Answer:
We can use Gauss's Law to find the electric field at a point outside the sphere. Let's take a spherical Gaussian surface with radius 15 cm centered at the center of the sphere. By symmetry, the electric field is constant and radial on this surface, so we can take it outside the integral:
∮E⋅dA = Qenc/ε0
where Qenc is the enclosed charge and ε0 is the permittivity of free space. The left-hand side is just E times the surface area of the sphere, which is 4πr^2, where r is the radius of the Gaussian surface. The enclosed charge is the total charge density times the volume enclosed by the surface, which is (4/3)πr^3 for a sphere:
E(4πr^2) = (4/3)πr^3 (30 nC/m^3)/ε0
We can solve for E:
E = r(30/ε0)/3
Plugging in r = 0.15 m and ε0 = 8.85×10^-12 F/m, we get:
E = (0.15 m)(30 nC/m^3)/(8.85×10^-12 F/m)/3 ≈ 2.01×10^6 N/C
So the magnitude of the electric field 15 cm from the center of the sphere is approximately 2.01×10^6 N/C.