Two engineering students John with a mass of 90 kg and Mary

Two engineering students, John with a mass of 90 kg and Mary with a mass of 44 kg, are 59 m apart. Suppose each has a 0.01% imbalance in the amount of positive and negative charge, one student being positive and the other negative. Find an approximation of the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.

Solution

90kg and 44kg of water spheres seperated by 59 m a part

in general 1 kg of mass of water contians equal number of positive and negative charges

if the postive charge is in excess with negative charge by 0.01% = 1 / 10,000

water is made up of two hydrogen atoms for every one oxygen atom (H₂O)

the molecular mass of water is 18.0153 atomic mass units, so 1 mole of water molecules has a mass of 18.0153 grams.

18.0153 gms of water contains 6.022 * 10^ 23 atoms ,

each atom contains equal number of positive and negative charges

for our problem we consider the difference betwen positive and negative charge is 0.01 % ,

ie for every 10,000 positive and negative charge 1 postive is in excess in 90 kg water sphere

simillarly for every 10,000 positive and negative charge 1 negative is in excess in 44 kg water sphere

number of positive or negative charge in 18.0153 gms of water = 6.022 *10^23

number of positive charge in 1 kg of water = (6.022 /18.0153) * 10^26

                                                            = 3.342714 * 10^26 positive charge

number of postive charges excess in every 1 kg = 3.342714 * 10^26 / 10,000

                                                                                   = 3.34271 * 10^22

number of positive charges excess in 90 kg = 90 * 3.34271 * 10^22

charge on 90 kg postive ions q1 = 90 * 3.34271 * 10^22 * q             q = 1.602 * 10^-19 C

simillarly number of negative charges excess in 44 kg = 44 * 3.34271 * 10^22 = q2

              charge on 44 kg negative ions q2 = 44 * 3.34271 * 10^22 * -q    , q = -1.602 * 10^-19 C

lets calculate the force of replusion between 90kg and 44kg water spheres

   F = k q1 * q2 / r ^2    ,k = 8.987 *10^9 N .m^2 C-1

       = (8.987 *10^9 ) * (90 * 3.34271 * 10^22 ) * (44 * 3.34271 * 10^22 ) * q ^2 / 59 ^2

      F = 2.93 * 10^17 N