Calculating Fraction of Electrons Removed from Iron Ball

In summary, to find the fraction of electrons removed from a 2.0 mm diameter iron ball charged to +48 nC, you first need to find the total number of electrons in the ball. This can be done by calculating the mass of the ball and converting it to moles, then multiplying by Avogadro's constant to find the number of atoms. From there, you can find the total number of electrons by multiplying by the number of electrons per atom. The resulting fraction is -9.32E-9, which represents the number of electrons removed over the total number of electrons in the ball. However, it is important to note that this answer may vary depending on the specific metal and density used in the calculation.
  • #1
glpg80
4
0

Homework Statement



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.


Homework 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

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 don't know how to start or tackle this problem :(
 
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  • #2


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


ideasrule said:
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?

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
 
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  • #4


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


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

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
 
Last edited:
  • #6


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 :(
 
Last edited:

What is the formula for calculating the fraction of electrons removed from an iron ball?

The formula for calculating the fraction of electrons removed from an iron ball is:
Fraction of electrons removed = (number of electrons removed / total number of electrons in the iron ball) * 100%.

What is the importance of calculating the fraction of electrons removed from an iron ball?

Calculating the fraction of electrons removed from an iron ball is important because it helps us understand the amount of charge that has been transferred to or from the iron ball. This can be useful in various scientific experiments and industries such as electrochemistry and metallurgy.

Can the fraction of electrons removed from an iron ball be negative?

No, the fraction of electrons removed from an iron ball cannot be negative. The fraction represents a ratio of the number of electrons removed to the total number of electrons, and this ratio cannot be negative.

What factors can affect the fraction of electrons removed from an iron ball?

The fraction of electrons removed from an iron ball can be affected by factors such as the type of material the ball is in contact with, the temperature, and the presence of any external electric fields. These factors can impact the transfer of electrons between the iron ball and its surroundings.

How can the fraction of electrons removed from an iron ball be experimentally determined?

The fraction of electrons removed from an iron ball can be experimentally determined by measuring the change in the electric charge of the ball before and after the transfer of electrons. This can be done using various instruments such as an electrometer or a Faraday cage.

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