More on Balancing Charges

A 100 pound boy and an 80 pound girl standing 2 meters apart suddenly have the number of electrons in each of their bodies increase by 1/100 of 1%. (a) What is the electrostatic force due to the unbalanced charge on each member of the pair? (b) What is the ratio of the electrostatic force between them to the gravitational force of the Earth on the boy? Assume the human body is composed entirely of water molecules (18 AMU with 10 electrons) and ignore the mass of electrons.

 
(a)  100 lb male x 0.454 Kg/lb = 45.4 Kg

       18 AMU/H2O x 1.67 x 10-27 Kg/AMU = 3.01 x 10-26 Kg/H2

       \ # H2O molecules in male = 45.4 Kg/3.01 x 10-26 Kg/H2O

                        = 1.51 x 1027 H2O molecules

      Each water molecule has 10 e-
       \ # e- in male = 1.51 x 1027 H2O mol. X 10 e-/mol

                               = 1.51 x 1028 e-

      1/100 of 1% = 0.01 x 0.01 = 10-4

       \ # e- added to male = 1.51 x 1028 e- x 10-4 = 1.51 x 1024 e-

    total unbalanced charge of male = -1.6 x 10-19 C/e- x 1.51 x 1024 e-

                        =  2.42 x 105 C         [where C=coulombs]

The unbalanced charge on the 80 pound female will simply scale as the ratio of their masses. 
\ total unbalanced charge of female = 80 lb female   x  2.42 x 105 C 
                                                                        100 lb male

                        = 1.93 x 105 C

\ electrostatic force between them at 2 meters:

FE =   KQ1Q2    = (9 x 109 N x m2/C2) x (2.42 x 105 C) x (1.93 x 105 C)
              R2                                         (2 m)2

     = 1.05 x 1020 N


(b) Fg = M1g = 45.4 Kg x 9.8 m/s2 

          = 444 N         [where N is Newtons]

Hence:


     FE = 1.05 x 1020 N    = 2.36 x 1017
      Fg           444 N

 

The conclusion is that, under certain circumstances, it is very important that electrical charges balance to extremely high accuracy. Otherwise electrical forces would easily overwhelm even the pull of Earth's gravity.