What Determines the Electric Field Inside a Spherical Conductor?

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SUMMARY

The discussion centers on the determination of the electric field inside a spherical conductor when a point charge is placed at its center. The correct answer to the posed question is D, which states that the electric field at a distance r from the center is given by E = q/4πε0r^2. Participants clarify that while the charge within the cavity generates an electric field, the conductor itself does not have a net electric field due to charge redistribution on its surfaces, which cancels any internal field.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and point charges
  • Knowledge of conductors and charge distribution
  • Basic equations of electrostatics, particularly E = kq/r^2
NEXT STEPS
  • Study Gauss's Law in detail to understand electric field calculations
  • Explore the concept of electric field lines and their behavior in conductors
  • Learn about the implications of charge distribution on conductors in electrostatic equilibrium
  • Investigate the effects of multiple charges on electric fields within conductors
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in electrostatics, particularly those studying electric fields and conductors in advanced physics courses.

pietastesgood
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Homework Statement



1. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer
radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric
field produced by the charge on the inner surface at a point in the interior of the conductor, a
distance r from the center, is:
A. 0
B. Q/4vπε0(R1)2
C. Q/4πε0(R2)2
D. q/4πε0r^2
E. Q/4πε0r^2

Answer is D.

Homework Equations



E=kq/r^2
flux=q/ε0

The Attempt at a Solution



I don't quite understand why there is an electric field within the conductor itself, which is what answer choice D is saying. Sure, there's a charge within the cavity, but since the net charge has to equal zero, there will be an equal but opposite charge on the inner cavity surface, which allows no flux to leave through a Gaussian surface within the conductor at a radius greater than the inner cavity.

Unless I'm actually just reading the problem wrong, and it's talking about just the charge on the inner surface, and ignoring the charge within the cavity. In that case, wouldn't the charge -q cause a field of magnitude q/4πε0(r-r1)^2? Any help would be appreciated!
 
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pietastesgood said:

Homework Statement



1. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer
radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric
field produced by the charge on the inner surface at a point in the interior of the conductor, a
distance r from the center, is:
A. 0
B. Q/4vπε0(R1)2
C. Q/4πε0(R2)2
D. q/4πε0r^2
E. Q/4πε0r^2

Answer is D.

Homework Equations



E=kq/r^2
flux=q/ε0

The Attempt at a Solution



I don't quite understand why there is an electric field within the conductor itself, which is what answer choice D is saying. Sure, there's a charge within the cavity, but since the net charge has to equal zero, there will be an equal but opposite charge on the inner cavity surface, which allows no flux to leave through a Gaussian surface within the conductor at a radius greater than the inner cavity.

Unless I'm actually just reading the problem wrong, and it's talking about just the charge on the inner surface, and ignoring the charge within the cavity. In that case, wouldn't the charge -q cause a field of magnitude q/4πε0(r-r1)^2? Any help would be appreciated!

Yeah, I think you have to read it carefully. There is no net electric field inside of the conductor. The charge at the center will create an electric field corresponding to answer D. The charges on the conductor will rearrange to cancel that. But the inner charge of -q is not a point charge. Be careful what field you think it makes.
 
Oops, you're right. If you take the inner charge -q and use Gauss' law, you get EA=q/ε0, and A=4πr^2, so E=q/4πε0r^2. Thanks for the help!
 

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