A sphere of radius 5.00 cmcm carries charge 3.00 nCnC. Calculate the electric-field magnitude at a distance 4.00 cmcm from the center of the sphere and at a distance 6.00 cmcm from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.

a) E = 8628.23 N/C

b) E = 7489.785 N/C

Explanation:

a) Given

R = 5.00 cm = 0.05 m

Q = 3.00 nC = 3*10⁻⁹ C

ε₀ = 8.854*10⁻¹² C²/(N*m²)

r = 4.00 cm = 0.04 m

We can apply the equation

E = Qenc/(ε₀*A)  (i)

where

Qenc = (Vr/V)*Q

If    Vr = (4/3)*π*r³  and  V = (4/3)*π*R³

Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³

then

Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C

We get A as follows

A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²

Using the equation (i)

E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)

E = 8628.23 N/C

b) We apply the equation

E = Q/(ε₀*A)  (ii)

where

r = 0.06 m

A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²

Using the equation (ii)

E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)

E = 7489.785 N/C

current flows in a wire