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UkoKoshka [18]
3 years ago
7

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

ic force in the earth’s electric field, which is typically 100 N/C, directed downward.
a) What is the ratio of the electric force on the bee to the bee's weight?
b) What electric field strength would allow the bee to hang suspended in the air? Express your answer with the appropriate units.
Physics
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

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t = 5.05 s

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This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

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