30: Problem 30.24

INTRO:
The two segments of the wire in the figure (Figure 1) have equal diameters but different conductivities σ₁ and σ₂. Current I passes through this wire.

Figure 1

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PART A:
If the conductivities have the ratio σ₂/σ₁=5, what is the ratio E₂/E₁ of the electric field strengths in the two segments of the wire?

- 1/25
- 5
- 25
- 1/5

SOLUTION:
Since the current I passes through the entire wire, and the diameters are equal... the current density for both sides is... 
J = I/A

This means that J₁ = J₂

using eq. (30.17) ... J = σE ...

J₁ = σ₁E₁
J₂ = σ₂E₂​

since J₁ = J₂... σ₁E₁ = σ₂E₂

by rearranging...
σ₂/σ₁ = E₁/E₂ = 5 (given)​

E₂/E₁ = [E₁/E₂]^-1

⇒ E₂/E₁ = 1/5​