Potential inside sphere with empty cavity

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SUMMARY

The discussion centers on calculating the electric field and potential for an insulating sphere with a uniform charge density ρ and a spherical cavity of radius R/2. The volume charge distribution is determined by subtracting the cavity's volume from the total sphere's volume, yielding a total charge of (7/6)πR³ρ. The electric field at point B is approached using Gauss' law, although the user expresses uncertainty regarding the setup due to the non-isotropic nature of the field. The user also seeks clarification on the proper method to account for the cavity's position when applying the electric field equation for a volume charge distribution.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Familiarity with electric field equations for volume charge distributions
  • Knowledge of potential energy concepts in electrostatics
  • Basic calculus for volume integration
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  • Study the application of Gauss' law for non-uniform charge distributions
  • Learn about electric field calculations for spherical charge distributions
  • Explore the concept of electric potential in relation to charge distributions
  • Investigate the effects of cavities in charged spheres on electric fields and potentials
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Students and professionals in physics, particularly those focusing on electrostatics, as well as educators seeking to enhance their understanding of electric fields and potentials in systems with non-uniform charge distributions.

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Homework Statement


An insulating sphere of radius R , centered at point A, has uniform chagre density ρ. A spherical cavity of radius R / 2 , centered at point C, is then cut out and left empty, see Fig.

(a) Find magnitude and direction of the electric field at points A and B.

(b) Find the potential at points A and B. Set V(r → ∞) = 0.

(c) Write down an algebraic solution (no integrals!) for E(r) and V (r) for the space outside the larger sphere, r > R. Choose r = 0 at point A, and the radius vector of point C as r = RC .

(question1) in the uploaded file


The Attempt at a Solution


(a)First I want to figure out the volume charge distribution, which I wanted to do by finding the the volume of the whole sphere minus the volume of the cavity.
\int_V \rho \cdot dVolume = \frac{4}{3}\pi \rho (R^3-\frac{R^3}{8})=\frac{7}{6}\pi R^3 \rho

Then to get E_B I wanted to use Gauss' law, but I am not sure how i would set that up because the E field isn't isotropic, so instead I am trying to use the equation for E-field of a volume charge distribution:
E(r)=k\int \frac{dq \hat{s}}{s^2}
Where i just calculated dq. s^2=R, but I am not sure what \hat{s} is
 

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thinking about it though, I am pretty sure that what i got for dq is wrong, because i don't think what i did takes into account that the cavity isn't centered at the origin
 
Treat it as one sphere minus another. Find the fields and potentials for each sphere and take the difference.
 

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