Electron Fraction Calculation based on Density, Avagadro's #, Electrostatic Constantby glpg80 Tags: avagadro, based, calculation, constant, density, electron, electrostatic, fraction 

#1
Apr1011, 08:02 PM

P: 4

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 = 1.6e19 C Electron Charge = 1.6e19 C K = (9e9 Nm^2)/(C^2) vector<E> = (Kq)/(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)(1.6E19) = 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 :( 



#2
Apr1011, 08:39 PM

HW Helper
P: 2,324

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?




#3
Apr1011, 09:12 PM

P: 4

V = (4/3)(pi)((0.001m)^3) = 4.19E9 (7870kg/m^3)(4.19E9m^3), meters cancel and 3.2976E5kg = mass how many atoms does it contain? (3.2976E5 kg)/(0.055845 kg/mol) = 1.84E6 mol (1.84E6 mol)(6.022E23 atoms/mol) = 1.11E18 atoms 



#4
Apr1011, 09:31 PM

HW Helper
P: 2,324

Electron Fraction Calculation based on Density, Avagadro's #, Electrostatic Constant
It would be, except you divided 3.2976E5 kg by 55.845, which is in g/mol.




#5
Apr1011, 09:39 PM

P: 4





#6
Apr1011, 09:56 PM

P: 4

Iron has:
29 electrons 29 protons 30 neutrons the iron ball was charged to +48E9 C (+48E9)/(1.6E19) = 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) = 9.32E9; no units. (web assign claims the correct answer for Iron is 3.24E11 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 :( 


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