Potential of a uniformyl charged sphere

Click For Summary
The discussion focuses on calculating the electric potential inside a uniformly charged sphere at a distance r from the center, where r is less than the sphere's radius R. The user attempts to derive the potential using the equation Va - Vb = ∫ E dr and identifies Vb as KQ/R. After integrating the electric field E inside the sphere, the user arrives at a potential expression that does not match the expected answer. The conversation emphasizes the need for simplification in the derived expression and suggests using algebraic manipulation to clarify the result. The user seeks assistance in resolving the discrepancy and simplifying their calculations.
ehabmozart
Messages
212
Reaction score
0

Homework Statement



The question asks to find the potential due to a uniformly charged sphere at r<R where R is the radius.


Homework Equations



Va-Vb= ∫ E dr from a → b

The Attempt at a Solution



My attempt was like this

Va is to be a the point of interest r.. Vb to be R... We know Vb to be KQ/R... Now E inside the sphere is KQr/R^3 .. After integration with respect to r we get (R^2-r^2)*Kq/2R^3 ... Thus the net potential at Va is Vb + the integral which is KQ/R + (R^2-r^2)*Kq/2R^3 ... However, this isn't the answer given.. Where was my mistake??
 
Physics news on Phys.org
How does it differ from the answer given?
Note that it is possible to simplify your expression.
 
mfb said:
How does it differ from the answer given?
Note that it is possible to simplify your expression.

I am not able to simplify.. Can you help me?
 
Just use (a-b)/c = a/c - b/c for (R^2-r^2)*Kq/(2R^3).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
968
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
888