Answer:
a = R/2
Step-by-step explanation:
Let R be the radius of the sphere and q = -2e be the charge in it. Let q₁ be the charge at radius a where the one of the point positive charges e is located. . Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πa³
q₁ = q(a/R)³ = -2e(a/R)³. The electric force due to q₁ at r is F₁ = kq₁q/a² = kq²(a/R)³/a² = k(-2e)²a/R³ = 4ke²a/R³.
Let h be the distance between the two point charges. The electric force due to the one point charge on the other is F₂ = ke²/h²
If the net force on either charge is zero, then
F₁ = F₂
4ke²a/R³ = ke²/h²
a = R³/4h²
Since h = 2a, since the charges are equidistant from each other,
a = R³/4(2a)² = R³/8a²
a = R³/8a²
a³ = R³/8
a = ∛(R³/8)
a = R/2