Answer: 27.21 V
Explanation:
The electric potential [tex]V_{E}[/tex] due to a point charge is expressed as:
[tex]V_{E}=k\frac{q}{r}[/tex]
Where:
[tex]k=9(10)^{9}\frac{Nm^{2}}{C^{2}}[/tex] is the electric constant
[tex]q=1.6(10)^{-19}C[/tex] is the electric charge of the hydrogen nucleus, which is positive
[tex]r=5.292(10)^{-11}m[/tex] is the distance
Rewritting the equation with the known values:
[tex]V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}[/tex]
Finally:
[tex]V_{E}=27.21 V[/tex]
