Electric field due to non uniformly charged spherical shell

Click For Summary
To calculate the electric field due to a non-uniformly charged spherical shell with a volume charge density of ρ = -kr, one should integrate over the volume of the sphere to determine the total charge enclosed. Gauss's law can then be applied to find the electric field, taking into account the azimuthal symmetry of the problem. The approach involves treating E.dA as E(dA) due to this symmetry. This method simplifies the calculations and leads to an accurate determination of the electric field. Ultimately, integrating the charge density is crucial for applying Gauss's law effectively.
lavesh
Messages
2
Reaction score
0
The volume charge density of spherical shell varies as ρ=-kr.If we have to calculate electric filed using gauss's law, can we treat as E. dA as E(dA) as there is azimuthal symmitry



Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
You want to integrate over the volume of the sphere to find the total charge enclosed. Then from there it's simple to find the electric field using Gauss's law.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
623
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
44
Views
4K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K