Conducting sphere with chrage in centre

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SUMMARY

The discussion centers on calculating the electric field, E, of a conducting sphere with a cavity containing a point charge, +Q, at its center. The conducting sphere has a radius R1, while the cavity has a radius R2, where R1 > R2. The participants clarify that while the electric field inside the cavity and outside the sphere exhibit similar radial dependence, they are not identical. The application of Gauss's Law is essential to derive the electric field accurately, rather than relying solely on logical assumptions.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric field concepts and properties of conductors
  • Knowledge of spherical symmetry in electric fields
  • Basic principles of charge induction in conductors
NEXT STEPS
  • Study the application of Gauss's Law in various geometries
  • Learn about electric field calculations for spherical conductors
  • Explore the concept of charge induction and its effects on electric fields
  • Investigate the differences between electric fields inside cavities and outside conductors
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This discussion is beneficial for physics students, educators, and anyone studying electrostatics, particularly those focusing on electric fields in conductive materials and spherical geometries.

Taylor_1989
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Homework Statement


Consider a conducting sphere of radius ##R_1##. Inside it there is a cavity, spherical in shape, with it origin at the same point as the conducting sphere and of radius ##R_2## ##(R_1>R_2)##. Assume there is a point charge equal to ##+Q## at the origin.

Use Gauss law to calculate the value of the electric field, E

Homework Equations

The Attempt at a Solution



I have no working because I am looking at this from a logical perspective and just want to know if my logic is correct.

my assumption is that this sphere with cavity is electrical neutral, so with the charge in the center being positive, then this will induce negative charge onto the inner shell, so then the electric field on the outside the shell will be the same, as the electric field insie the cavity and the, region between the outter surface and the cavity will be 0

Is this logic correct
 
Last edited:
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Taylor_1989 said:
... then the electric field on the outside the shell will be the same, as the electric field insie the cavity ...
Mostly correct except that the field inside the cavity has the same radial dependence as the field outside the sphere but it cannot be construed to be same.

Now what? You are asked to find the electric field using Gauss's Law. Logic alone will not do that.
 

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