Calculating Honeybee's Electrostatic Charge: A Step-by-Step Guide

[hotstuff]

Measurements show that a honeybee in active flight can acquire an electrostatic charge as great as 93.0 pC.
Q1 is How many electrons must be transferred to produce this charge?
my proposed ans using the formula f=k e
after rearranging equation e= 93pC * ./8.99 *10^9 = ?(the ans)
but i do not know what pC unit is and not sure if doing right. pls point me in the right direction. thanks

 

[geosonel]

Measurements show that a honeybee in active flight can acquire an electrostatic charge as great as 93.0 pC.
Q1 is How many electrons must be transferred to produce this charge?
my proposed ans using the formula f=k e
after rearranging equation e= 93pC * ./8.99 *10^9 = ?(the ans)
but i do not know what pC unit is and not sure if doing right. pls point me in the right direction. thanks

$$\mbox{93 picoCoulombs} \ = \ 93 \, \times \, 10^{-12} \ Coulombs$$
$$\mbox{charge on 1 electron} \ = \ 1.602 \, \times \, 10^{-19} \ Coulombs$$
$$\mbox{number electrons transferred} \ = \ \frac{93 \, \times \, 10^{-12} \ Coulombs}{1.602 \, \times \, 10^{-19} \ Coulombs}$$

 

[thepatientmental]

The unit pC stands for picocoulombs, which is a unit of electric charge equal to 10^-12 coulombs. To solve this problem, we can use the formula Q = Ne, where Q is the total charge in coulombs, N is the number of electrons, and e is the elementary charge (1.6 x 10^-19 coulombs). So, to find the number of electrons needed to produce an electrostatic charge of 93.0 pC, we can rearrange the equation to N = Q/e. Plugging in the values, we get N = 93.0 pC / (1.6 x 10^-19 C) = 5.81 x 10^17 electrons. This means that approximately 581 trillion electrons must be transferred to produce an electrostatic charge of 93.0 pC on a honeybee in active flight. Hope this helps!