How Will Charge Distribute on a Spherical Conductor Near a Positive Charge?

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SUMMARY

The charge distribution on a spherical conductor near a positive charge results in positive charges accumulating on the side closest to the external positive charge. This occurs because the electric field inside the conductor must remain zero, necessitating that the induced charges on the conductor's surface counteract the external electric field. The method of image charges is a useful technique for solving such problems, providing a visual and mathematical framework for understanding charge distribution in electrostatics.

PREREQUISITES
  • Understanding of electrostatics principles
  • Familiarity with electric fields and charge interactions
  • Knowledge of conductors and their properties in electrostatics
  • Concept of the method of image charges
NEXT STEPS
  • Study the method of image charges in detail
  • Explore electric field calculations using Gauss's Law
  • Learn about induced charge distribution in conductors
  • Investigate the behavior of electric fields in different geometries
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding charge distribution phenomena in conductive materials.

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Homework Statement



The problem shows a picture of a spherical conductor that is placed some distance away from a positive charge.

It wants me to draw in how I think the charge will be distributed.

Homework Equations



E=(KQq)/r^(2)

The Attempt at a Solution



My original thoughts were that the negative charge would be on the side closest to the positive charge because opposite charges attact.

But then

I remembered that the E field inside of a conductor is always zero. So then I figured that the positive charges would be on the side closest to the positive charge because the e field of the charges on the right side of the sphere need to be facing the opposite direction as the e field from the outside charge in order for the e fields to cancel each other out inside the sphere.

Is this right?

It just throws me off a little bit because I know that opposite charges attract.
 
Last edited:
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I noticed my post didn't make much sense there towards the end but I just edited so hopefully it makes sense now if anyone was having trouble understanding what I was saying.
 

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