Question

A 60 kg person is standing on a thin, conductive disk (the person is wearing thick rubber shoes, so the person is electrically separate from the conductive disk). The disk is connected to a VGG and the person places a hand on an identical, but separate VGG. If each of the VGG are activated (turned on) at the same instant and both remove electrons at a rate 500 determine: (a) the amount of time the person would have to hold onto the VGG to float 1 mm above the conductive disk; (b) the amount of mass that the person will gain or lose during this time. Did the person gain or lose this mass

Answer 1

Answer:

a)   t = 3.2 10¹⁴ s , b)    Δm = 1,456 10⁻¹³ kg, mass decreases

Explanation:

In this exercise the removed electrons leave a positive charge on each disk, as the two charges are of the same sign they repel each other, so using Newton's equilibrium equation

                    $F_{e}$ –W = 0

                    F_{e} = W = mg

Electric force is

                    F_{e} = k q₁ q₂ / r²

Since the two generators remove 500 e/s, q₁ = q₂ = q

                    F_{e} = k q² / r²

                   k q² / r² = m g

                   q = √ m g r²/ k

                   q = √ (60 9.8 0.001² / 8.99 10⁹)

                   q = √ 6.54 10⁻⁴

                   q = 2.56 10⁻² C

This is the charge on each disk

                  q = # _electron  e

                  # _electron = q / e

                  # _electron = 2.56 10⁻² / 1.6 10⁻¹⁹

                  # _electron = 1.6 10¹⁷  electrons

Let's use a rule of proportions to find the accumulation time, if the rate is 500 e/s

                  t = 1.6 10¹⁷  1/500

                  t = 3.2 10¹⁴ s

b) how electrons are losing their mass decreases

                 Δm = me # _electron

                 Δm = 9.1 10⁻³¹  1.6 10¹⁷

                 Δm = 1,456 10⁻¹³ kg