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Electric Fields 
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#1
Jun1610, 01:16 AM

P: 190

1. The problem statement, all variables and given/known data
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? 2. Relevant equations E = F on q/q+ 3. 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? 


#2
Jun1610, 01:30 AM

HW Helper
P: 6,207

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/s^{2} 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 ? 


#3
Jun1610, 01:37 AM

P: 190

weight?



#4
Jun1610, 02:30 AM

HW Helper
P: 6,207

Electric Fields



#5
Jun1610, 02:48 AM

P: 190

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? 


#6
Jun1610, 02:30 PM

HW Helper
P: 6,207

Yes that is correct.



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