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

AI Thread Summary
Gauss' Law is essential for understanding the charge distribution in a coaxial cable, where the inner conductor and the inner surface of the outer conductor must have equal charge per unit length due to the constant electric field within the shell. To calculate the electric field at the surface of the inner wire, the relationship between electric field (E) and charge density (λ) can be utilized, specifically using the equation E = λ/(2π(ε0)r). The voltage between the conductors can be expressed in terms of the separation and λ, requiring integration of the electric field. The approach to finding the field between the conductors involves integrating the electric field equation with respect to the distance from the axis. Understanding these principles is crucial for solving the problem effectively.
KarlsBarkley
Messages
1
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
Last edited by a moderator:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top