Electric Fields


by Mitchtwitchita
Tags: electric, fields
Mitchtwitchita
Mitchtwitchita is offline
#1
Jun16-10, 01:16 AM
P: 191
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?
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rock.freak667
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#2
Jun16-10, 01:30 AM
HW Helper
P: 6,210
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 ?
Mitchtwitchita
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#3
Jun16-10, 01:37 AM
P: 191
weight?

rock.freak667
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#4
Jun16-10, 02:30 AM
HW Helper
P: 6,210

Electric Fields


Quote Quote by Mitchtwitchita View Post
weight?
Right, so EQ=mg. What is E?
Mitchtwitchita
Mitchtwitchita is offline
#5
Jun16-10, 02:48 AM
P: 191
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?
rock.freak667
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#6
Jun16-10, 02:30 PM
HW Helper
P: 6,210
Yes that is correct.


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