Question

An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group of 23 electrons be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?

Answer 1

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

$\frac{kq_1q_2}{r^2} = mg$

here we have

$q_1 = e$

$q_2 = 23 e$

$m = 9.11 \times 10^{-31} kg$

also we know that

$e = 1.6 \times 10^{-19} C$

so we will have

$\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)$

$\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}$

$r = 24.35 m$

so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.