Answer:
[tex]v = 7793150 m/s[/tex]
Explanation:
First, we are going to calculate the electrical potential in the point middle between the two charges
Remember that the electrical potential can be calculated as:
[tex]v = \frac{kQ}{r}[/tex]
Where [tex]k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }[/tex]
and it is satisfy the superposition principle, thus
[tex]v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}[/tex]
[tex]v = 222.73v[/tex]
The electrical potential at 10 cm from charge 1 is:
[tex]v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}[/tex]
[tex]v = 395.44 v[/tex]
Since the work - energy theorem, we have:
[tex]q\Delta v = \frac{mv^{2} }{2}[/tex]
where q is the electron's charge and m is the electron's mass
Therefore:
[tex]v = \sqrt{\frac{2q\Delta v}{m} }[/tex]
[tex]v = 7793150 m/s[/tex]
