Irodov 3.162: Where Current Cannot Cross – Amit & Deepali Sinha Foundation

Problem Irodov 3.162

A conductor of resistivity $\rho$ is in contact with a dielectric of permittivity $\varepsilon$ . At a point $A$ of the conductor’s surface, the electric displacement in the dielectric has magnitude $D$ ; the vector $\vect D$ points away from the conductor at an angle $\alpha$ to the surface normal. Find the surface charge density on the conductor at $A$ , and the current density inside the conductor near $A$ .

The boundary between the conductor (resistivity ρ, below) and the dielectric (ε, above). At the point A the displacement D leaves the conductor at the angle α to the normal. — The boundary between the conductor (resistivity $\rho$ , below) and the dielectric ( $\varepsilon$ , above). At the point $A$ the displacement $\vect D$ leaves the conductor at the angle $\alpha$ to the normal.

Given

resistivity $\rho$
permittivity $\varepsilon$
$D$ at $A$
angle $\alpha$ to the normal
steady current (assumed)
dielectric does not conduct (assumed)

Find

the surface charge density

\sigma

A

and the current density

j

near

A

Solution

Step 1No current crosses the surface

The current is steady and the dielectric carries none, so no charge can pile up at the surface: the normal component of $\vect j$ in the conductor vanishes at the boundary. By Ohm’s law in local form, $\vect j = \vect E/\rho$ , so does the normal component of the field just inside the conductor:

E_{n,\mathrm{in}} = \rho\, j_n = 0 \tag{1}

Step 2The surface charge from the normal displacement

Gauss’s theorem for $\vect D$ on a flat pillbox straddling the surface at $A$ gives the free surface charge as the jump of the normal displacement. Inside the conductor the normal displacement $D_{n,\mathrm{in}}$ is proportional to $E_{n,\mathrm{in}}$ , so it is zero by (1); outside it is $D\cos\alpha$ :

\sigma = D_n - D_{n,\mathrm{in}}

\sigma = D\cos\alpha

(2)

Step 3The current from the tangential field

Just inside the conductor near A the field E and the current density j run along the surface, in the sense of the tangential part of D; the free charge at A is positive. — Just inside the conductor near $A$ the field $\vect E$ and the current density $\vect j$ run along the surface, in the sense of the tangential part of $\vect D$ ; the free charge at $A$ is positive.

The field is conservative, so its tangential component is continuous across the boundary. In the dielectric it is the tangential part of $\vect D$ divided by $\varepsilon_0\varepsilon$ , and by (1) it is the whole field just inside the conductor:

E_{\mathrm{in}} = E_t = \frac{D\sin\alpha}{\varepsilon_0\varepsilon}

Ohm’s law, $j = E_{\mathrm{in}}/\rho$ , then gives the current density, directed along the surface:

j = \frac{D\sin\alpha}{\varepsilon_0\varepsilon\rho}

(3)

Check

At $\alpha = 0$ the displacement is purely normal, so the tangential field vanishes and $j = 0$ : the surface is then an equipotential, and $\sigma = D$ is the electrostatic result for a conductor’s surface. At $\alpha = 90^\circ$ the displacement lies along the surface, $\sigma = 0$ and $j$ is largest. Dimensions: $D/(\varepsilon_0\varepsilon)$ is a field in V/m, and dividing by $\rho$ in $\Omega\cdot$ m gives A/m $^2$ .

Remark

The surface charge does two jobs at once: it ends the normal displacement of the dielectric and it bends the current lines inside the conductor so that they run along the surface. This is why the field outside a current-carrying wire is not normal to it, as it would be on a conductor at rest: its tangential part is the field that drives the current.