• Support PF! Buy your school textbooks, materials and every day products Here!

Resistance between ball and bowl

  • Thread starter Piglet1024
  • Start date
  • #1
1. A metallic semi-spherical bowl is filled with salty water. A metallic ball is suspended at the center of the bowl, so that it is half submerged in the water. The outer radius of the ball is a, and the inner radius of the bowl is b. The conductivity of the water is [tex]\sigma[/tex]. Calculate the resistance between the ball and the bowl.



2. R=[tex]\frac{L}{\sigma A}[/tex]



3. So I thought that for A I could use the difference between the surface area of the bowl and the ball, but then I thought well what if I need to find the area of the cross sections because the original thought seemed too simplistic. L, I feel, is obviously (b-a) but I could be wrong. This is more of a conceptual question, but did I think of the correct method above or did I completely miss the point?
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,623
1,257


Your method won't work. That equation only applies to situations with a constant cross-sectional area, like a cylindrical wire. In this problem, you discovered that the meaning of A wasn't clear. When there's a question as to what the variables in an equation should equal, that's often a sign that you're not using the right equation.

What I would do is assume there's a current I flowing from the ball to the bowl. From that, you should be able to figure out the electric field E(r) and calculate the potential difference V between the ball and bowl. The resistance will then be V/I.
 

Related Threads on Resistance between ball and bowl

Replies
1
Views
3K
Replies
3
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
13
Views
8K
  • Last Post
Replies
3
Views
755
  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
5
Views
5K
Top