Capacitance of three concentric shells

  • Thread starter Thread starter Kashmir
  • Start date Start date
  • Tags Tags
    Capacitance
Click For Summary
To calculate the capacitance between three concentric shells, charge Q is applied to the outer shell and -Q to the innermost shell. The potential V is determined by integrating the electric field, resulting in V=kq[1/a -1/c]. The capacitance C is then calculated as C=Q/V, and this method is confirmed to be correct as the middle shell acts as an equipotential surface, not affecting the capacitance. An alternative scenario is proposed, involving a thick concentric shell with a specific wall thickness, which could provide a more complex challenge. Overall, the approach to finding capacitance remains valid without the need for integration.
Kashmir
Messages
466
Reaction score
74
Homework Statement
A friend of mine sent me this problem about finding the capacitance.
We have three concentric shells of radius a, b, c. And we've to find the capacitance between x and y.

I need help.

Thank you
Relevant Equations
C=Q/V
IMG-20230727-WA0003.jpg


I want to calculate the capacitance of this system between the points x&y.
So suppose I give a charge Q to the outermost shell and -Q to the innermost shell. To find the capacitance C, I try to find the potential V between the outermost shell and innermost shell .
To find V ,I integrate the electric field and find it out to be V=kq[1/a -1/c] where k=1/(4 *pi*epsilon). Then I can find C =Q/V.
Is this approach correct.
 
Last edited:
Physics news on Phys.org
This approach is correct. You should get the capacitance between the outer and inner shells the usual way as if the middle shell were not there. The middle shell is affecting nothing because it is an equipotential surface at the potential that would be there if there were vacuum between the shells.

A slightly more interesting problem might be to find the capacitance if a thick concentric shell were placed in between having wall thickness, say ##d=\frac{1}{2}(c-a)## with equal vacuum gaps between conductors.
 
  • Like
Likes Kashmir and berkeman
kuruman said:
This approach is correct. You should get the capacitance between the outer and inner shells the usual way as if the middle shell were not there. The middle shell is affecting nothing because it is an equipotential surface at the potential that would be there if there were vacuum between the shells.

A slightly more interesting problem might be to find the capacitance if a thick concentric shell were placed in between having wall thickness, say ##d=\frac{1}{2}(c-a)## with equal vacuum gaps between conductors.
Thank you. I’ll try your question :)
 
You don't need to do any integration. Just write down the potential at radius a due to the two shells. You know that the potential due to the outer shell there is the same as it is at radius c.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 82 ·
3
Replies
82
Views
16K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K