Question

A 0.09 g honeybee acquires a charge of +23 pc while flying. The earth's electric field near the surface is typically (100 N/C, downward). What is the ratio of the electric force on the bee to the bee's weight? Multiply your answer by 10 before entering it below.

Answer 1

Answer:

$2.6\times 10^{-5}$

Explanation:

Given:

• $m$ = mass of the honeybee = 0.09 g = $9\times 10^{-5}\ kg$

• $q$ = charge on the honeybee = 23 pC = $2.3\times 10^{-11}\ C$

• $E$ = electric field near the surface of earth = 100 N/C

Assume:

• $g$ = acceleration due to gravity = $9.8\ m/s^2$

• $W$ = weight of the honeybee

• $F$ = electric force on the honeybee

• $R$ = ratio of the electric force and the weight of the honeybee

We know that

$F = qE\,\,\, and\,\,\, W = mg\\\therefore R = \dfrac{F}{W}\\\Rightarrow R = \dfrac{qE}{mg}\\\Rightarrow R = \dfrac{2.3\times 10^{-11}\ C\times 100 N/C}{9\times 10^{-5}\ kg\times 9.8\ m/s^2}\\\Rightarrow R = 2.6\times 10^{-6}$

So, the ration of the electric force on the bee to its weight is $2.6\times 10^{-6}$.

On multiplying this ration by 10, the ratio becomes $2.6\times 10^{-5}$.