A solid insulating sphere has a radius A and net charge of +Q. The insulating sphere has a uniform charge density. a conducting sphere of inner radius B and outer radius C surrounds it.
Use Gauss's Law to find the Electic Field for R<A
Use Gauss's Law to find the Electic Field for A<R<B
Use Gauss's Law to find the Electic Field for B<R<C
Use Gauss's Law to find the Electic Field for R>C
What is the electric potential for R < A
What is the electric potential for R = A
1) For R<A
Choose a sphere of radius R<A as your Gaussian surface. Let p be the charge density of the insulator, and let V' be the volume of your Gaussian sphere. Then the charge enclosed by the Gaussian surface is:
Q_in = pV'
where V' = (4/3)*pi*R^3
Since the magnitude of the electric field is constant everywhere on the spherical Gaussian surface and is normal to the surface at each point, gauss' law gives:
Electric flux through surface = E*SurfaceArea
With full formulas and good explanations, the problems are solved.