Electric Potential of Charge Distribution

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SUMMARY

The discussion focuses on calculating the electric potential and field due to two point charges, -q and q/3, located at the origin and at the point (a,0,0), respectively. The key equation for the electric field is provided as E = (q/(4πε₀))[-1/r² + (q/3)/(r² + a² - 2ra cos(α))]. The user seeks assistance in finding the potential function and confirming that the V = 0 equipotential surface is a sphere. The solution involves summing the potentials from both charges rather than performing integrals.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with point charge interactions
  • Knowledge of spherical coordinates and equipotential surfaces
  • Basic calculus for integration (though not required for this specific problem)
NEXT STEPS
  • Study the principle of superposition in electric fields
  • Learn about the concept of equipotential surfaces in electrostatics
  • Explore the mathematical derivation of electric potential from electric fields
  • Investigate the properties of spherical charge distributions
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in understanding electric potential and field calculations involving point charges.

metgt4
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Homework Statement



Two point charges, -q and q/3, are situated at the origin and at the point (a,0,0) respectively. At what point on the x-axis does the Electric Field Vanish? Find the potential function and show that the V = 0 equipotential surface is a sphere


Homework Equations



I have found that the electric field can be described as:

E = \frac{q}{4pi\epsilon_{o}}[\frac{-1}{r^2}\widehat{r}+\frac{q}{3(r^2 + a^2 - 2racos\alpha}\widehat{r'}]

where alpha is the angle between r and r', but I'm not sure how to find the potential field. I know that the potential is the negative integral of the electric field, but I get stuck when trying to integrate the potential due to the positive charge. I'm sure that the solution is something SUPER obvious that I simply am overlooking in my sleep deprived state, but any hints or help would be greatly appreciated.

Thanks!
Andrew
 
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You don't need to do any integrals. For any point, find the potential there for each point charge and sum them.
 

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