Irodov 3.78: Field Bends Away Inside Glass – Amit & Deepali Sinha Foundation

Problem Irodov 3.78

A point $A$ lies on a curved boundary between glass (permittivity $\varepsilon = 6.0$ ) and vacuum. Just outside the glass near $A$ the electric field has magnitude $E_0 = 10.0$ V/m and makes an angle $\alpha_0 = 30^\circ$ with the normal $\vect n$ to the boundary at $A$ , which points from the glass into the vacuum. Find the field strength $E$ just inside the glass near $A$ , the angle $\alpha$ between $\vect E$ and $\vect n$ , and the surface density of the bound charge at $A$ .

The curved boundary between glass (below) and vacuum. At A the vacuum field E_0 makes the angle α_0 with the local normal n, which points out of the glass. — The curved boundary between glass (below) and vacuum. At $A$ the vacuum field $\vect E_0$ makes the angle $\alpha_0$ with the local normal $\vect n$ , which points out of the glass.

Given

$E_0 = 10.0$ V/m
$\alpha_0 = 30^\circ$
glass, $\varepsilon = 6.0$ (table value)

Find

E

and

\alpha

in the glass at

A

, and the bound charge density

\sigma'

there

Solution

Step 1Two matching conditions at the boundary

Only the immediate neighborhood of $A$ matters, so the curved boundary acts as its tangent plane there. The circulation of $\vect E$ around a thin rectangle straddling the boundary is zero, so the tangential component is continuous.

E\sin\alpha = E_0\sin\alpha_0 \tag{1}

There is no free charge on the boundary, so Gauss’s theorem for $\vect D$ over a flat pillbox makes the normal component of $\vect D$ continuous. The permittivity of glass is a tabulated material constant, $\varepsilon = 6.0$ . With $D = \varepsilon_0\varepsilon E$ in the glass and $D = \varepsilon_0 E_0$ in vacuum:

\varepsilon E\cos\alpha = E_0\cos\alpha_0 \tag{2}

Step 2The direction of the field in the glass

Dividing (1) by (2) eliminates both magnitudes.

\frac{E\sin\alpha}{\varepsilon E\cos\alpha} = \frac{E_0\sin\alpha_0}{E_0\cos\alpha_0}

Clearing $\varepsilon$ and evaluating with $\tan 30^\circ = 0.577$ :

\tan\alpha = 6.0\times 0.577 = 3.46

\tan\alpha = \varepsilon\tan\alpha_0, \quad \alpha \approx 74^\circ

(3)

The field in the glass leans much further from the normal than the field in vacuum does, almost along the boundary.

The field on both sides of A. In the glass E arrives at the larger angle α and is about half as long as E_0; the + just below A is the bound charge there. — The field on both sides of $A$ . In the glass $\vect E$ arrives at the larger angle $\alpha$ and is about half as long as $\vect E_0$ ; the $+$ just below $A$ is the bound charge there.

Step 3The magnitude of the field in the glass

Squaring (1) and (2) and adding them removes the angle $\alpha$ .

\begin{aligned} E^2\sin^2\alpha + E^2\cos^2\alpha &= E_0^2\sin^2\alpha_0 + \frac{E_0^2\cos^2\alpha_0}{\varepsilon^2}\\ E^2 &= \frac{E_0^2}{\varepsilon^2}\left(\cos^2\alpha_0 + \varepsilon^2\sin^2\alpha_0\right) \end{aligned}

Numerically, with $\cos^2\alpha_0 = 0.75$ and $\sin^2\alpha_0 = 0.25$ :

E = \frac{10.0}{6.0}\sqrt{0.75 + 36\times 0.25} = 1.667\times 3.12 \text{ V/m}

E = \frac{E_0}{\varepsilon}\sqrt{\cos^2\alpha_0 + \varepsilon^2\sin^2\alpha_0} \approx 5.2 \text{ V/m}

(4)

Step 4The bound charge at

A

Gauss’s theorem for $\vect E$ over the same pillbox counts all charge, free and bound. The only charge on the boundary is the bound charge $\sigma'$ , and the outward normals of the pillbox are $\vect n$ on the vacuum face and $-\vect n$ on the glass face.

\varepsilon_0\left(E_0\cos\alpha_0 - E\cos\alpha\right) = \sigma'

Substituting $E\cos\alpha$ from (2):

\sigma' = \varepsilon_0\left(E_0\cos\alpha_0 - \frac{E_0\cos\alpha_0}{\varepsilon}\right)

Numerically, $\sigma' = 8.85\times10^{-12}\times(5/6)\times 10.0\times 0.866$ C/m $^2$ :

\sigma' = \frac{\varepsilon_0(\varepsilon - 1)}{\varepsilon}E_0\cos\alpha_0 \approx 64 \text{ pC/m}^2

(5)

It is positive: the normal component of $\vect E$ jumps up on leaving the glass, so field lines start on the surface there.

Check

The two components of the glass field, rebuilt from the answers, reproduce both conditions, tangential on the first line and normal on the second.

\begin{aligned} 5.20\sin 73.9^\circ &= 5.00 \text{ V/m} = 10.0\sin 30^\circ \text{ V/m}\\ 5.20\cos 73.9^\circ &= 1.44 \text{ V/m} = \frac{10.0\cos 30^\circ}{6.0} \text{ V/m} \end{aligned}

With $\varepsilon = 1$ all three answers collapse to no boundary at all: $\alpha = \alpha_0$ , $E = E_0$ and $\sigma' = 0$ .

Remark

For a static field the boundary law is $\tan\alpha/\tan\alpha_0 = \varepsilon$ , and $\vect E$ bends away from the normal as it enters the medium of higher permittivity. Light crossing into glass bends toward the normal, under a sine law with the refractive index; the two laws describe different things, a static field line and a wavefront’s ray, and must not be confused. The surface charge is the polarization’s normal component, $\sigma' = P_n$ , with $P_n = \varepsilon_0(\varepsilon-1)E\cos\alpha$ in the glass.