Anonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r≤r ρ(r)=0 for r≥r where ρ0=3q/πr3 is a positive constant.

A)
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
which when integrated from 0 to r is 
total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
and when r = R our total charge is 
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
and after substituting ρ0 = 3Q / πR³ we have 
total charge = Q ◄ 

B) E = kQ/d² 
since the distribution is symmetric spherically 

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
so 
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is 
and after substituting for ρ0 is 
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4)) 
which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² 

answer; how far out of place which the object finished relative to the starting location;

helium

oxygen

neon

sodium