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 

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

i believe the answer is electric potential.

explanation: