The energy needed to construct a charged hollow spherical shell with finite thickness

Click For Summary
The discussion focuses on calculating the energy required to construct a charged hollow spherical shell with inner radius r1 and outer radius r2, given a charge density p. The user attempts to divide the sphere into infinitesimal shells and derives the charge for each shell as 4πR^2*p*dr. They encounter a challenge with the dr^2 term in their integral and express concern that their method does not account for previously created shells. The conversation emphasizes the need to correctly integrate the work done in adding a thin shell of charge q to the existing thick shell. A proper approach involves considering the cumulative effect of all shells during the integration process.
Avital
Messages
1
Reaction score
0
Homework Statement
I am trying to solve the relevant integral
Relevant Equations
I found that the energy needed to create a shell with radius R and charge Q is U=kQ^2/2R
So in the probleme is given a sphere with inner radius of r1 and outher radius r2 and elecric charge density p.

i tried to devide the sphere to shells with radius dr. And got that the charge for each shell is 4piR^2*p*dr
when plugin this expression to the work needed to create one shell i get dr^2.

so first of all how in general i can deal with dr^2 in integrals and secondly i guess that the way iam solving it is not taking into acount the shells created before so what is the right approach. Thanks.
 
Physics news on Phys.org
Consider the work done in adding a thin shell of charge q to an existing thick shell of charge density p.
 
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 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 14 ·
Replies
14
Views
13K
  • · Replies 5 ·
Replies
5
Views
4K