1. Identify the Goal:
Calculate the excess charge accumulated at the interface of two sections of a strip-line conductor with different resistivities (ρ_A and ρ_B) carrying a constant current I.

2. Current Density & Electric Field:
Since the cross-section A is uniform, the current density J is constant throughout:

J = I / A

Using Ohm’s law (E = ρJ), the electric fields in the two sections are:

• Section A: E_A = ρ_A (I / A)
• Section B: E_B = ρ_B (I / A)

3. Applying Gauss’s Law:
Construct a Gaussian surface (pillbox) enclosing the interface. The net electric flux Φ leaving the interface is:

Φ = (E_B – E_A) · A

According to Gauss’s Law, the enclosed charge Q is ε₀Φ.

4. Calculation:
Substitute the expressions for electric field:

Q = ε₀ [ (ρ_B I / A) – (ρ_A I / A) ] · A
Q = ε₀ I (ρ_B – ρ_A)