How Do You Calculate Current Density Between Coaxial Cylinders?

In summary, we are given two long, coaxial metal cylinders with a separation of a material of conductivity sigma and dielectric constant epsilon. The inner cylinder has a radius of a, the outer cylinder has a radius of b, and both have a length of L. The inner conductor is held at a potential Vo with respect to the outer one. We are asked to find the current density J(r) between the two cylinders. Two methods are attempted, with the first one yielding the equation J(r) = V*sigma/(r*ln(b/a)) and the second one yielding the equation J(r) = V*sigma/(r*ln(b/a)), which are both considered correct.
  • #1
RyanP
17
0

Homework Statement


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Two long, coaxial metal cylinders are separated by a material of conductivity sigma and dielectric constant epsilon. The radius of the inner cylinder is a, the radius of outer cylinder is b, and the length of both is L.

Suppose that the inner conductor is held at a potential Vo with respect to the outer one. What is the current density J(r) between the two?

Homework Equations


J(r) = I/A = sigma*E
C=Q/V
R=s/(A*sigma)

The Attempt at a Solution


First attempt:
I found resistance of the dielectric to be ln(b/a)/(2pi*L*sigma). As a function of radius r, R(r) = ln(r/a)/(2pi*L*sigma).

Current = V/R = Vo / (ln(b/a)/(2piLsigma)) = Vo*2pi*L*sigma / ln(r/a) as a function of r.
Then divide by Area=2pi*L*r to get current density J(r) = Vo*sigma / (r*ln(r/a)). Second attempt:
Vo = Q/C. I found capacitance of the configuration to be 2pi*epsilon_o*L*epsilon / ln(b/a) using the expression for resistance. Then Q = Vo*2pi*L*epsilon*epsilon_o / ln(b/a).
EA = Q/epsilon_o, so E = Vo*epsilon / (r*ln(b/a)). Using J(r) = E*sigma, I get J(r) = sigma*Vo*epsilon / rln(b/a).

The difference in the two is the presence of epsilon (dielectric constant) in the numerator, and b rather than r in the ln term of the denominator. Are either of these approaches correct?
 
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  • #2
Can bound charges support current flow?
Better stick with method 1.
 
  • #3
RyanP said:
Current = V/R = Vo / (ln(b/a)/(2piLsigma))
Looks reasonable (but I've not checked it in detail).
RyanP said:
= Vo*2pi*L*sigma / ln(r/a) as a function of r.
That I don't understand. The same total current has to flow through any given radius, no?
 
  • #4
Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?
 
  • #5
RyanP said:
Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?
Sounds right, looks right ... I'm buying!
 
  • #6
RyanP said:
Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?
Looks right to me too.
 

1. What are coaxial metal cylinders?

Coaxial metal cylinders are two cylindrical metal tubes that are positioned one inside the other, with a common axis. They are often used for conducting electricity or transmitting signals.

2. What are the advantages of using coaxial metal cylinders?

Coaxial metal cylinders provide better signal transmission and noise reduction compared to other types of cables. They also have a longer lifespan and are less susceptible to interference.

3. How do you connect coaxial metal cylinders?

To connect coaxial metal cylinders, the inner conductor of one cylinder is inserted into the outer conductor of the other cylinder. This creates a tight connection that allows for efficient signal transmission.

4. What are the common applications of coaxial metal cylinders?

Coaxial metal cylinders are commonly used in telecommunication systems, cable television, and internet connections. They are also used in medical equipment, radar systems, and military applications.

5. Are there different types of coaxial metal cylinders?

Yes, there are different types of coaxial metal cylinders based on their size, impedance, and insulation materials. Some common types include RG-6, RG-11, and RG-58, which are used for different applications and have varying levels of signal loss.

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