Modification to metal sphere problem

  • Thread starter Thread starter cherrios
  • Start date Start date
  • Tags Tags
    Metal Sphere
Click For Summary
SUMMARY

The discussion centers on calculating the electric field between a metal sphere and its surrounding conducting shell. The metal sphere has a radius of x and carries a charge of 1 micro Coulomb, while the conducting shell has an inner radius of y and an outer radius of Z, with a total charge of 4 micro Coulombs. By applying Gauss's law, the electric field E can be determined using the formula E(r) = q_{in}/(4πε₀r²), where q_{in} is the charge on the inner sphere. The surface charge density on the inner and outer surfaces of the conducting shell can also be derived from these calculations.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Knowledge of electric field concepts and calculations
  • Familiarity with spherical symmetry in charge distributions
  • Basic principles of charge density and its implications
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Explore the concept of electric field due to spherical charge distributions
  • Learn about surface charge density calculations for conductors
  • Investigate the effects of induction in electrostatic problems
USEFUL FOR

Physics students, electrical engineers, and anyone studying electrostatics or conducting materials will benefit from this discussion.

cherrios
Messages
8
Reaction score
0
This problem is somewhat similar to the one I had posted yesterday.

There is a metal spehre, radius=x, that it surrounded by a conducting shell (also spherical) that has an inner radius=y and outer radius=Z. the total charge is 4 micro Coloumbs--> 1 micro Coloumb on the inner sphere, and the rest is distributed in the shell

1)Find electric field between the outer surface of the metal sphere and the inner radius of the conducting shell.

Would I need to take a Gaussian surface between the outer surface of the metal sphere and inner radius of the conducting shell? And also, how would I find the surface charge density on the inner and outer surfaces of the conducting shell?
 
Physics news on Phys.org
Assuming that you have no induction and uniform chage density, the outer shell will not contribute to the electric field, since it has no charge between it and inner shell. You simply have [tex]q_{in} = 1\mu C[/tex], and by Gauss's law for a spherical shell
[tex]\int \vec{E}.d\vec{A} = \int EdA = E \int dA = E(4 \pi r^2) = \frac{q_{in}}{\epsilon_0}[/tex], which will yield E(r).
 

Similar threads

Replies
5
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
5
Views
4K
Replies
23
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K