Electron Fraction Calculation based on Density, Avagadro's #, Electrostatic Constant

glpg80

1. The problem statement, all variables and given/known data

A 2.0 mm diameter iron ball is charged to +48 nC. What fraction of its electrons have been removed? The density of iron is 7,870 kg/m3.


2. Relevant equations

Q = Ne; N = number of atoms in 1 mole, e is atomic charge, Q = point charge

N = Avagadro's Constant = 6.02E23

Proton Charge = 1.6e-19 C

Electron Charge = -1.6e-19 C

K = (9e9 Nm^2)/(C^2)

vector<E> = (K|q|)/(r^2)

A=4(pi)r^2

D=2r

55.845 is the atomic mass of Iron

3. The attempt at a solution

I know that either the surface area of a sphere, or the Area of the spherical metallic ball comes into play but i do not know which one? i know that electrons will go to the surface of the ball or protons will go to the surface of a ball, and that the conductor internally has an electric field of 0 N/C.

I also know that the density of the iron ball comes into play but have failed at connecting density with charge or even if there is a connection.

I know that it will be a fraction of two things over one another since it is asking for a comparison.

I know that 48nC is 3E11 electrons.

i attempted to solve for E as a vector but got an answer that is way off by more than double.

Q = Ne (6.02E23)(55.845)(1.6E-19) = 5.38E6

I then treated the ball and some other point source as a capacitor using the area of the surface of the sphere, plugging and chugging from this point on but got an incorrect number.

i dont know how to start or tackle this problem :(

ideasrule

You just need to find out how many electrons there are in total. So how heavy is a 2.0 mm iron ball? How many atoms does it contain? How many electrons are there per atom?

glpg80

ok, so mass and volume relate to density. i need to find the volume of a 2mm iron ball.

V = (4/3)(pi)((0.001m)^3) = 4.19E-9

(7870kg/m^3)(4.19E-9m^3), meters cancel and 3.2976E-5kg = mass

how many atoms does it contain?

(3.2976E-5 kg)/(0.055845 kg/mol) = 1.84E-6 mol

(1.84E-6 mol)(6.022E23 atoms/mol) = 1.11E18 atoms

ideasrule

It would be, except you divided 3.2976E-5 kg by 55.845, which is in g/mol.

glpg80

ok, fixed the improper units conversion, kg cancels and mol becomes the correct unit which is then used to calculate the number of atoms, which is 1.11E18

glpg80

Iron has:

29 electrons
29 protons
30 neutrons

the iron ball was charged to +48E-9 C

(+48E-9)/(-1.6E-19) = -3E11 Number of Electrons from the Electrical Charge

To account for the Electrons in an atom of iron; (1.11E18 atoms)(29 Electrons per atom) = 3.219E19 Electrons total

present it as a fraction= (-3E11)/(3.219E19) = -9.32E-9; no units.

(web assign claims the correct answer for Iron is 3.24E-11 with no units. Web Assign changes the metals and the Densities each time a new browser is opened and it is now getting late - i will have to do this over with another example)

i still need help in solving this please. i do not know where i am messing up. i am stuck :(