Respuesta :
Explanation:
Point charge = [tex]q_{1} = +3.00 \times 10^{-9} C[/tex]
[tex]q_{2} = +2.00 \times 10^{-9} C[/tex]
Separation by distance = s = 50.0 cm = 0.5 m (as 1 m = 100 cm)
Electric potential energy of electron at 0.10 m = [tex]-9 \times 10^{9} \times (1.6 \times 10^{-19}) \times 10^{-9}[\frac{3.00}{0.1} + \frac{2.00}{0.4}][/tex]
Electric potential energy of electron at 0.10 m = [tex]-5.04 \times 10^{-17}[/tex] J
Change in P.E. will be calculated as follows.
P.E = [tex]2.16 \times 10^{-17} J[/tex]
K.E = [tex]2.16 \times 10^{-17} J [/tex]
Therefore, we will calculate the speed as follows.
Speed (v) = [tex]\sqrt{\frac{2KE}{m}}[/tex]
v = [tex]\sqrt{[\frac{2 \times 2.16 \times 10^{-17}}{9.1 \times 10^{-31}}]}[/tex]
= [tex]6.890 \times 10^{6}[/tex] m/s
when electron is 10.0 cm from the +3.00 nC charge then its speed is [tex]6.890 \times 10^{6}[/tex] m/s.
