A nonconducting sphere has radius R = 3.77 cm and uniformly distributed charge q = +3.92 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V at radial distance from the center (a) r = 2.70 cm and (b) r = R? (Hint: See an expression for the electric field.)

a)   V = 6.52 10⁻⁵ V, b) Vb = 9.34 10⁻⁵ V

Explanation:

b) When using Gauss's law all the charge is considered concentrated in the inside of the sphere, for which the electrice field is

             Q = 3.92 10⁻¹⁵ C

             E = k Q / r²

Electric potential and electric field are related

          E = - dV / dr

         dV = - E dr

          Vb-Va = - k Q (1 / rb -1 / ra)

          Vb = - 8.99 10⁸ 3.92 10⁻¹⁵ 1 / 0.0377

          Vb = 9.34 10⁻⁵ V

a) In this case we must know the electric field for the interior of the sphere, use Gauss's law, the charge inside is

             ρ = Q / V

              = ρ (4/3 π r³)

             Ф = E A = Q_{int} /ε₀

            E 4π r² = ρ 4/3 π r³ /ε₀

            E = ρ r / 3ε₀

            dV = ρ / 3ε₀ ∫ r dr

            V = ρ / 3ε₀ r²

We evaluate between the center of the sphere V = 0 and the point R

            V = ρ / 3ε₀ ½ (R²)

            V = Q 3 / 4π R³ 3ε₀   (R²) = k Q 1 / 2R

            V = 8.99 10⁸ 3.92 10⁻¹⁵ 1 / 0.027 2

             V = 6.52 10⁻⁵ V

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more