Answer:
E ’= E / 8
therefore the correct answer is A
Explanation:
Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface
Фi = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
E (4πr²) = q_{int} / ε₀
density is
ρ = q_{int} / V
q_{int} = ρ V = ρ 4/3 π r³
we substitute
E (4π r²) = ρ 4/3 π r³ /ε₀
E = 1 /3ε₀ ρ r
let's change the density by
ρ = Q / V = Q / (4/3 π R³)
E = 1 / 4πε₀ Q r / R³
if we now distribute the same charge on a sphere of radius R' = 2R
E ’= 1 / 4pieo Q r / (2R)³
E ’= 1 / 4ft Qr / R³ ⅛
E ’= E / 8
therefore the correct answer is A