Potential inside sphere with empty cavity

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
bfusco
Messages
126
Reaction score
1

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.
[tex]\int_V \rho \cdot dVolume = \frac{4}{3}\pi \rho (R^3-\frac{R^3}{8})=\frac{7}{6}\pi R^3 \rho[/tex]

Then to get [itex]E_B[/itex] 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:
[tex]E(r)=k\int \frac{dq \hat{s}}{s^2}[/tex]
Where i just calculated dq. [itex]s^2=R[/itex], but I am not sure what [itex]\hat{s}[/itex] is
 

Attachments

Physics news on Phys.org
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