Answer:
the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.
Explanation:
As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron
So we have
[tex]\frac{kq_1q_2}{r^2} = mg[/tex]
here we have
[tex]q_1 = e[/tex]
[tex]q_2 = 23 e[/tex]
[tex]m = 9.11 \times 10^{-31} kg[/tex]
also we know that
[tex]e = 1.6 \times 10^{-19} C[/tex]
so we will have
[tex]\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)[/tex]
[tex]\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}[/tex]
[tex]r = 24.35 m[/tex]
so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.
