Coulombs Law and electrostatic force

  • Thread starter mb85
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Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters. The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 8.9 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1, then to sphere 2, and finally removed. What is the magnitude of the electrostatic force F' that now acts on sphere 2?

so when sphere 3 touches sphere one, the charge is transfered by 1/2. so the force is then 4.45.
when sphere 3 then touches sphere 2 i thought u added the charges? so 4.45N + 8.9N? which is 13.35N?

i think im missing something...
 

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Andrew Mason
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mb85 said:
Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters. The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 8.9 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1, then to sphere 2, and finally removed. What is the magnitude of the electrostatic force F' that now acts on sphere 2?

so when sphere 3 touches sphere one, the charge is transfered by 1/2. so the force is then 4.45.
when sphere 3 then touches sphere 2 i thought u added the charges? so 4.45N + 8.9N? which is 13.35N?

i think im missing something...
The principle here is that the electrons in the spheres when in contact will move until they are all at equal potential. Since the spheres are identical, what does that tell you about the distribution of charge between the two touching spheres?

AM
 
  • #3
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simple concept i did not realize and i was jumping the gun and not establishing the charge distribution. But the overall charge(q) is 3/8. So then its 3/8 of the initial force. i got the answer. thanks!
 
  • #4
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i maybe missing something obvious, but i dont see why it is 3/8.
 

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