Electric Field Uniformity in a Sphere with Cavity

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SUMMARY

The discussion centers on the electric field uniformity within a spherical cavity cut from a uniformly charged solid sphere. It is established that the electric field in the cavity is uniform and has a magnitude of kqd/r^3, where d is the distance between the centers of the spheres. The principle of superposition is applied to calculate the electric field by considering the contributions from both the solid sphere and the cavity. The conclusion confirms that the electric field vectors cancel out at points within the cavity, leading to a consistent field strength throughout.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with the principle of superposition in electrostatics
  • Knowledge of spherical coordinates and geometry
  • Basic calculus for evaluating electric field contributions
NEXT STEPS
  • Study the principle of superposition in electrostatics
  • Learn about electric field calculations for non-uniform charge distributions
  • Explore the concept of electric field lines and their implications
  • Investigate the effects of different geometries on electric field uniformity
USEFUL FOR

Students and educators in physics, particularly those focusing on electrostatics, as well as engineers and researchers working with electric fields in various applications.

phantom113
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Homework Statement


A charge q is uniformly distributed over the volume of a solid sphere of radius R. A spherical cavity is cut out of this solid sphere and the material and its charge are discarded. Show that the electric field in the cavity will then be uniform, of magnitude kqd/r^3, where d is the distance between the centers of the spheres.


Homework Equations


principle of superposition
E=kq/r^2


The Attempt at a Solution


I'm not really sure where to start with this one. I know you can take the electric fields of the two spheres and add them together, but the cavity has no charge enclosed. Maybe I'm just reading the problem incorrectly. A push in the right direction would be terrific.

Thanks.
 
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I noticed that it is easy to calculate the E field at one particular point - the center of the small sphere. The total E is the E due to the big sphere (with no hole) minus the E due to the small sphere of charge.

Perhaps you can extend the method to any point in the small sphere . . .
 
OK, thanks I got the equation to work(had to assume that one part of edge of the cavity was also the edge of the sphere to get the radius of the cavity). Is the reason that it's a uniform field that the field vectors at a random point not in the middle are in opposite directions and end up canceling to the same value as the field at the middle?
 

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