Problem 1. Electric force vs gravitational force. (a) Find the ratio of the elec
ID: 251439 • Letter: P
Question
Problem 1. Electric force vs gravitational force. (a) Find the ratio of the electric force to the gravitational force for the electron and proton in a hydrogen atom. Is gravitation important for understanding atomic structure? (b) Suppose that the Earth and the moon each had a net charge of Q. What value of Q would be necessary for the resulting electric repulsion to exactly cancel their gravitational attraction? If the charge were due to an excess of electrons, how many electrons would this require (on each body)? How much would these electrons weight?
Explanation / Answer
a) Fe = k*q^2/r^2
= 9*10^9*(1.6*10^-19)^2/(5.29*10^-11)^2
= 8.23*10^-8 N
Fg = G*me*mp/r^2
= 6.67*10^-11*9.1*10^-31*1.67*10^-27/(5.29*10^-11)^2
= 3.62*10^-47
Fe/Fg = 8.23*10^-8/(3.62*10^-47)
= 2.27*10^39
here, Fe >> Fg. so, gravitation is not important for understanding atomic structure
b)
Apply, G*Me*Mm/d^2 = k*Q^2/d^2
G*Me*Mm = k*Q^2
==> Q = sqrt(G*Me*Mm/k)
= sqrt(6.67*10^-11*5.98*10^24*7.35*10^22/(9*10^9))
= 5.707*10^13 C
c) no of elctrons that the earth and the moon should have, N = Q/e
= 5.707*10^13/(1.6*10^-19)
= 3.57*10^32 elctrons
d) weight of electrons, m = N*me
= 3.57*10^32*9.1*10^-31
= 324.9 kg
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