Respuesta :
Answer:
[tex]2.6\times 10^{-5}[/tex]
Explanation:
Given:
- [tex]m[/tex] = mass of the honeybee = 0.09 g = [tex]9\times 10^{-5}\ kg[/tex]
- [tex]q[/tex] = charge on the honeybee = 23 pC = [tex]2.3\times 10^{-11}\ C[/tex]
- [tex]E[/tex] = electric field near the surface of earth = 100 N/C
Assume:
- [tex]g[/tex] = acceleration due to gravity = [tex]9.8\ m/s^2[/tex]
- [tex]W[/tex] = weight of the honeybee
- [tex]F[/tex] = electric force on the honeybee
- [tex]R[/tex] = ratio of the electric force and the weight of the honeybee
We know that
[tex]F = qE\,\,\, and\,\,\, W = mg\\\therefore R = \dfrac{F}{W}\\\Rightarrow R = \dfrac{qE}{mg}\\\Rightarrow R = \dfrac{2.3\times 10^{-11}\ C\times 100 N/C}{9\times 10^{-5}\ kg\times 9.8\ m/s^2}\\\Rightarrow R = 2.6\times 10^{-6}[/tex]
So, the ration of the electric force on the bee to its weight is [tex]2.6\times 10^{-6}[/tex].
On multiplying this ration by 10, the ratio becomes [tex]2.6\times 10^{-5}[/tex].
