A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r (A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E

 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 = /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

a, c, d, e are all applying about success strategies.