Capacitance of three coaxial metal tubes

Click For Summary
SUMMARY

The discussion focuses on calculating the capacitance per unit length of three coaxial metal tubes with radii a, b, and c, where a wire connects the innermost and outermost tubes. Participants clarify that the two tubes act as equipotentials due to the conducting wire, which influences charge distribution. The potential difference between the outermost and innermost tubes is derived from the potentials of the intermediate tube, leading to a simplified calculation of capacitance. The key takeaway is that the capacitance can be determined by analyzing the potential differences between the tubes.

PREREQUISITES
  • Understanding of electrostatics and capacitance concepts
  • Familiarity with coaxial cable geometry
  • Knowledge of potential difference and equipotential surfaces
  • Basic proficiency in solving electrostatic problems
NEXT STEPS
  • Study the derivation of capacitance for coaxial cylinders
  • Learn about the concept of equipotential surfaces in electrostatics
  • Explore the relationship between charge distribution and potential difference
  • Investigate the application of Gauss's Law in cylindrical coordinates
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone studying electrostatics or capacitance in cylindrical geometries.

boardbox
Messages
16
Reaction score
0

Homework Statement



Find the capacitance per unit length of three long coaxial metal tubes, with radii
a < b < c . A wire connects the innermost and outermost tubes (radii a and c).

Homework Equations





The Attempt at a Solution



I'm a little confused as to how I should set this up. What confused me is the wire that runs between the innermost and outermost tube. My thought is that it just makes the two tubes one big conductor. If that's the case and I put some charge on the conductor, how does it get distributed? Could I just put a charge on it and say the charge on the innermost tube is the area of the innertube over the area of the whole conductor times the charge?
 
Physics news on Phys.org
boardbox said:
Could I just put a charge on it and say the charge on the innermost tube is the area of the innertube over the area of the whole conductor times the charge?

I don't think you can assume that. However, if the two cylinders are connected by a conducting wire, you can say that they are equipotentials.
 
Alright well let me think this out for a second.

What I'm after is the potential from c to a, after that the problem is simple. That's going to be V(c) - V(a). Well V changes, so taking two steps V(c) - V(b) + V(b) - V(a). If V(c) = V(a) then I get zero potential. That seems a little silly to me, also means I'm dividing by zero in the next step, so I don't think that's right.

Here's another thought and this strikes me as a bit less silly. Say I have some charge on the innermost cylinder and some other charge on the middle cylinder. I could find the potential difference between the two. Now I know the outermost cylinder has the same potential at the innermost one, which means that the potential difference between it and the middle cylinder is the same as the difference between the middle cylinder and inner cylinder. So the potential of the set is just two times the potential between the inner and middle cylinder.
 
boardbox said:
What I'm after is the potential from c to a, after that the problem is simple.

Why do you say that? What is the applicable definition of capacitance here?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
6K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K