Calculating Electric Potential for a Non-Negligible Thickness Toroid

Click For Summary
The discussion revolves around calculating the electric potential of a toroid with a non-negligible thickness. There is confusion regarding the linear charge density, specifically why it is expressed as λ = q/(2πR) despite the toroid's width of 2a. Participants highlight that the toroid's larger area implies a potentially lower linear charge density. The challenge of determining the charge distribution for a metallic toroid is acknowledged as a significant complexity in the calculations. Overall, the conversation emphasizes the need for a revised approach to account for the toroid's thickness in electric potential calculations.
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For A.1 of this problem,
1675405830601.png

The solution is
1675399497646.png

However, I have a doubt about the linear charge density ##\lambda##.

I don't understand how ##\lambda = \frac {q}{2\pi R} ## since this is not a thin ring, but has a non-negligible width of ##2a##

I think that the toroid has a larger area than thin circle with a circumference ##2\pi R## so linear charge density should be less than that expression.

EDIT: How would we calculate the electric potential if the thickness was not neglected?

Many thanks !

Problem from:
https://www.ipho2021.lt/uplfiles/Th2.pdf
https://www.ipho2021.lt/uplfiles/Th2-Solution.pdf
 

Attachments

  • 1675399479240.png
    1675399479240.png
    36.3 KB · Views: 109
Last edited by a moderator:
Physics news on Phys.org
1675414550671.png
 
  • Like
Likes member 731016
BvU said:
Thank you for your reply @BvU!

However, how would we calculate the electric potential if the thickness was not neglected?

Many thanks!
 
Callumnc1 said:
Thank you for your reply @BvU!

However, how would we calculate the electric potential if the thickness was not neglected?

Many thanks!
Since it is metallic, your first challenge would be to figure out the charge distribution. Good luck with that.
 
  • Like
Likes member 731016, nasu and BvU
haruspex said:
Since it is metallic, your first challenge would be to figure out the charge distribution. Good luck with that.
Thank you for your reply @haruspex! Yeah that seems hard!
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
64
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
2K