Electric field between two wires

In summary, the conversation revolves around a transmission line consisting of two parallel conducting wires embedded in a plastic material. The calculations discussed include the net electric field between the wires, the potential difference between the wires, and the capacity per unit length of the transmission line. The individual electric fields of each wire are also calculated. The conversation ends with a request for help and a reminder of an upcoming physics test.
  • #1
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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  • #2
Bump? I really need help here! I got a physics test in 2 days.
 

1. What is an electric field?

An electric field is a physical quantity that describes the effect of electric forces on charged particles. It is a vector quantity, meaning it has both magnitude and direction.

2. How is the electric field between two wires calculated?

The electric field between two wires can be calculated using Coulomb's Law, which states that the electric field is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

3. What factors affect the strength of the electric field between two wires?

The strength of the electric field between two wires is affected by the magnitude of the charges on the wires, the distance between the wires, and the medium in which the wires are located.

4. Can the electric field between two wires be manipulated?

Yes, the electric field between two wires can be manipulated by changing the distance between the wires or by altering the charges on the wires. This can be done by using a power source or by changing the material of the wires.

5. What are some real-world applications of the electric field between two wires?

The electric field between two wires has many applications in technology, such as in electronic circuits, antennas, and electric motors. It is also used in medical devices, such as MRI machines, to create images of the human body.

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