Calculating Electric Field of a Solid Sphere with Nonuniform Charge Density

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SUMMARY

The discussion focuses on calculating the electric field of a solid insulating sphere with a nonuniform charge density defined as σ = ar, where a is a constant. The charge contained within a radius r (where r < b) is determined to be Q_r = πar^4. For specific values of a = 2 x 10^-6 C/m^4 and b = 1 m, the electric field E at r = 0.6 m is calculated using the formula E = Q_r * r / (4πE_0R^3), leading to an initial incorrect result of 4378 N/C. The error was identified as a miscalculation in the application of the charge formula.

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  • Familiarity with spherical coordinates and volume elements
  • Knowledge of Coulomb's law and electric field equations
  • Basic calculus for integrating charge density
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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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