Magnitude of charge on oil drop (electric field)

  • Thread starter kbyws37
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  • #1
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A positively charged oil drop of mass 1.00 × 10−15 kg is placed in the region of a uniform electric field between two oppositely charged, horizontal plates. The drop is found to remain stationary under the influence of the Earth's gravitational field and the uniform electric field of 0.680 × 104 N/C. What is the magnitude of the charge on the drop?



I am not sure where to start. I am given a mass so would I start with

U = -(G(m1)(m2))/r

Thanks in advance
 

Answers and Replies

  • #2
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If the drop is stationary, what can you say about the net force on the drop?
 
  • #3
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would force = 0?
 
  • #4
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do you mean net force?
 
  • #5
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Sorry, I'm still confused.
Wouldn't gravity be the force that allows it to drop?
Which equation would be good to use?
 
  • #6
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The force of gravity acts down on the drop, but the drop is charged and it sits (stationary) between two charged plates. Since it is stationary, what can you say about the acceleration and net force on it?
 
  • #7
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There is no acceleration and thus net force = 0
 
  • #8
340
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Good, now what two forces act on the drop?
 
  • #9
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gravity and drag force?
 
  • #10
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Drag force??
A positive charge is in an electric field.
 
  • #11
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gravity and the weight of the drop
 
  • #12
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those are the same character!(force)
I gave you a big hint.
 

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