[Electromagnetic Fields] Electrical conductivity of a material

[MrPacane]

Homework Statement

Two inflnitely long, straight, parallel Wires are embedded in an infmite medium of conductivity 2 Sm^-1. The wires are identical, with a round cross-section of radius 0.25 cm. The centres of the wires are 1 cm apart. Find the conductance per metre between the wires.

A) 0.76 Sm^-1
B) 1.19 Sm^-1
C) 3.05 Sm^-1
D) 4.77 Sm^-1

Homework Equations

J = σE
∫D∙ds = ∫ρdv = Q or ∇∙D = ρ
D = εE

The Attempt at a Solution

I have no idea where to start from :( ...

 

[anathema]

Dear student,

Thank you for your question. I am happy to help you with your problem.

To solve this problem, we will use the equations you have listed. First, we need to find the electric field between the two wires. Since they are infinitely long and parallel, the electric field will be constant and directed towards each other. We can use the formula for the electric field between two parallel plates, which is E = V/d, where V is the potential difference and d is the distance between the plates.

In this case, the potential difference between the wires is the same, and we can calculate it using Ohm's law: V = IR, where I is the current and R is the resistance. We know that the resistance of each wire is given by R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area. Since the wires are identical, we can simplify this to R = ρL/πr^2, where r is the radius of the wire.

Now, we can substitute this into our equation for the potential difference: V = (IρL)/(πr^2). Since the wires are in an infinite medium, we can assume that the current is the same in both wires, and we can calculate it using Ohm's law again: I = V/R = (Vπr^2)/ρL.

Finally, we can substitute this into our equation for the electric field: E = V/d = [(Vπr^2)/ρL]/d = (Vπr^2)/(ρLd). This gives us the electric field between the wires.

Next, we can use the equation J = σE to calculate the current density between the wires. Since the wires are in an infinite medium, we can assume that the current density is the same throughout the medium. Substituting our calculated electric field and the given conductivity of 2 Sm^-1, we get J = (2 Sm^-1)(Vπr^2)/(ρLd).

Finally, we can calculate the conductance per metre between the wires using the equation G = I/V = J/σ. Substituting our calculated current density and the given conductivity, we get G = [(2 Sm^-1)(Vπr^2)/(ρLd)]/2 Sm^-1 = (Vπr^2)/(ρLd).