What is the Net Force on a Helium Nucleus Due to Proton and Electron Charges?

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SUMMARY

The net force on a helium nucleus due to a proton and an electron is calculated using Coulomb's law. The proton is positioned at x = 1.6 nm and the electron at y = 0.85 nm. The calculated forces are F_pHe = -0.180i nN from the proton and FeHe = 0.638j nN from the electron. The confusion arose from incorrect unit conversions, specifically using millimeters instead of nanometers, leading to significantly smaller force values.

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A proton is on the x-axis at x= 1.6nm. An electron is on the y-axis at y=0.85nm. Find the net force the two exert on a helium nucleus (charge + 2e) at the orgin.

there's a solution from the book, but i don't really understand it:
k = 9X10^9 N*m^2 and e=1.6*10^-19

Coulomb force of the proton on the helium nucleus is F_pHe = k(e)(2e)(-i)/(1.6mm)^2 = -0.180i*nN (i get 1.8X10^-28 instead)

for the y:
FeHe= k(-e)(2e)(-j)/(.85nm)^2 = 0.638j nN (i get 6.377 X 10^-28)

the answer would be -0.180i*nN + 0.638j nN right? i just don't know why my numbers are so small
 
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Whatupdoc said:
A proton is on the x-axis at x= 1.6nm. An electron is on the y-axis at y=0.85nm.
there's a solution from the book, but i don't really understand it:
k = 9X10^9 N*m^2 and e=1.6*10^-19
Coulomb force of the proton on the helium nucleus is F_pHe = k(e)(2e)(-i)/(1.6mm)^2 = -0.180i*nN (i get 1.8X10^-28 instead)...
i just don't know why my numbers are so small


The distances given was in nm-s (10^-9 m) but you used mm-s (10^-3). or you even have forgotten that 10^-3, too.

ehild
 
(9X10^9 N*m^2)(1.6*10^-19)(2(1.6*10^-19))(-i)/(1.6mm)^2 = 1.8X10^-28

right? i did use 10^9
 
no, for the distance you just used 1.6 m .
enter the distance into your calculator as (1.6E-9 [m])^2

and be sure to use the EXP button (or EE button) ,
NOT the ambiguous sequence / 1.6 x 10 ^-9 ^2 !
 

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