Irodov 3.76: Bound Charge Ignores the Shape – Amit & Deepali Sinha Foundation

Problem Irodov 3.76

A conductor of arbitrary shape carries a charge $q$ and is completely enclosed by a layer of uniform dielectric of permittivity $\varepsilon$ in contact with it; the layer’s outer surface, also of arbitrary shape, borders vacuum. Find the total bound charge on the inner surface of the dielectric and the total bound charge on its outer surface.

A conductor of arbitrary shape carrying the charge q, wholly surrounded by a dielectric of permittivity ε that touches it; outside the dielectric is vacuum. — A conductor of arbitrary shape carrying the charge $q$ , wholly surrounded by a dielectric of permittivity $\varepsilon$ that touches it; outside the dielectric is vacuum.

Given

charge $q$ on the conductor
uniform dielectric of permittivity $\varepsilon$ enclosing it, in contact

Find

the total bound charges

q'_{\mathrm{in}}

on the inner and

q'_{\mathrm{out}}

on the outer surface of the dielectric

Solution

Step 1Gauss’s theorem for D inside the dielectric

Take any closed surface $S$ that lies entirely inside the dielectric and encloses the conductor. The only free charge inside it is the conductor’s $q$ , so whatever the shapes:

\oint_S \vect D\cdot d\vect S = q \tag{1}

A closed surface S inside the dielectric, round the conductor. The bound charge is negative on the inner surface and positive on the outer one, equal in total. — A closed surface $S$ inside the dielectric, round the conductor. The bound charge is negative on the inner surface and positive on the outer one, equal in total.

Step 2No bound charge in the bulk

In the uniform dielectric $\vect P = \varepsilon_0(\varepsilon - 1)\vect E$ and $\vect D = \varepsilon_0\varepsilon\vect E$ , so the polarization is a fixed fraction of $\vect D$ :

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

The bulk bound charge density is $-\nabla\cdot\vect P$ . With $\varepsilon$ uniform and no free charge inside the dielectric, $\nabla\cdot\vect D = 0$ there:

\rho' = -\nabla\cdot\vect P = -\frac{\varepsilon - 1}{\varepsilon}\,\nabla\cdot\vect D = 0

So between the conductor and $S$ the only bound charge is $q'_{\mathrm{in}}$ , on the inner surface.

Step 3The bound charge on the inner surface

The bound charge enclosed by $S$ is minus the flux of $\vect P$ through it. By (2) and (1):

q'_{\mathrm{in}} = -\oint_S \vect P\cdot d\vect S = -\frac{\varepsilon - 1}{\varepsilon}\oint_S \vect D\cdot d\vect S

q'_{\mathrm{in}} = -\frac{\varepsilon - 1}{\varepsilon}\,q

(3)

Step 4The bound charge on the outer surface

Bound charge is charge displaced inside neutral molecules, so the dielectric as a whole stays neutral. With no bulk bound charge, the two surface charges cancel:

q'_{\mathrm{out}} = -q'_{\mathrm{in}}

q'_{\mathrm{out}} = +\frac{\varepsilon - 1}{\varepsilon}\,q

(4)

Check

Gauss’s theorem for $\vect E$ over the same $S$ counts free and bound charge together. With $\vect E = \vect D/(\varepsilon_0\varepsilon)$ inside the dielectric and (1), the enclosed total must be $q/\varepsilon$ , and (3) gives exactly that:

q + q'_{\mathrm{in}} = q - \frac{\varepsilon - 1}{\varepsilon}\,q = \frac{q}{\varepsilon}

For $\varepsilon = 1$ both bound charges vanish, as they must with no medium.

Remark

The answer depends on nothing but $q$ and $\varepsilon$ , however the conductor and the layer are shaped: the local density $\sigma' = -\sigma(\varepsilon - 1)/\varepsilon$ at a conductor-dielectric interface, summed over the conductor’s surface, gives $q'_{\mathrm{in}}$ directly. Outside the dielectric the total charge is $q + q'_{\mathrm{in}} + q'_{\mathrm{out}} = q$ , but the field there is not that of $q$ alone unless the conductor and the layer are concentric spheres: $q'_{\mathrm{out}}$ is spread over the outer surface in a way the shapes decide.