How Does Gauss' Law Apply to Coaxial Cable Charge Distribution?

In summary, the conversation discusses the use of a high voltage coaxial cable to supply power to an x-ray generator. The cable has an inner wire with a radius of 1 mm and an outer shield with a radius of 10 mm. Both have the same charge density per unit length, but opposite charges. Using Gauss' Law, it is explained that the inner conductor and inner surface of the outer conductor must have the same charge per unit length. The field at the surface of the inner wire is calculated using cylindrical symmetry. The symbolic form of the field between the two conductors is then written as a function of distance from the axis. The conversation also discusses solving for the field inside a conductor and using the voltage and distance to determine the
  • #1
KarlsBarkley
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Homework Statement



A high voltage coaxial cable is used to supply power to an x-ray generator. The cable consists of an inner wire of radius r=1 mm and a thin hollow outer conductor of radius R= 10 mm. The inner wire and the outer shield have the same charge density per unit length of , but opposite charges. Assume that the insulator between the conductors is air. The inner conductor is held at electric potential Vi and the outer conductor is grounded.

a) Use Gauss’ Law to explain why the inner conductor and the inner surface of the outer conductor must have the same charge per unit length.
b) Calculate the field at the surface of the inner wire. (Remember to use cylindrical symmetry.)
c) Write the symbolic form of the field between the two conductors as a function of distance from the axis.


Homework Equations


Eq #1: ε0E(2πrL)=λL
Eq #2: E=λ/(2π(ε0)r)



The Attempt at a Solution



A) I honestly, don't know how to use Gauss' Law to prove that the two surfaces have to have the same charge per unit length. However, I do know they have to be the same because the electric field is the constant throughout the inside of the shell. I just don't understand how to represent this mathematically. The only thing I could think of was to use Eq #1 because we know every variable except E and λ. But we do know that E1=E2 and therefore λL should be equal for both sides.

B) The only thing I could think of is to solve for E in terms of λ. I just don't know if that's a reasonable answer or if I should be able to derive λ from some obscure equation I haven't hunted down yet.

C) because I am so lost with the other two I have no idea how to even approach this question. My best guess is to take eq #2 and integrate it with respect to r

Any help would be amazing and much appreciated.
 
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  • #2
KarlsBarkley said:
A) I honestly, don't know how to use Gauss' Law to prove that the two surfaces have to have the same charge per unit length. However, I do know they have to be the same because the electric field is the constant throughout the inside of the shell. I just don't understand how to represent this mathematically. The only thing I could think of was to use Eq #1 because we know every variable except E and λ. But we do know that E1=E2 and therefore λL should be equal for both sides.

Consider that it is known that the field inside a conductor is zero. Suppose you were to draw a gaussian surface at a radius that puts it inside the material of the outer conductor? What's the net flux through the gaussian surface if the field must be zero? What does Gauss's law say about the charge contained within?

B) The only thing I could think of is to solve for E in terms of λ. I just don't know if that's a reasonable answer or if I should be able to derive λ from some obscure equation I haven't hunted down yet.
You're given information about the voltage between the conductors. You should be able to solve for that voltage in terms of the separation, and λ using the form of the field that you've written (an integration is required).

Since you know the voltage and the distances involved, you can determine λ and then revisit the equation for the field, with a value to plug in.

C) because I am so lost with the other two I have no idea how to even approach this question. My best guess is to take eq #2 and integrate it with respect to r

I think that once you've handled (b) you'll have (c) in hand.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html" that might help you.
 
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Related to How Does Gauss' Law Apply to Coaxial Cable Charge Distribution?

1. What is Gauss' Law and how does it relate to coaxial cables?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field to the charge enclosed by a closed surface. It applies to all types of electric fields, including the field inside a coaxial cable.

2. How does the electric field behave inside a coaxial cable?

The electric field inside a coaxial cable is constant and points radially outward from the inner conductor towards the outer conductor. This is due to the symmetry of the cable and the fact that the charges on the inner and outer conductor are equal and opposite.

3. How does the electric field outside a coaxial cable behave?

Outside the coaxial cable, the electric field lines are curved and point radially inward towards the cable. This is because the charges on the inner and outer conductor create an electric field that extends beyond the boundaries of the cable.

4. What is the relationship between charge and electric field strength in a coaxial cable?

According to Gauss' Law, the electric field strength inside a coaxial cable is directly proportional to the charge enclosed by the inner conductor. This means that as the charge increases, so does the electric field strength.

5. How is Gauss' Law used to calculate the electric field inside a coaxial cable?

Gauss' Law is used to calculate the electric field inside a coaxial cable by integrating the electric field over a closed surface surrounding the inner conductor. This allows us to determine the electric field strength at any point inside the cable.

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