A 0.15g honeybee acquires a charge of 20pCwhile flying. The electric field near

Question

A 0.15g honeybee acquires a charge of 20pCwhile flying. The electric field near the surface of the earth is typically 100N/C , directed downward.

Part A

What is the ratio of the electric force on the bee to the bee's weight?

Express your answer using two significant figures.

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Part B

What electric field strength and direction would allow the bee to hang suspended in the air?

Express your answer using two significant figures.

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Part C

What would be the necessary electric field direction for the bee to hang suspended in the air?

What would be the necessary electric field direction for the bee to hang suspended in the air?

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Explanation / Answer

Part A

F = q .E( force/charge) = 20x10-9 C x 100N/C =2x10-6 N
W = 0.15 x10-3 x 9.8 = 1.47 x 10-3 N

F/W = 1.36 x 10-3

Part B
F = W
qE = mg

20x10-9 C x E = 0.15x10-3 x 9.8

E = 1.47 x 10-3 / 20x10-9 = 7.35 x104 N/C

Part C
=> Upward......positive charge...so, the electic field shall push upwards to hang in air...

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