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Potential inside sphere with empty cavity

  • Thread starter bfusco
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  • #1
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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.
[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 Im not sure how i would set that up because the E field isnt isotropic, so instead Im 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 im not sure what [itex] \hat{s} [/itex] is
 

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  • #2
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thinking about it though, im pretty sure that what i got for dq is wrong, because i dont think what i did takes into account that the cavity isn't centered at the origin
 
  • #3
haruspex
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Treat it as one sphere minus another. Find the fields and potentials for each sphere and take the difference.
 

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