Irodov 3.165: Charge Where the Resistivity Jumps – Amit & Deepali Sinha Foundation

Problem Irodov 3.165

Two cylindrical conductors with the same cross-section but different resistivities, $\rho_1$ and $\rho_2$ , are joined end to end. A current $I$ flows through them from conductor 1 into conductor 2. What charge sits at the boundary between them?

The two cylinders joined end to end, resistivities ρ_1 and ρ_2, with the current I flowing from conductor 1 into conductor 2. — The two cylinders joined end to end, resistivities $\rho_1$ and $\rho_2$ , with the current $I$ flowing from conductor $1$ into conductor $2$ .

Given

resistivities $\rho_1$ , $\rho_2$
equal cross-sections
current $I$ from $1$ to $2$
permittivity of both conductors that of vacuum (assumed)

Find

the charge

q

at the boundary

Solution

Step 1The field on each side of the boundary

Call the common cross-section $S$ . The current is steady and the cross-sections are equal, so the current density $j = I/S$ is the same in both conductors and runs along the axis. By Ohm’s law in local form, $\vect E = \rho\vect j$ , the fields on the two sides differ:

\begin{gathered} E_1 = \frac{\rho_1 I}{S}\\ E_2 = \frac{\rho_2 I}{S} \end{gathered} \tag{1}

Step 2Gauss’s theorem across the boundary

A flat closed surface (dashed) straddling the boundary, drawn for ρ_2 > ρ_1: the field leaving through the face in conductor 2 is stronger than the field entering in conductor 1, so the boundary carries positive charge. — A flat closed surface (dashed) straddling the boundary, drawn for $\rho_2 > \rho_1$ : the field leaving through the face in conductor $2$ is stronger than the field entering in conductor $1$ , so the boundary carries positive charge.

Enclose the boundary in a flat closed surface whose two faces, of area $S$ , lie just inside conductor 1 and conductor 2. Both conductors have the permittivity of vacuum, as metals are taken to have, and the field is along the axis, so only the two faces carry flux. With (1):

\frac{q}{\varepsilon_0} = E_2 S - E_1 S = (\rho_2 - \rho_1)I

Solving for the charge:

q = \varepsilon_0(\rho_2 - \rho_1)I

(2)

For $\rho_2 > \rho_1$ the charge is positive; for $\rho_2 < \rho_1$ it is negative.

Check

Equal resistivities give $q = 0$ : one uniform conductor has no charge inside. Dimensions: $\varepsilon_0\rho$ is (F/m)( $\Omega\cdot$ m) = s, so $\varepsilon_0\rho I$ is a charge in coulombs. A perfect conductor 1, $\rho_1 = 0$ , has no field inside, and the charge becomes $\varepsilon_0 E_2 S$ by (1): the surface density $\varepsilon_0 E_2$ from which the field lines of conductor 2 start, the familiar result for the face of a metal.

Remark

The charge does not depend on the cross-section: it is fixed by the current and the two resistivities, since it is this charge that makes the field jump to match the change of resistivity. Every junction of different metals in a circuit carries such a charge, and so do the surfaces of a current-carrying wire, whose surface charge shapes the field that drives the current along the bends.