Calculating Electric Field of a Solid Sphere with Nonuniform Charge Density

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To calculate the electric field of a solid sphere with nonuniform charge density, the charge contained within radius r is given by Q_r = πar^4. For a sphere with a = 2 x 10^-6 C/m^4 and radius b = 1 m, the electric field E at r = 0.6 m can be derived using the formula E = Q_r * r / (4πE_0R^3). Substituting the values results in E = (π(2 x 10^-6)(0.6)^5) / (4π(8.85 x 10^-12)(1)^3), which simplifies to 4378 N/C. The calculation needs to ensure that Q_r is correctly used in the formula for E.
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Consider a solid insulating sphere of radius b with nonuniform charge density \sigma = ar where a is a constant. Find the charge Q_r contained within the radius r, when r < b. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4\pi r^2.
I got this part, the answer was Q_r = \pi ar^4.
The second part says if a = 2 x 10^-6 C/m^4 and b= 1 m, find E at r= 0.6 m. Answer in units of N/C.
I used the equation E= Qr/4\pi E_0 R^3
Plugging in the previous answer for Q gave me E= (\pi ar^4)r/ 4\pi E_0 R^3
so \pi ar^5/4\pi E_0 B^3 since R=B in this problem.
then \pi (2 x 10^-6)(.6)^5/ 4\pi (8.85 x 10^-12)
This gave me 4378 N/C... which is wrong.
Help please?
 
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E= Qr/4\pi E_0 R^3

Did you mean Q_r?
 
Yes, sorry I meant Q_r * r.
 
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