Question (a)
To calculate the electrostatic force, Coulomb's Law is applied.
"F_{E}=K\\frac{q_{1}q_{2}}{r^{2}}"
Where:
Numerically evaluating
"F_{E}=9*10^{9}Nw\\frac{m^{2}}{C^{2}}*\\frac{-1*10^{-16}C*-1*10^{-16}C}{(0.01m)^{2}}\\\\ F_{E}=9*10^{-19}Nw"
The electrostatic force between the two drop is
"\\boxed{F_{E}=9*10^{-19}Nw}"
Question (b)
The charge of an electron is:"e^{-}=-1.60*10^{-19}C"
The net charge in each drop is "q=-1*10^{-16}C"
Therefore the number of electrons in each sphere is
"N_{e^{-}}=\\frac{-1*10^{-16}C}{-1.60*10^{-19} C}=625electrons"
The number of electrons in each drop is.
"\\boxed{N_{e^{-}}=625electrons}"
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