Answer:
The total charge on the sphere is 46.11 x 10⁻³ C
Explanation:
Given:
charge density of innermost section, σ = −5.0C/m³
radius of the innermost section, r = 6.0cm
charge density of the outer layer ,σ = +8.0 C/m³
radius of the outer layer, r =12.0cm
volume of sphere is given as ![= \frac{4}{3}.\pi r^3](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%7D%7B3%7D.%5Cpi%20%20r%5E3)
Charge enclosed by the innermost section = volume x charge density
volume ![= \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.06^3) = 9.05 X 10^{-4} m^3](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%20r%5E3%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%20%280.06%5E3%29%20%3D%209.05%20%20X%2010%5E%7B-4%7D%20m%5E3)
Enclosed charge, q₁ = −5.0C/m³ X 9.05 x 10⁻⁴ m³
= - 4.53 x 10⁻³ C
Charge at the outer surface
Volume = ![\frac{4}{3}\pi [r_2{^3} - r_1{^3}] = \frac{4}{3}\pi [0.12{^3} - 0.06{^3}] = 0.00633 {m^3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%20%5Br_2%7B%5E3%7D%20-%20r_1%7B%5E3%7D%5D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%20%5B0.12%7B%5E3%7D%20-%200.06%7B%5E3%7D%5D%20%3D%200.00633%20%7Bm%5E3%7D)
Enclosed charge, q₂ = +8.0 C/m³ X 0.00633 m³
= 50.64 x 10⁻³ C
Total charge on the sphere; Q = q₁ + q₂
= - 4.53 x 10⁻³ C + 50.64 x 10⁻³ C
= 46.11 x 10⁻³ C
Therefore, the total charge on the sphere is 46.11 x 10⁻³ C