Irodov 3.79: Flux and Circulation at a Boundary – Amit & Deepali Sinha Foundation

Problem Irodov 3.79

A uniform isotropic dielectric of permittivity $\varepsilon$ fills the half-space below a plane surface; above it is vacuum. Just above the surface the field is $\vect E_0$ , of magnitude $E_0$ , pointing out of the dielectric at an angle $\theta$ to the surface’s normal $\vect n$ . The field is uniform on each side of the surface. Find:

the flux of $\vect E$ through a sphere of radius $R$ whose center lies on the surface;
the circulation of $\vect D$ around a closed rectangular path $\Gamma$ that straddles the surface, its plane perpendicular to the surface and containing $\vect E_0$ : two sides of length $l$ run along the surface just above and just below it, the vacuum side traversed in the direction of the tangential component of $\vect E_0$ and the dielectric side in the opposite direction, and the two short sides crossing the surface are negligibly short.

The dielectric below the surface, the vacuum field E_0 at the angle θ from the normal n, the sphere of radius R centered on the surface, and the contour Γ with its long sides of length l along the surface. — The dielectric below the surface, the vacuum field $\vect E_0$ at the angle $\theta$ from the normal $\vect n$ , the sphere of radius $R$ centered on the surface, and the contour $\Gamma$ with its long sides of length $l$ along the surface.

Given

permittivity $\varepsilon$
vacuum field $E_0$ at angle $\theta$ to $\vect n$
uniform fields on both sides
sphere radius $R$ centered on the surface
contour $\Gamma$ , sides $l$ along the surface

Find

(a)

\oint\vect E\cdot d\vect S

over the sphere; (b)

\oint\vect D\cdot d\vect r

around

\Gamma

Solution

Step 1The field inside the dielectric

There is no free charge on the surface. The tangential component of $\vect E$ is continuous, and Gauss’s theorem for $\vect D$ over a flat pillbox makes the normal component of $\vect D$ continuous, with $D = \varepsilon_0 E_0$ in vacuum and $D = \varepsilon_0\varepsilon E$ in the dielectric:

E_t = E_0\sin\theta \tag{1}

\varepsilon E_n = E_0\cos\theta \tag{2}

Here $E_t$ and $E_n$ are the components of the field in the dielectric along the surface and along $\vect n$ .

Step 2(a) The flux of E through the sphere

The upper hemisphere lies in the uniform vacuum field. The flux of a uniform field through a closed surface is zero, so its flux through the hemisphere equals its flux through the flat disc of radius $R$ that closes it, the disc’s normal being $\vect n$ :

\Phi_{\mathrm{up}} = \pi R^2 E_0\cos\theta

The lower hemisphere lies in the uniform dielectric field, and the same argument, with the disc oriented along $-\vect n$ , the side the lower hemisphere faces, gives with (2):

\Phi_{\mathrm{down}} = -\pi R^2 E_n = -\frac{\pi R^2 E_0\cos\theta}{\varepsilon}

Adding the two halves:

\oint\vect E\cdot d\vect S = \frac{\varepsilon - 1}{\varepsilon}\,\pi R^2 E_0\cos\theta

(3)

It is positive for $\varepsilon > 1$ : more field lines leave the sphere through the top than enter through the bottom.

Step 3(b) The circulation of D around the contour

The short sides contribute nothing, so only the two long sides count. On the vacuum side the path runs along the tangential component of $\vect E_0$ , where $D_t = \varepsilon_0 E_0\sin\theta$ ; on the dielectric side it runs the opposite way, where, by (1), $D_t = \varepsilon_0\varepsilon E_0\sin\theta$ :

\oint_\Gamma\vect D\cdot d\vect r = \varepsilon_0 E_0\sin\theta\cdot l - \varepsilon_0\varepsilon E_0\sin\theta\cdot l

\oint_\Gamma\vect D\cdot d\vect r = -\varepsilon_0(\varepsilon - 1)\,l\,E_0\sin\theta

(4)

Traversing $\Gamma$ the other way round flips the sign.

Check

The flux in (a) must equal the bound charge inside the sphere over $\varepsilon_0$ , since there is no free charge. The bound surface density is the normal polarization of the dielectric at its surface, $\sigma' = \varepsilon_0(\varepsilon - 1)E_n$ , and the sphere cuts out a disc of area $\pi R^2$ :

\frac{\sigma'\pi R^2}{\varepsilon_0} = (\varepsilon - 1)\,\frac{E_0\cos\theta}{\varepsilon}\,\pi R^2

which is (3). For $\varepsilon = 1$ both answers vanish, as they must with no boundary.

Remark

The circulation of $\vect E$ around the same $\Gamma$ is zero, because $E_t$ is the same on both sides; that of $\vect D$ is not, because $D_t$ jumps by the factor $\varepsilon$ . So $\vect D$ is not a potential field even in electrostatics: the curl of $\vect D$ is the curl of $\vect P$ , which is concentrated on the surface wherever the tangential polarization jumps. The flux of $\vect D$ counts free charge only and the circulation of $\vect E$ is always zero; it is the flux of $\vect E$ and the circulation of $\vect D$ that feel the boundary.