An insulating sphere of radius 5.00 cm , centered at the origin, has a uniform volume charge density 3.05 uC/m^3 . There is a spherical cavity cut out of its center of radius 2.00 cm.(adsbygoogle = window.adsbygoogle || []).push({});

Well i know to use Gauss' Law:

Φ = ∫E∙dA = Q/ε

where Φ is the electric flux, E is the electric field, dA is a differential area on the closed surface S with an outward facing surface normal defining its direction, Q is the charge enclosed by the surface, and ε is the electric constant.

1.) What is the electric field at 1.43 cm ?

I found that to be zero.

2.) What is the electric field inside the spherical shell at 2.86 cm ?

Q = (4π/3)*(0.0286^3-0.0200^3)*3.05x10^-6C

The surface area integrated over is: S = 4π*0.0286^2. Hence:

E = Q/εS = (0.0286^3-0.0200^3)*3.05x10^-6/(3ε*0.028...

for surface area i get 1.03*10^-2, for Q i get 4.188*1.53*10^-5*3.05*10^-6= 1.95*10^-10

Then i divide Q/S which is 1.90*10^-12. but i keep getting it wrong. what am i doing wrong?

3.) What is the electric field outside the spherical shell at 6.75 cm ?

Q = (4π/3)*(0.0500^3-0.0200^3)*3.05x10^-6C

The surface area integrated over is: S = 4π*0.0675^2. Hence:

E = Q/εS = (0.0500^3-0.0200^3)*3.05x10^-6/(3ε*0.067...

I did the same as part two equations and i get 3.89*10^-9 which is wrong also. Help!

3. The attempt at a solution[/

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# Homework Help: Charged Sphere with Cavity

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