Answer:
[tex]8.4\cdot 10^{-19}C[/tex]
Explanation:
The electron remains suspended is the electric force acting on it (upward) is equal in magnitude to the force of gravity (downward) acting on it.
The electric force on the electron is given by:
[tex]F_E=\frac{keq}{r^2}[/tex]
where:
k is the Coulomb's constant
[tex]e=1.6\cdot 10^{-19}C[/tex] is the magnitude of the electric charge
q is the unknown positive charge
r = 11.62 m is the distance between the electron and the charge
The force of gravity on the electron is
[tex]F_G=mg[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Equating the two forces and solving for q, we find:
[tex]\frac{keq}{r^2}=mg\\q=\frac{mgr^2}{ke}=\frac{(9.11\cdot 10^{-31})(9.8)(11.62)^2}{(9\cdot 10^9)(1.6\cdot 10^{-19})}=8.4\cdot 10^{-19}C[/tex]
