Two materials, for example silver and copper, of same cross section can be welded into a composite wire. When a voltage is applied across this wire, a stable current i whose value is determined by Ohm¡¦s Law is formed. Since the wire has uniform cross section, the current densities in two different materials have to be the same. But their conductivities£m1 and£m2 of the two conductor types are different and thus the electric fields within each of the two conductors are different. According to Gauss¡¦s Law, the change from one electric field to another has to be generated by some net charge accumulated at their interface. Show that this amount of charge is equal to
i/(4£k) * [1/£m2- 1/£m1]
assuming£m2>£m1
To show that the amount of charge at the interface between the two materials is equal to i/(4πε) * [1/σ2 - 1/σ1], we need to apply Gauss's Law and rely on some principles of electrostatics.
Let's start by considering a small section of the composite wire where the two materials meet. We'll assume that the cross-sectional area of this interface is A.
According to Ohm's Law, the current i flowing through the wire is determined by the voltage applied across it and its total resistance R. In this case, since the wire is made up of two different materials, we can express the resistance as R = ∆x/(Aσ), where ∆x is the thickness of the interface and σ is the conductivity of the material.
Now, let's evaluate the electric fields within each of the two conductors. We'll assume that the electric field in material 1 is E1 and in material 2 is E2.
For material 1:
Ohm's Law states that E1 = σ1 * J1, where J1 is the current density in material 1.
For material 2:
Similarly, E2 = σ2 * J2, where J2 is the current density in material 2.
Since the wire has a uniform cross section, the current densities J1 and J2 must be the same. Therefore, we can say that J1 = J2 = i/A.
Next, using Gauss's Law, we know that the net electric field across a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε). In this case, the closed surface we're considering is at the interface between the two materials.
The charge enclosed at the interface is given by Q = (∆x/A) * i.
Note: (∆x/A) represents the volume of the interface multiplied by the charge density, which is equal to the current i passing through the interface.
Therefore, the net electric field across the interface is E = Q / (Aε) = (∆x/A) * i / (Aε) = (∆x / A²ε) * i.
We can equate this electric field to the difference in electric fields between the two materials:
(∆x / A²ε) * i = E2 - E1
Substituting the expressions for E1 and E2, we get:
(∆x / A²ε) * i = σ2 * J2 - σ1 * J1
Since J1 = J2 = i/A, we have:
(∆x / A²ε) * i = σ2 * (i/A) - σ1 * (i/A)
Simplifying further:
(∆x / A²ε) = (σ2 / A) - (σ1 / A)
(∆x / Aε) = σ2 - σ1
Finally, multiplying both sides by A and rearranging, we get:
∆x / ε = A * (σ2 - σ1)
Recognizing that ∆x / ε is the amount of charge on the interface, we have:
Q = A * (σ2 - σ1)
Therefore, the amount of charge at the interface is equal to Q = A * (σ2 - σ1), which can be rewritten as Q = i/(4πε) * [1/σ2 - 1/σ1] under the assumption σ2 > σ1.
Hence, the amount of charge at the interface between the two materials is equal to i/(4πε) * [1/σ2 - 1/σ1].