Insulating Spherical Charged Solid with a Cavity

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SUMMARY

The discussion focuses on calculating the electric field inside an insulating spherical charged solid with a cavity containing a point charge of -2.09×10-6 C. The charge density of the solid is 7.36×10-4 C/m3, and the electric field is evaluated at a distance of 9.48 cm from the center of the cavity. The solution emphasizes the use of Gauss's Law and superposition, concluding that the electric field inside the solid is independent of the charge density and the cavity size.

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  • Basic principles of electrostatics
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Homework Statement



A point charge of -2.09*10^(-6) C is located in the center of a spherical cavity of radius 6.54 cm inside an insulating spherical charged solid. The charge density in the solid is 7.36×10^(-4) C/(m^3) .

Calculate the magnitude of the electric field inside the solid at a distance of 9.48 cm from the center of the cavity.



Homework Equations



[tex]\Phi[/tex]= EA = Qenc/[tex]\epsilon[/tex]0

E = Q/(4(pi)(epsilon sub 0)(r^2)


The Attempt at a Solution




Bear with me, latex is a bit beyond my capabilities at the moment. I scanned my work. The answer didn't look right and of course, it was not. It also seemed strange not to use the radius of the cavity and even the given charge distribution, but I'm not sure how to incorporate them.
 

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Nobody wants to take a whack at it?
 
If you imagine the case of cavity being at the center of insulated sphere, the field is zero by symmetry. Use superposition to solve the problem. The fiels is indeed independent of the charge density of the sphere and the size of the cavity.

Surendranath
 

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