsilon = 1$ there is no dielectric, and the boxed field becomes Coulomb’s law,

E = \frac{q}{2\pi\varepsilon_0\cdot 2\, r^2} = \frac{q}{4\pi\varepsilon_0 r^2},

while $D_1 = D_2 = q/4\pi r^2$ . The total flux of $\vect D$ is also right: $(D_1 + D_2)\cdot 2\pi r^2 = q$ for the boxed $D_1$ and $D_2$ .

Remark

The field is that of a charge $2q/(1+\varepsilon)$ in vacuum, so the dielectric adds a bound charge $-(\varepsilon-1)q/(\varepsilon+1)$ to $q$ . None of it lies on the plane: $\vect P = \varepsilon_0(\varepsilon-1)\vect E$ is radial, hence parallel to the boundary, so $P_n$ and with it $\sigma'$ vanish there, and $\nabla\cdot\vect P$ vanishes in the bulk. The whole bound charge sits at the charge itself, in the half of its surroundings that is dielectric.