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Electrostatic Force Question

  • Thread starter craig.16
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  • #1
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Homework Statement


During a collision between two initially uncharged polystyrene balls of radius 2 mm, some electrons are transferred from one of the balls to the other. In what direction does the resulting electrostatic force between the balls act?

The magnitude of the force between the balls is 0.02 N when their centres are separated by 7 mm. Assuming that the charge is uniformly distributed over the surface of each of the balls, determine the number of electrons that was transferred from one ball to the other in the collision.

As polystyrene does not conduct electricity, it is likely that the charge is in fact concentrated at the points on the two balls which came into contact in the collision. Determine the number of electrons transferred in the collision under this assumption, assuming also that the balls are free to rotate.


Homework Equations


F=(1/4(pi)e0)*(q1q2/r2)


The Attempt at a Solution


For the first part I put the resulting electrostatic force between the balls is acting inward since one ball is now positively charged after losing an electron whereas the other is negatively charged as its gained an electron. Is this correct?

For the 2nd part I used the equation above to get:
0.02=(8.99*109)(q2/(7*10-3)2)
q2=1.09*10-16
q=1.04*10-8
no of electrons=(1.04*10-8)/(1.6*10-19)=6.53*1010
is this correct?

Finally for the 3rd part I havent got a clue where to start so any hints on how to begin it would be grateful, thanks.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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well for I and II part you are right!!!!

For the third part even i have no idea, i'll try and get back on this!!!
 
  • #3
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Good to know I have the first two parts right at least thanks. :smile:
 
  • #4
gneill
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The charge on each ball will be on the surface in one location, in some small area (where contact was made). What does that tell you about the distance between the charges for this situation?
 
  • #5
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The distance would be zero but that cant work out because you cant divide by zero.
 
  • #6
gneill
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The distance would be zero but that cant work out because you cant divide by zero.
Really? How do you work that out from the given information?
 
  • #7
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Honestly It was first instinct barely any thought put into it. The only thing I can think of is a distance of 4mm as its the total of both radius of each ball. But that still doesnt make sense considering its where contact was made and the center of each ball is clearly not where contact was made. I know this answer is probably really obvious but for some reason I cant think of what it will be, sorry.
 
  • #8
gneill
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Honestly It was first instinct barely any thought put into it. The only thing I can think of is a distance of 4mm as its the total of both radius of each ball. But that still doesnt make sense considering its where contact was made and the center of each ball is clearly not where contact was made. I know this answer is probably really obvious but for some reason I cant think of what it will be, sorry.
To quote the original problem, "their centres are separated by 7 mm". Their radii are 2mm each. The distance between their surfaces is therefore...?
 
  • #9
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Thanks for that. I knew I had a problem with over complicating the working out and not thinking about the basics but I didn't know I was this bad. I'll have to remind myself for future reference to start thinking at the most basic level and work my way up instead of just assuming its harder than what it actually might turn out to be.

o the distance between the surfaces is 3mm and I need to work out how many electrons are transferred. Do I simply use the same equation (couloumbs law) with the same values only the radius is now 3mm instead of 7mm and one of the charges is the answer I got in the first part to work out what the other charge is for the other ball? Then do the same process as in part 2 to work out the number of electrons?
 
  • #10
gneill
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o the distance between the surfaces is 3mm and I need to work out how many electrons are transferred. Do I simply use the same equation (couloumbs law) with the same values only the radius is now 3mm instead of 7mm and one of the charges is the answer I got in the first part to work out what the other charge is for the other ball? Then do the same process as in part 2 to work out the number of electrons?
Think of it as a whole new problem. You've got distance and force, and you want to find the charges. The charges are equal and opposite.
 
  • #11
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Just had another go at this question today and worked out where I went wrong on the last part. Thanks for the help though gneill:smile:
 

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