Asked by Erin
Calculate the electrical attraction that a proton in a nucleus exerts on an orbiting electron if the two particles are 1.13×10^-10 m apart.
Answers
Answered by
GK
Use Coulomb's Law:
F = k*Q1*Q2 / r^2
k = 9.00x10^9 N.m^2/C^2 (Coulomb's constant)
Q1 = Q2 = 1.60x10^-19C (Elementary charge)
r = given in the question (distance)
<b>Note</b>: The charge of a proton is equal to the charge of an electron except for the algebraic sign.
F = k*Q1*Q2 / r^2
k = 9.00x10^9 N.m^2/C^2 (Coulomb's constant)
Q1 = Q2 = 1.60x10^-19C (Elementary charge)
r = given in the question (distance)
<b>Note</b>: The charge of a proton is equal to the charge of an electron except for the algebraic sign.
Answered by
erin
I tried that and got 1.8x10^8 but it wasn't the right answer
Answered by
bobpursley
You didn't do it right. Just doing the orders of magnitude only..
F=E9*E-19*E-19/E-20 ( I squared the distance)
F=E(9-38+20) appx 10^-9 Newtons.
F=E9*E-19*E-19/E-20 ( I squared the distance)
F=E(9-38+20) appx 10^-9 Newtons.
Answered by
Erin
what does E mean? Could you explain it another way because I'm confused.
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