Question

An electron is in a vacuum near Earth’s surface and located at$\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$on a vertical yaxis. At what value of yshould a second electronbe placed such that its electrostatic force on the first electron balancesthe gravitational force on the first electron?

Short answer

$y=-5.1m$

Step-by-step solution

Given

Electron is in a vacuum near Earth’s surface and located at y = 0

Understanding the concept

Use Coulomb’s law to solve the problem.

Calculate the value of y

with,$F=m_{e}g$

$y^2=\frac{k{e}^2}{m_{e}g}=\frac{(9.00\times {10}^{9}N{m}^2/C^2){(1.60\times {10}^{-19}C)}^2}{(9.11\times {10}^{-31}kg)(9.8m/s^2)}$

Which leads to$y=\pm 5.1m$ . We choose $y=-5.1m$since the second electron must be below the first one, so that the repulsive force (acting on the first) is in the direction opposite to the pull of Earth’s gravity.