Two small aluminum spheres, each having a mass of 0.0750 kg, are separated by 90

Question

Two small aluminum spheres, each having a mass of 0.0750 kg, are separated by 90.0 cm. (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 g/mol, and its atomic number is 13.)(b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00 104 N (roughly one ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Explanation / Answer

(a)
No of Electrons each sphere contain, n = (Mass/Molar Mass) * Avogadro's No * Atomi Number
n = 0.0750/(26.982 * 10^-3) * 6.023 * 10^23 * 13
n = 2.176 * 10^25

(b)
Distance, r = 90.0 cm = 0.9 m
Let the Charge to be removed/added = q

F = k*qq/r^2
1.0 * 10^4 = (9.0 * 10^9 * q^2)/ 0.9^2
Solving for q
q = 9.48 * 10^-4 C

Charge of 1 electron = 1.6 * 10^-19 C
No of electrons = (9.48 * 10^-4 C) / (1.6 * 10^-19 C)
No of electrons to be removed = 5.925 * 10^15 electrons


(c)
Fraction of all the electrons , = 5.925 * 10^15/(2.176*10^25)
Fraction of all the electrons , = 2.72 * 10^-10

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