Calculating point charges and ratio of electrostatic force to weight

AI Thread Summary
A honeybee in active flight can acquire an electrostatic charge of approximately 93 pC, which requires the transfer of about 5.8e11 electrons. When two bees with this charge are separated by 12 cm, the magnitude of the electrostatic force between them is calculated to be 5.399e-9 N. The ratio of this electrostatic force to the weight of a 0.14 gm bee was attempted but initially calculated incorrectly as 3.856e-13. It was clarified that electrostatic force and weight are fundamentally different forces, although both follow the inverse square law. Understanding this distinction is crucial for accurate calculations in related physics problems.
zooboodoo
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Homework Statement


Measurements show that a honeybee in active flight can acquire an electrostatic charge on the order of 93 pC.

a)How many electrons much be transferred to produce this charge?
Ne = *
b) Supposing two bees, both with this charge, are separated by a distance of 12 cm. What is the magnitude of the electrostatic force between the these two bees? (You may treat the bees as point charges.)
FE= N *
5.399e-9 OKc) What is ratio of this electrostatic force to the weight of 0.14 gm bee?
FE/Fweight =
3.783e-7 NO


Homework Equations





The Attempt at a Solution



I got through the first two parts, but in order to calculate the ratio of the electro static force to the weight I simply tried Fe/Fweight, 5.39e-9 / .00014 = 3.856e-13
Is there a specific relationship between the forces that I should know before trying to proceed?
thanks
 
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zooboodoo said:
Is there a specific relationship between the forces that I should know before trying to proceed?
Nope. They are completely different kinds of force (except for the fact that, in freshman physics, they both obey the inverse square law).
 
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