Question

Two small alum:num spheres. each having mass 00250 kg, are separated by 80.0 cm.
(a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26 {282 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 X 104 N (roughly 1 ton)? Assume that
the spheres may be treated as point Charges.
(c) What fraction of all the electrons in each sphere does this represent?

Answer 1

(a)

Mass of aluminum sphere =0.0250 kg= 25 g

As 1 atom of aluminum contains 13 electrons

So 1 mole of Al will contain 13 N_A electrons

Number of mole of aluminum given = 25/13 =1.92 moles

Electron contained =1.92*13=25 N_A

(b) $F=\dfrac{q^2}{4\pi\epsilon r ^2}$

$1\times10^4=\dfrac{q^2}{0.8^2}\times9\times10^9$

$q=\dfrac{0.8\times10^{-5/2}}{3}C=0.27\times10^{-5/2}C$

1 electron carries 1.6x10^-19 C charge

then number of electrons needed to constitute the above charge $=\dfrac{0.27\times10^{-5/2}}{1.6\times10^{-19}}=0.16\times10^{33/2}=0.5\times10^{16}$

(c) This fraction is very small and of order 10^-6 as electrons that constitute the charge are of order 10¹⁶

and electrons contained by the sphere is of order 10²³