Question

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04% imbalance in the amount of positive and negative charge, one student being positive and the other negative. Estimate the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.

Answer 1

`6.8370869499* 10^(20)\ N`

Step-by-step explanation:

`N_A` = Avogadro's number =
`6.022* 10^(23)`

e = Charge of electron =
`1.6* 10^(-19)\ C`

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

`q=imbalance* n* e`

Total number of protons and electrons in each sphere

`n=(mN_AZe)/(M)`

`q=imbalance* (mN_AZe)/(M)`

`q_1=0.0004* (96* 6.022* 10^(23)* 18* 1.6* 10^(-19))/(0.018)\\\Rightarrow q_1=3699916.8\ C`

`q_2=0.0004* (48* 6.022* 10^(23)* 18* 1.6* 10^(-19))/(0.018)\\\Rightarrow q_1=1849958.4\ C`

Electrical force is given by

`F=(kq_1q_2)/(r^2)\\\Rightarrow F=(8.99* 10^(9)* 3699916.8* 1849958.4)/(30^2)\\\Rightarrow F=6.8370869499* 10^(20)\ N`

The electrostatic force of attraction between them is
`6.8370869499* 10^(20)\ N`