Electric field between two wires

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SUMMARY

The discussion focuses on calculating the electric field, potential difference, and capacitance per unit length of a transmission line consisting of two parallel conducting wires with charges +λ and -λ. The dielectric constant of the surrounding material is εr = 2.1. The net electric field between the wires is derived using the formula Et = (1/(4πε))*((Qa/r^2)+(Qb/(b-r)^2)). The potential difference is calculated using ΔV = -∫E dr = -E(b-a), and the capacitance per unit length is determined as Cl ≃ (2∏ε/ln(b/a)), yielding a numerical value of approximately 6.7*10^-11 Farads per unit length for b = 4 mm and a = 0.7 mm.

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depredador93
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Hello PF!

I've been having a hard time with this exercise, I did what I could but I highly doubt I got it right since I missed a few lessons :/. Anyway, here it is:

A transmission line consists of two parallel conducting wires of length
l and radius a separated by a distance b (center to center). The left wire is connected to the positive terminal of a generator acquiring a charge +λ [C/m]. Likewise, the right wire is connected
to the negative terminal acquiring a charge −λ [C/m]. The two wires are
completely embedded into a plastic material of dielectric constant εr = 2.1.

Perform the following calculations:

1. Calculate the net electric field for a point between the two wires whose
distance with respect the left wire is r, closer to the right wire.
Calculate the electric field inside each wire. Assume that the distance
between wires is large enough so they do not influence each other.

2. Calculate the potential difference between the two wires.

3. Calculate the capacity per unit of length of the transmission line. Show
that for b >> a this is Cl ≃ (∏ε/ln(b/a)).
Calculate the numerical value if b = 4 mm and a = 0.7 mm.
---------------------------------------------------------------------------------------------
1. What I tried to do in the first one is to calculate the total electric field created by the wires:
Et = Ea + Eb, being a the left wire and b the right one, using vectors of course. This is my result:
Et = (1/(4πε))*((Qa/r^2)+(Qb/(b-r)^2)). I doubt that it's that easy.

2. Here all I could come up with is ΔV = -∫E dr = -E(b-a).

3. I used C = (2∏εL)/(ln(b/a)) = (2∏*2.1*(8.85*10^-12)*L)/(ln(0.004/0.0007) = 6.7*10^-11 Farads per unit length. About the second part for for b >> a I am completely clueless.

Thanks a lot!
 
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