A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C/m3. The outer layer has a uniform charge density of +8.0C/m3 and extends from an inner radius of 6.0 cm to an outer radius of 12.0 cm.
A) Determine the electric field for 0 B) Determine the electric field for 6.0cm C) Determine the electric field for 12.0cm

Q = 68π*(0.06³) C

Explanation:

Coulomb's law for any spherically symmetric charge distribution, says that if charge is within the Gaussian surface of radius r, it contributes as if it were at the center, and if charge is outside the surface, it does not contribute at all. This fact is considered to solve the following conditions;

A) for 0<r<6 E1(r) is given by

E1(r) = (0 →r)k∫[(4πx² ρ1dx)/(r²)] = [{(4πk ρ1)/(r²)}*(r³ /3)] = [{(4πk ρ1)/3}*r] = [4/3)*π*9*10⁹*(-5)*r] V/m

E1 = -(60π*10⁹)*r V/m

B) for 6<r<12 E2(r) is given by  

E2(r) =-[{(4/3)π*5*9*10⁹(0.06)³}/r²] + (6 →r)k∫[(4πx² ρ2dx)/(r²)] or

E2 = -[{129560000π}/r²]+ [{(4πk ρ2)/(r²)}*{(r³ /3) -0.06³/3)] or

= -[{129560000π}/r²]+ [{(4πk*8)/(3r²)}*{(r³-0.06³]

C) 12cm<r<50cm is given by

E3(r) = kQ/r², where  

Q = (4π/3)*[-5*0.06³ + 8*(0.12³ -0.06³)] = (4π/3)*[8*(0.12³)-13*0.06³)] = (4π/3)*(0.06³)*[(64-13)] Core

Q = 68π*(0.06³) C

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg