Answer:
The net force on the helium nucleus at the origin is equal to [tex]78.79\times 10^{-11}[/tex]
Explanation:
Here let the force exerted by the proton be [tex]F_{1}[/tex] and force exerted by electron be [tex]F_{2}[/tex].
⇒The net force becomes [tex]\sqrt{F1^{2}+F2^{2} }[/tex]
Now lets calculate [tex]F_{1}[/tex] and [tex]F_{2}[/tex]
F = [tex]\frac{KQq}{r^{2} }[/tex] ; where for [tex]F_{1}[/tex] r=[tex]1.2nm[/tex] and for [tex]F_{2}[/tex] r=[tex]O.8nm[/tex]
upon calculating we get [tex]F_{1}[/tex]=[tex]32\times 10^{-11}[/tex] and [tex]F_{2}[/tex]=[tex]72\times 10^{-11}[/tex]
⇒F = [tex]78.79\times 10^{-11}[/tex]