Basic Gaussian Surface Conceptual Problem

Click For Summary
SUMMARY

The discussion centers on the application of Gauss's Law to determine the charge induced on the surface of a hollow sphere within a solid copper cube. The user correctly identifies that the electric field (E field) must be zero within the conductor, leading to the conclusion that the total charge encapsulated by the Gaussian surface is zero. Consequently, a charge of -Q is induced on the surface of the void surrounding the charge Q at the center of the hollow sphere. This understanding is confirmed by other participants in the discussion.

PREREQUISITES
  • Understanding of Gauss's Law
  • Knowledge of electric fields in conductors
  • Familiarity with charge distribution in electrostatics
  • Concept of Gaussian surfaces
NEXT STEPS
  • Study the implications of Gauss's Law in different geometries
  • Learn about electrostatic shielding in conductors
  • Explore the behavior of electric fields in hollow conductors
  • Investigate the relationship between charge distribution and electric potential
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in electrostatics and the behavior of electric fields in conductive materials.

fysicsandphol
Messages
6
Reaction score
0
So I'm pretty sure I have the right answer I just want to make sure I am getting the idea.

There is a large solid copper cube. At the center of the cube there is a hollow sphere of radius a. At the center of the hollow sphere there is a charge Q.
What is the total charge induced on the surface of the void?

To solve it, I set up a gaussian surface enclosing the hollow sphere. Because copper is a conductor, (ignoring other forces), the E field must be zero everywhere within the solid copper, otherwise ions would move to balance the E field (right?). Therefore the closed integral of E dot dA will always be zero. This implies the charge encapsulated must be 0. This implies that there must be a total charge of -Q on the surface of the void. Is this right?
 
Physics news on Phys.org
Yes, that's correct.
 
thanks
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
421
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K