Question

Charge of uniform volume density ρ = 2.10 µC/m³ fills a nonconducting solid sphere of radius 5.60 cm. What is the magnitude of the electric field (a) 1.80 cm and (b) 8.80 cm from the sphere's center?

Answer 1

(a) r = 1.80 cm = 0.018 m

Now, volume of the surface enclosing the charge, V =(4/3)πr³ = (4/3)π x 0.018³ = 0.244 x 10^-4 m³
Total charge = ρV = 2.10x10^-6 x 0.244x10^-4 = 0.5124x10^-10 C
E = (9x10^9)(0.5124x10^-10)]/(0.018)² = 1423.3 N/C

(b) Total volume of the sphere, V =(4/3)πr³ = (4/3)π x 0.056³ = 0.7355x10^-3m³
Total charge = ρV = 2.1x10^-6 x 0.7355x10^-3 = 1.54x10^-9 C

Therefore, by symmetry (or using Gauss's Law) this behaves as a point charge, so at a distance of 8.80 cm:
E = (9x10^9)(1.54x10^-9)]/(0.088)² = 1790 N/C