Calculating the Electric Potential from an Electric Field problem

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SUMMARY

The discussion focuses on calculating the electric potential (V) from an electric field (E) for a nonconducting sphere with a radius of 2.31 cm and a uniformly distributed charge of +3.50 fC. The electric potential at the center of the sphere is defined as Vo = 0. For radial distances of 1.45 cm and 2.31 cm, the calculated potentials are -5.37 x 10^-4 V and -0.001 V, respectively. The confusion arises regarding the integration variable ds, where it is suggested that ds equals r/2, which differs from the initial assumption that ds equals r.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of electrostatics, particularly for nonconducting spheres
  • Proficiency in using Coulomb's law and the concept of charge distribution
NEXT STEPS
  • Study the derivation of electric potential from electric field equations
  • Learn about the integration of electric fields in spherical coordinates
  • Explore the concept of charge distribution in nonconducting materials
  • Review the application of Coulomb's law in calculating electric fields
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone studying electrostatics, particularly those tackling problems involving electric potential and electric fields in nonconducting materials.

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Homework Statement



A nonconducting sphere has radius R = 2.31cm and uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere's center to be Vo = 0. What is V at radial distance (a) r = 1.45cm and (b) r = R?

Homework Equations



V = the negative integral from i to f of E * ds

For a, E is kqr/R^3.

For b, E is kq/r^2

The Attempt at a Solution



When I plug all this in I get -5.37 x 10^-4 V and -.001V, respectively. However, when I look at the answer explanation in the student manual, it seems to indicate that ds (or in this case rs) = r/2. Why is it r/2? How is the book getting that for ds? At first I thought ds was equal to r, but I guess I'm not grasping something. Could someone explain why this is so? I basically have the right answer, but only if I divide both my answers by 2.
 
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what are your limits of integration and what's your formula for V after you integrate?
 

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