A 2.0 mm diameter aluminum ball is charged to +58 nCWhat fraction of its electrons have been removed? The density of aluminum is 2,700 kg/m3.
Since the atomic weight of aluminum is also very close to 27, the 2700 kg is very close to 100,000 molesTherefore, each 1 m^3 of aluminum is 100,000 times Avogadro's number of atomsNumber of atoms in each 1 m^3 will be 6.0 x 10^23 x 100000 6.0 x 10^28Number of electrons in each 1 m^3 of aluminum will be 27 x 6 x 10^28 1.6 x 10^30Volume of a 2.0-mm-diameter ball is (4/3)(pi)(0.002 m)^3 3.35 x 10^(-8) m^3 Number of electrons in this volume will be (3.35 x 10^(-8) m^3)(1.6 x 10^30/m^3) 5.36 x 10^22 How many electrons are required for a charge of -58 nC? That would be 58 x 10^(-9) C/(1.6 x 10^(-19) C) 3.6 x 10^11 So the fraction removed is 3.6 x 10^11/(5.36 x 10^22) about 7 x 10^(-12), that is to say, 7 trillionths, or 1 out of every 150 billionNot a whole lot!!!
