Overlapping spheres of charge: finding the E field between them

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SUMMARY

The discussion focuses on calculating the electric field (E field) in the overlapping region of two spheres with uniform charge densities. The first sphere has a positive charge density ρ and radius R, while the second sphere, shifted by a distance d, has a negative charge density -ρ and the same radius R. Using Gauss' Law, the participant derived the E field inside a single sphere as E = ρr / (3ε). However, when combining the E fields from both spheres, the participant concluded that the total E field is zero, raising concerns about the consistency of the radius used in the Gaussian surface.

PREREQUISITES
  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric field calculations for spherical charge distributions
  • Knowledge of vector addition in physics
  • Concept of charge density and its implications in electric fields
NEXT STEPS
  • Review the application of Gauss' Law for multiple charge distributions
  • Study the concept of electric field superposition in overlapping charge regions
  • Explore vector calculus as it applies to electric fields and charge distributions
  • Investigate the implications of charge density on electric field strength and direction
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This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and professionals involved in teaching or applying concepts of electric fields and charge distributions.

jerro
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Homework Statement



You have two spheres. The first is centered at the origin and as uniform positive charge density ρ and radius R. The second is shifted up a distance d, and it has uniform negative charge density -ρ and radius R.

Find the E field in the region of overlap.


Homework Equations



Gauss' Law


The Attempt at a Solution



I first found an expression for the E field inside a single sphere.

∫E dA = [itex]\frac{Q}{\epsilon}[/itex]
E(4[itex]\pi[/itex]*[itex]r^{2}[/itex]) = [itex]\frac{\rho*(4/3)\pi*r^{3}}{\epsilon}[/itex]
E=[itex]\frac{\rho*r}{3\epsilon}[/itex]

Now, for extending the case to include both spheres.

I add the E field from one to the E field of the other, giving [itex]\frac{\rho*r}{3\epsilon}[/itex] - [itex]\frac{\rho*r}{3\epsilon}[/itex], which gives zero.

I'm not sure that this is correct, I'm feeling weary of r, the radius of the Gaussian surface, and whether it is the same for both spheres.
 
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r is the distance to the center of the sphere, and that is different for those spheres.
In addition, both E and r are vectors.
 

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