Electric Fields near the surface of the earth

AI Thread Summary
The discussion focuses on calculating the electric force on a honeybee with a charge of +23 pC in an electric field of 100 N/C. The electric force on the bee is determined to be 2.3 x 10^-9 N, resulting in a ratio of the electric force to the bee's weight of 2.3 x 10^-5 N/kg. For the bee to hang suspended in the air, the electric field strength required is calculated to be approximately 4.3 x 10^7 N/C, directed upward. The solution emphasizes that the electric force must equal the bee's weight for suspension. Overall, the calculations demonstrate the relationship between electric fields and forces acting on charged objects.
Mitchtwitchita
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Homework Statement



A 0.10 g honeybee acquires a charge of +23 pC while flying.

(a) The electric field near the surface of the Earth is typically 100 N/C, directed downward. What is the ratio of the electric force on the bee to the bee's weight?

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


Homework Equations



E = F on q/q+


The Attempt at a Solution



(a) E = fon q/q+
Therefore, F = Eq
= (100 N/C)(23 x 10^-12 C)
= 2.3 x 10^-9 N

Therefore, (2.3 x 10^-9)/(1.0 x 10^-4 kg)
= 2.3 x 10^-5 N/kg

(b) I don't really know how to get this one started. I think it would be directed upward though. Can someone please help me with part b and let me know if I was correct on part a?
 
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For a) you want to get a ratio.

weight = mg. So you need to multiply 1.0 x 10-4 kg by 9.81 m/s2

For b), if the bee is suspended, the resultant force is zero. So the electric force should be the same as the bee's what ?
 
weight?
 
Mitchtwitchita said:
weight?

Right, so EQ=mg. What is E?
 
Ah. Therefore E = mg/q
=[(1.0 x 10^-4 kg)(9.81 m/s)]/(23 x 10^-12 C)
=4.3 x 10^7 N/C?
 
Yes that is correct.
 
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