The magnitude of the electric field must be [tex]5.59\cdot 10^{-11} N/C[/tex]
Explanation:
In order for the electron to be in equilibrium, the force of gravity acting on the electron must be equal to the force due to the electric field.
The force of gravity on the electron located on the Earth's surface is:
[tex]F_G = mg[/tex]
where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron mass
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The force due to the electric field is
[tex]F_E = qE[/tex]
where
[tex]q=1.6\cdot 10^{-16}C[/tex] is the electron charge
E is the magnitude of the electric field
Since the two forces must be balanced,
[tex]F_G = F_E[/tex]
So we find:
[tex]mg=qE\\E=\frac{mg}{q}=\frac{(9.11\cdot 10^{-31})(9.81)}{1.6\cdot 10^{-19}}=5.59\cdot 10^{-11} N/C[/tex]
Learn more about gravitational force:
brainly.com/question/1724648
brainly.com/question/12785992
And electric force:
brainly.com/question/8960054
brainly.com/question/4273177
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