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StephenDoty
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Two small aluminum spheres, each of mass 0.0250kg, are separated by 80cm or .8m. If the atomic mass of aluminum is 26.982 grams per mole and its atomic number is 13 then how many electrons does each sphere contain? How many electrons would have to be removed form one sphere and added to the other to cause an attractive force of 1.00 X 10^4 N? What fraction of all the electron in one of the spheres does this represent? The spheres may be treated as point charges.
0.05kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom = 9.43117 X 10^25 electrons
If an electron is removed from one sphere it become +e and if an electron is added to the other sphere it becomes -e so the number of electrons that must be subtracted and added = (Ne) and (-Ne).
Thus
F=k(Ne)(-Ne)/.8m^2
but since the force = k * the magnitude of the product of the charges / r^2
1.00 X 10^4 N= k(Ne)^2/0.64m^2
(1.00X10^4N)*(.64m^2)/k = (Ne)^2
(1.00X10^4N)*(.64m^2)/k = (N^2) * (1.6X10^-19)^2
(1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2] = N^2
sqrt((1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2]) =N
9.90243 X 10^7 electrons = N
The ratio is 9.90243 X 10^7 / 9.43117 X 10 ^25
Did I do this correctly?
Thank you.
Stephen
0.05kg * 1000g/1kg * 1 mole/26.982g * (6.023 X 10^23) atoms/ 1 mole * 13 electrons / 1 atom = 9.43117 X 10^25 electrons
If an electron is removed from one sphere it become +e and if an electron is added to the other sphere it becomes -e so the number of electrons that must be subtracted and added = (Ne) and (-Ne).
Thus
F=k(Ne)(-Ne)/.8m^2
but since the force = k * the magnitude of the product of the charges / r^2
1.00 X 10^4 N= k(Ne)^2/0.64m^2
(1.00X10^4N)*(.64m^2)/k = (Ne)^2
(1.00X10^4N)*(.64m^2)/k = (N^2) * (1.6X10^-19)^2
(1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2] = N^2
sqrt((1.00X10^4N)*(.64m^2)/[k* (1.6X10^-19)^2]) =N
9.90243 X 10^7 electrons = N
The ratio is 9.90243 X 10^7 / 9.43117 X 10 ^25
Did I do this correctly?
Thank you.
Stephen