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glpg80
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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 :(