What Is the X-Component of the Electric Field in Coaxial Cylindrical Conductors?

In summary, the problem involves two coaxial cylindrical conductors, one with a radius of 2 cm and a total charge of +8nC, and the other with an inner radius of 6 cm, an outer radius of 7 cm, and a total charge of -16nC. The task is to calculate the x-component of the electric field at a point P located at the midpoint of the length of the cylinders, 4 cm from the origin and making an angle of 30 degrees with the x-axis. Using Gauss's law, the electric field is calculated to be 2.84245 e 11, and it travels radially outward. The surface charge and linear charge density do not need to be calculated,
  • #1
heartofaragorn
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0

Homework Statement


Two coaxial cylindrical conductors are shown. The inner cylinder has radius a = 2 cm, length 10 m, and carries a total charge of Q inner = +8nC. The outer cylinder has an inner radius b = 6 cm, outer radius c= 7 cm, length 10 m, and carries a total charge of Q outer = -16nC. What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r= 4 cm from the origin and makes an angle of 30 degrees with the x-axis?


Homework Equations


Surface charge density = 2pi * radius a * length
Linear charge density = 2pi * radius a * surface charge density
Electric field = 2 * k * linear charge denisty / r


The Attempt at a Solution


I drew a Gaussian sphere at the radius r = 4 according to where point P lies and tried to determine the charge inside, which I think may be where I went wrong. There is an enclosed cylinder of +8 nC that has a radius of 2 that falls completely within the Gaussian sphere; however, I cannot see that the outer charge affects this particular sphere since it is within the 2nd cylinder. I calculated the surface charge density to be 2.5133 using the radius of 0.04 m, then plugged that into the linear charge density formula to receive an answer of 0.63165. I then tried plugging that into the formula for the electric field given by my prof and received an answer of 2.84245 e 11. I took that and mulitplied by cos 30 degrees to account for the diagnol line upon which P lies. I cannot get the right answer (interactive online problem) and I'm getting the feeling that I'm either using the wrong charge, wrong radius, or wrong formulae...or a combination of all 3! Please help me! Thanks! :rolleyes:
 
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  • #2
So what direction does the electric field, from Gauss's law, travel in? I'm guessing this is where are getting tripped up.

Also, do you have a picture? I can't tell why they even give you an outer cylinder, something you mentioned also, maybe the problem creator is just trying to throw you off. Could you repost your work but in terms of symbols rather than numbers?
 
  • #3
I'll try attaching the image, but as for the symbols, I don't have anything that will display them. My computer is a little too old to have packages like Microsoft Equation or anything like that. If it's really a problem, I'll try seeing if I can download something. Also, I'm really not sure what direction the electric field travels in!
 

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  • #4
Your Gaussian cylinder at (radius r, say) totally encloses the inner conductor. What's Gauss's law? Set up the equation and solve for E. (No need to compute surface charge!)

The charge on the outer conductor is irrelevant. The field at r = 4 cm will be radially outward.

You can just type in your equations without using symbols, or you can use Latex. (See https://www.physicsforums.com/showthread.php?t=8997".)
 
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  • #5
The interactive help that comes with the problem stated that I had to calculate the surface charge in order to calculate the linear charge. Also, do I have the right formula for E? And if I don't need the surface charge, from where do I get the linear charge?
 
  • #6
heartofaragorn said:
The interactive help that comes with the problem stated that I had to calculate the surface charge in order to calculate the linear charge.
That makes no sense to me. Using the data given, you can calculate the linear charge density directly.
Also, do I have the right formula for E?
Yes. (Use Gauss's law to verify it, if you're unsure.)
And if I don't need the surface charge, from where do I get the linear charge?
You are given the total charge and the length. (Assume it's uniformly distributed.)
 
  • #7
YEA! It worked! Thanks so much!
 

Related to What Is the X-Component of the Electric Field in Coaxial Cylindrical Conductors?

1. What are coaxial cylindrical conductors?

Coaxial cylindrical conductors are two cylindrical conductors that share a common axis and are separated by an insulating material. The inner conductor carries the signal, while the outer conductor serves as a shield to reduce external interference.

2. How are coaxial cylindrical conductors used in science and technology?

Coaxial cylindrical conductors are commonly used in electronic circuits, communication systems, and transmission lines. They are also used in medical imaging devices such as MRI machines, and in high-frequency applications such as radio telescopes.

3. What are the advantages of using coaxial cylindrical conductors?

One major advantage of coaxial cylindrical conductors is their ability to reduce external interference, allowing for a more reliable and accurate transmission of signals. They also have a lower resistance compared to other types of conductors, resulting in less power loss.

4. What are the differences between coaxial cylindrical conductors and twisted pair cables?

Coaxial cylindrical conductors have a single inner conductor surrounded by an outer conductor, while twisted pair cables have two insulated conductors twisted together. Coaxial cylindrical conductors have a higher bandwidth and can transmit signals at longer distances, while twisted pair cables are more prone to interference.

5. How do you calculate the characteristic impedance of coaxial cylindrical conductors?

The characteristic impedance of coaxial cylindrical conductors can be calculated using the formula Zc = (138*log(b/a))/(√εr), where a is the radius of the inner conductor, b is the radius of the outer conductor, and εr is the dielectric constant of the insulating material between the conductors.

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