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

`2.6* 10^(-5)`

Step-by-step explanation:

Given:

• 
  `m` = mass of the honeybee = 0.09 g =
  `9* 10^(-5)\ kg`

• 
  `q` = charge on the honeybee = 23 pC =
  `2.3* 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 = (F)/(W)\\\Rightarrow R = (qE)/(mg)\\\Rightarrow R = (2.3* 10^(-11)\ C* 100 N/C)/(9* 10^(-5)\ kg* 9.8\ m/s^2)\\\Rightarrow R = 2.6* 10^(-6)`

So, the ration of the electric force on the bee to its weight is
`2.6* 10^(-6)`.

On multiplying this ration by 10, the ratio becomes
`2.6* 10^(-5)`.