Question 21.66: An electron is in a vacuum near Earth’s surface and located ......
An electron is in a vacuum near Earth’s surface and located at y = 0 on a vertical y axis. At what value of y should a second electron be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?
With F = m _{e} g, Eq. 21-1 leads to
\vec{F}=k \frac{q_1 q_2}{r^2} \hat{ r } (Coulomb’s law), (21-1)
y^2=\frac{k e^2}{m_e g}=\frac{\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right)\left(1.60 \times 10^{-19} \, C \right)^2}{\left(9.11 \times 10^{-31} \,kg \right)\left(9.8 \,m / s ^2\right)}
which leads to y = ± 5.1 m. We choose y = -5.1 m 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.