Irodov 3.75: A Dielectric Skin Screens the Conductor

Problem Irodov 3.75

A conductor carrying surface charge density $\sigma$ is in contact with a dielectric of permittivity $\varepsilon$ . Show that the bound charge on the dielectric’s surface where it touches the conductor has the surface density $\sigma' = -\sigma(\varepsilon - 1)/\varepsilon$ .

A small piece of the interface: the conductor below carries the surface charge σ, the dielectric of permittivity ε lies above it, and n is the normal out of the conductor (drawn for σ > 0). — A small piece of the interface: the conductor below carries the surface charge $\sigma$ , the dielectric of permittivity $\varepsilon$ lies above it, and $\vect n$ is the normal out of the conductor (drawn for $\sigma > 0$ ).

Given

surface charge density $\sigma$ on the conductor
permittivity $\varepsilon$ of the dielectric in contact

Find

show that the bound charge density on the interface is

\sigma' = -\sigma(\varepsilon - 1)/\varepsilon

Solution

Step 1The displacement just outside the conductor

Inside the conductor $\vect E = 0$ , so $\vect D = 0$ there too. The tangential component of $\vect E$ is continuous across any boundary and vanishes inside, so just outside the field, and with it $\vect D$ , is normal to the surface. Gauss’s theorem for $\vect D$ over a flat pillbox of face area $\Delta S$ straddling the surface counts only the free charge $\sigma\,\Delta S$ it encloses, and only the outer face carries flux:

D_n\,\Delta S = \sigma\,\Delta S

so with $\vect n$ the normal out of the conductor,

\vect D = \sigma\,\vect n \tag{1}

Step 2The polarization of the dielectric there

In a linear isotropic dielectric $\vect D = \varepsilon_0\varepsilon\vect E$ , so the polarization is the part of $\vect D$ beyond the vacuum term:

\vect P = \vect D - \varepsilon_0\vect E = \vect D - \frac{\vect D}{\varepsilon} = \frac{\varepsilon - 1}{\varepsilon}\,\vect D

Substituting (1):

\vect P = \frac{\varepsilon - 1}{\varepsilon}\,\sigma\,\vect n \tag{2}

Step 3The bound charge on the dielectric’s face

The bound surface charge of a dielectric is the normal component of $\vect P$ along the dielectric’s own outward normal. At the face that touches the conductor, the dielectric’s outward normal points into the conductor, that is along $-\vect n$ . From (2):

\sigma' = \vect P\cdot(-\vect n)

\sigma' = -\frac{\varepsilon - 1}{\varepsilon}\,\sigma

(3)

The bound charge is opposite in sign to the conductor’s charge and smaller in magnitude.

The polarization P points away from the conductor, and the bound charge σ' just inside the dielectric is opposite to σ and sparser (drawn for ε = 3, where |σ'| = 2σ/3). — The polarization $\vect P$ points away from the conductor, and the bound charge $\sigma'$ just inside the dielectric is opposite to $\sigma$ and sparser (drawn for $\varepsilon = 3$ , where $|\sigma'| = 2\sigma/3$ ).

Check

Gauss’s theorem for $\vect E$ , which counts free and bound charge together, gives the same result independently. The field just outside is $D/(\varepsilon_0\varepsilon)$ , so the pillbox sees the total charge

\sigma + \sigma' = \varepsilon_0 E_n = \frac{\sigma}{\varepsilon}

and the boxed $\sigma'$ satisfies it. For $\varepsilon = 1$ there is no bound charge, as there is no medium.

Remark

The two layers together carry $\sigma/\varepsilon$ : the dielectric screens all but a fraction $1/\varepsilon$ of the conductor’s charge, which is why the field next to the conductor drops by the factor $\varepsilon$ and why filling a capacitor raises its capacitance by the same factor. As $\varepsilon$ grows without bound, $\sigma'$ tends to $-\sigma$ and the screening becomes complete, as it would be for a second conductor.