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

Electrostatic boundary condition

  • Thread starter wakko101
  • Start date
68
0
The question : Consider a thin spherical shell of radius R with a uniform charge density sigma. If a very small piece of this surface were removed, leaving a small hole, what would the electric field be at a point just above/below the hole?

Relevent info : the field due to the patch of surface (when intact) is the same as that of an infinite plane just below and above the surface (because if you're just above or below, the essentially flat surface looks like an infinite plane).

Does this mean that the field just above and below the surface is due solely to the patch itself, in which case the field would be 0? Or must one still take into account any contribution from all other points. Then again...if we're thinking about the points just above and below, would the leftover contributions be essentially perpendicular to the surface, in which case, it is 0 again? Did that even make sense?

This problem is confusing me, any hints/advice would be appreciated.

Cheers,
W. =)
 

Answers and Replies

979
1
Try thinking of the hole as a point charge of the opposite charge...
 
pam
455
1
This is an old classic problem.
When the surface is whole, the field on one side is E, the field on the other side is zero.
Using symmetry, tells you that E/2 is due to the small piece.
 

Related Threads for: Electrostatic boundary condition

  • Last Post
Replies
2
Views
6K
Replies
14
Views
967
Replies
6
Views
848
Replies
0
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
4
Views
658
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
Top