Electric field of a point outside a charged sphere

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SUMMARY

The electric field outside a charged conducting sphere with a negative net charge and a positive point charge at its center can be determined by summing all charges present. The relevant equation is E=kq/r^2, where 'q' represents the total charge, including both the negative charge of the sphere and the positive charge at the center. If the negative charge equals the positive charge, the electric field outside the sphere is zero. This conclusion is supported by Gauss' law and Coulomb's law, which illustrate the symmetry of the electric field in this scenario.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with Gauss' law
  • Knowledge of charge interactions in conductors
  • Basic proficiency in using the equation E=kq/r^2
NEXT STEPS
  • Study Gauss' law applications in electrostatics
  • Explore electric field calculations for different charge configurations
  • Investigate the properties of conductors in electrostatic equilibrium
  • Learn about the implications of charge neutrality in electric fields
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Students of physics, electrical engineers, and anyone interested in electrostatics and electric field calculations.

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If a conducting sphere has a negative net charge and there is a positive point charge at the center of the sphere, what charge would you use to find the Electric Field at a point outside the sphere?

The equation is E=kq/r^2.

Would I use the net charge of the sphere, or do I need to calculate a new charge due to the positive one at the center?
 
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Are you talking about a hollow conducting sphere of net negative charge, with a positive charge inside the hollow? Luckily, the answer is very simple, just add up all the charges (including the positive one in the middle) and use this in the equation.

For example, if the net negative charge on the conducting sphere is exactly equal to -1 times the positive charge in the middle, then the charges cancel, and there is no electric field outside the sphere.

To find out why it works this way, you could either do an integral and Coulomb's law, or use Gauss' law, using the symmetry of the problem.
 

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