Problem 23.27 Honeybees acquire a charge while flying due to friction with the a
ID: 1529554 • Letter: P
Question
Problem 23.27
Honeybees acquire a charge while flying due to friction with the air. A 110 mg bee with a charge of + 13 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
Part A
What is the ratio of the electric force on the bee to the bee's weight?
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Part B
What electric field strength would allow the bee to hang suspended in the air?
Express your answer with the appropriate units.
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Part C
What electric field direction would allow the bee to hang suspended in the air?
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Problem 23.27
Honeybees acquire a charge while flying due to friction with the air. A 110 mg bee with a charge of + 13 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.
Part A
What is the ratio of the electric force on the bee to the bee's weight?
F/W =SubmitMy AnswersGive Up
Part B
What electric field strength would allow the bee to hang suspended in the air?
Express your answer with the appropriate units.
E =SubmitMy AnswersGive Up
Part C
What electric field direction would allow the bee to hang suspended in the air?
What electric field direction would allow the bee to hang suspended in the air? upward downwardSubmitMy AnswersGive Up
Explanation / Answer
A)F(electric) = q E
F(electric) = 13 x 10^-12 x 100 = 13 x 10^-10 N
F(gravity) = mg = 0.00011 x 9.81 = 0.00108 N
Ratio = F(electric)/F(gravity) = 13 x 10^-10/0.00108 = 1.21 x 10^-6
Hence, F(electric)/F(gravity) = 1.21 x 10^-6
B)For this to occur,
e E = m g
E = mg/e = 0.00108/13 x 10^-12 = 8.31 x 10^7 N/C
Hence, E = 8.31 x 10^7 N/C
C)The direction of E field should be upwards to balance the downward gravitational force.
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