How does the electrostatic potential at point B compare to point A?

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SUMMARY

The electrostatic potential at point B, located at a distance of 2r from the center of a charged sphere with net charge +Q, is the same as the electrostatic potential at point A, which is at a distance r from the sphere. This conclusion is drawn from the formula for electric potential, V = kQ/r, where the potential depends solely on the charge of the sphere and the distance from it, not on the test charges introduced. Therefore, regardless of the test charges (+q and +2q), the potential remains constant at both points.

PREREQUISITES
  • Understanding of electrostatic potential and its calculation using V = kQ/r
  • Familiarity with the concept of test charges in electrostatics
  • Knowledge of the principle that electric potential is independent of the test charge
  • Basic grasp of Coulomb's law and electric fields
NEXT STEPS
  • Study the implications of electric potential in different configurations of charge distributions
  • Learn about the concept of equipotential surfaces in electrostatics
  • Explore the relationship between electric potential and electric field strength
  • Investigate the effects of varying distances on electric potential in spherical charge distributions
USEFUL FOR

Students studying electrostatics, physics educators, and anyone seeking to deepen their understanding of electric potential and its applications in electrostatic problems.

LonghornDude8
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Homework Statement


A test charge +q is brought to a point A a distance r from the center of a sphere having a net charge +Q. Next, a test charge +2q is brought to a point B a distance 2r from the center of the sphere.
Compared with the electrostatic potential at A, the electrostatic potential at B is
A. smaller
B. the same
C. greater


Homework Equations


V=kq/r
U=kq1q2/r


The Attempt at a Solution


Because you are solving for electric potential I assume that you forget about the charge of the test charges and therefore you would get A. smaller. But if you use the charge of the test charges, you get b. the same. I'm just looking for someone to agree or disagree with me
 
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LonghornDude8 said:
Because you are solving for electric potential I assume that you forget about the charge of the test charges and therefore you would get A. smaller.
I agree.
But if you use the charge of the test charges, you get b. the same. I'm just looking for someone to agree or disagree with me
I disagree. As you said, the electric potential set up by the charged sphere is the same at all points in space regardless of how much additional charge you bring in.
 
Ya, what I said first was right. I just got a little confused because they gave extra info... I hate when that happens.

"I know one thing - that I know no thing." - lol, paradox
 

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