Question

A 0.10 g honeybee acquires a charge of +23 pC while flying, due to friction of the wings against the air. We will try and understand why this is advantageous. A. The electric field near the surface of the earth is typically 100 N/C, directed downward. What is the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight?

Answer 1

The ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶

Step-by-step explanation:

Given;

mass of the honeybee, m = 0.1 g

charge acquired by the honeybee, Q = 23pC = 23 x 10⁻¹² C

the electric field near the earth's surface, E = 100 N/C

The magnitude of the electric force on the bee is given by;

F = QE

F = (23 x 10⁻¹²)(100) = 23 x 10⁻¹⁰ N

The weight of the bee is given by;

W = mg

W = 0.1 x 10⁻³ x 9.8

W = 9.8 x 10⁻⁴ N

The the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is given by;

`(F)/(W) = (23*10^(-10))/(9.8*10^(-4)) = 2.35 *10^(-6)`

Therefore, the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶