Question

1.Suppose two bees, each with a charge of 93.7 pC, are separated by a distance of 1.21 cm. Treating the bees as point charges, with the magnitude of the electrostatic force experienced by the bees. Compare this force to the weight of a 0.112 g bee. By what factor is the force smaller than the bee? 

Answer 1

given charge of each bee=93.7 pc

1pc=10^-12coloumbs

distance between two bees=1.21cm=1.21/100=0.0121m

According to Columb's law F=1/4πε₀ * Q₁*Q₂/r²=[(9.0 x 10⁹ N•m²/C²)*(93.7*10^-12coloumbs)*(93.7*10^-12coloumbs)] / (0.0121m)² F=0.0000005397=5397*10^-10Newtons

here Q₁ and Q₂ are charges of bee, and r is the distance between the bee's

weight of bee with mass 0.122g=mass*g=0.122*9.8m/s²*10^-3=1.1956*10^-3Newtons

weight of the bee is greater than the electrostatic force between the two bees because the gravitational pull is greater than the force of attraction between two charged bees.