Answer to Question #103722 in Physics for issac
2020-02-24T13:18:25-05:00
Calculate the magnitude of Electrostatic force between two human beings
each having mass 100Kg and are 1m apart with 1% excess of electron.
(Molecular mass of H2O= 18gms)
Assume Human Body is made up of 100% water.
1
2020-02-25T10:24:14-0500
N = m μ N A = 100 0.018 6.02 ⋅ 1 0 23 = 3.34 ⋅ 1 0 27 N=\frac{m}{\mu }N_A=\frac{100}{0.018 }6.02\cdot 10^{23}=3.34\cdot 10^{27} N = μ m N A = 0.018 100 6.02 ⋅ 1 0 23 = 3.34 ⋅ 1 0 27
Q = 0.01 N e = 0.01 ( 3.34 ⋅ 1 0 27 ) ( 1.6 ⋅ 1 0 − 19 ) = 5.34 ⋅ 1 0 6 C Q=0.01Ne=0.01(3.34\cdot 10^{27})(1.6\cdot 10^{-19})=5.34\cdot 10^{6}C Q = 0.01 N e = 0.01 ( 3.34 ⋅ 1 0 27 ) ( 1.6 ⋅ 1 0 − 19 ) = 5.34 ⋅ 1 0 6 C
F = k Q 2 r 2 F=k\frac{Q^2}{r^2} F = k r 2 Q 2
F = 9 ⋅ 1 0 9 ( 5.34 ⋅ 1 0 6 ) 2 1 2 = 2.6 ⋅ 1 0 23 N F=9\cdot 10^{9}\frac{(5.34\cdot 10^{6})^2}{1^2}=2.6\cdot 10^{23}N F = 9 ⋅ 1 0 9 1 2 ( 5.34 ⋅ 1 0 6 ) 2 = 2.6 ⋅ 1 0 23 N
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