A sphere of radius 8.0 cm carries a uniform volume charge density ρ = 500 nC/m³. What is the electric field at r = 3.0 cm?

Question

Quizplus · Accepted Answer

First, we need to find the charge enclosed within the sphere of radius 3.0 cm. Since the charge density is uniform, we can use
ρ = Q/V ,
where V is the volume of the sphere. Thus,
Q = ρV = (500 nC/m3) (4/3π(0.03 m)3) = 0.566 nC.
Now, we can use Gauss's Law to find the electric field at r = 3.0 cm. Since the sphere has spherical symmetry, we choose a spherical Gaussian surface with radius 3.0 cm. The electric field is then given by
E = Qenc/ε0A,
where Qenc is the charge enclosed by the Gaussian surface, ε0 is the permittivity of free space, and A is the area of the Gaussian surface. Since the electric field is radial, the area of the Gaussian surface is simply 4πr2. Thus,
E = (0.566 nC)/(8.85×10-12 N-1C2)(4π(0.03 m)2) = 564.7 N/C ≈ 565 N/C.

A Sphere of Radius 8