Answer:
[tex]4.9\cdot 10^{-6}[/tex]
Explanation:
The Coulomb force on the bee is:
[tex]F_E=qE[/tex]
where
[tex]q=40 pC=40\cdot 10^{-12} C[/tex] is the charge of the bee
[tex]E=120 N/C[/tex] is the magnitude of the electric field
Substituting into the formula,
[tex]F_E=(40\cdot 10^{-12} C)(120 N/C)=4.8\cdot 10^{-9} N[/tex]
The gravitational force on the bee is
[tex]F_G = mg[/tex]
where
[tex]m=0.10 g=1\cdot 10^{-4}kg[/tex] is the bee's mass
[tex]g=9.8 m/s^2[/tex] is the gravitational acceleration
Substituting into the formula,
[tex]F_G = mg=(1\cdot 10^{-4}kg)(9.8 m/s^2)=9.8\cdot 10^{-4} N[/tex]
So, the ratio between the two forces is
[tex]\frac{F_E}{F_G}=\frac{4.8\cdot 10^{-9} N}{9.8\cdot 10^{-4} N}=4.9\cdot 10^{-6}[/tex]
