Question

Show that the total amount of charge at the junction of the two materials in figure is $\epsilon_0 I (\frac{1}{\sigma_2} - \frac{1}{\sigma_1})$, where $I$ is the current flowing through the junction, and $\sigma_1$ and $\sigma_2$ are the conductivities for the two materials.

Answer 1

Q = \epsilon_0 I \left(\frac{1}{\sigma_2} - \frac{1}{\sigma_1}\right)

Answer 2

1. Step 1: Express Electric Field in Each Material

Since the current $I$ is continuous and the cross-sectional area $A$ is the same, the current density is

$$J = \frac{I}{A}.$$

By Ohm’s law in differential form, $J = \sigma E$, so the electric fields are

$$E_1 = \frac{I}{A\,\sigma_1} \quad \text{and} \quad E_2 = \frac{I}{A\,\sigma_2}.$$

2. Step 2: Find the Surface Charge Density

The discontinuity of the perpendicular component of $\mathbf{E}$ at the junction gives rise to a surface charge density $\sigma_s$ via Gauss’s law:

$$\sigma_s = \epsilon_0 \left(E_2 - E_1\right).$$

Substituting the values of $E_1$ and $E_2$:

$$\sigma_s = \epsilon_0 \left(\frac{I}{A\,\sigma_2} - \frac{I}{A\,\sigma_1}\right) = \epsilon_0 \frac{I}{A}\left(\frac{1}{\sigma_2} - \frac{1}{\sigma_1}\right).$$

3. Step 3: Total Charge at the Junction

The total charge $Q$ stored at the junction is the charge density multiplied by the area $A$:

$$Q = \sigma_s \, A = \epsilon_0 I \left(\frac{1}{\sigma_2} - \frac{1}{\sigma_1}\right).$$

Thus, the total amount of charge at the junction is

$$Q = \epsilon_0 I \left(\frac{1}{\sigma_2} - \frac{1}{\sigma_1}\right).$$