physic 難題 11

(a) I presume that the force here refers to electrostatic force, not gravitational force.

Charge on an electron = 1.6 x 10^ -19 C
Hence, force between two electrons 0.7 m apart
= (9x10^9).(1.6x10^-19)^2/0.7^2 N = 4.702x10^-28 N

(b) Let n be the number of electrons in the 70 kg human body, hence no. of protons = no. of neutrons = n
Total mass of electrons = (9.1x10^-31)n
Total mass of protons plus neutrons = 2n.(1.67x10^-27)
Thus, 70 = (9.1x10^-31)n + 2n.(1.67x10^-27)
n = 2.095x10^28
Mass of 2% of electrons = 0.02 x (2.095x10^28) x (9.1x10^-31) kg
= 3.8x10^ -4 kg = 0.38 g

(c) Using the value of n from (b) and the force calculated from (a),
excessive force of repulsion = 0.02 x (2.095x10^28) x (4.702 x 10^-28) N
= 0.197 N
It seems that the student's claim is not right.