What is the speed of the moving charge when it's 1.0cm from the fixed charge?

AI Thread Summary
The discussion centers on calculating the speed of a moving charged particle when it is 1.0 cm away from a fixed charge. Initial calculations using Coulomb's law to find the force and then applying Newton's second law led to confusion due to an imaginary solution. Participants suggested using energy conservation principles instead, as the acceleration is not constant in this scenario. One user reported arriving at a speed of 402.5 m/s, while another calculated around 300 m/s, highlighting discrepancies in their approaches. The conversation emphasizes the importance of correctly applying physics principles to solve the problem effectively.
pat666
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Homework Statement


Two identical 25g particles each carry 5.0uC of charge. One is held fixed, and the other is placed 1.0mm away and released
Find the speed of the moving charge when it's 1.0cm from the fixed charge

Homework Equations





The Attempt at a Solution


I thought that using F=Q1Q2K/d^2 I could find the repelling force which is 2.25*10^5 N. then F=ma so a=2.25*10^5/0.025 =9*10^6m/s^2. then using kinematics v^2 =u^2+2as
0=u^2+2*9*10^6*0.009 the trouble is that that is unsolvable (imaginary solution) Its extremely possible that my procedure is entirely wrong, I was just making it up on the fly... Help!
 
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pat666 said:

The Attempt at a Solution


I thought that using F=Q1Q2K/d^2 I could find the repelling force which is 2.25*10^5 N. then F=ma so a=2.25*10^5/0.025 =9*10^6m/s^2. then using kinematics v^2 =u^2+2as
0=u^2+2*9*10^6*0.009 the trouble is that that is unsolvable (imaginary solution) Its extremely possible that my procedure is entirely wrong, I was just making it up on the fly... Help!

The charge is held and then released, meaning that the initial velocity is zero.
 
argh thanks - stupid little mistake
 
Hey, I solved it but the answer I got (402.5m/s) is still wrong?
 
pat666 said:
Hey, I solved it but the answer I got (402.5m/s) is still wrong?

Recheck your force calculation, I don't think your force should be that high (in the order of 105)
 
I rechecked it and got the same F=9E9*5E-6*5E-6/.001^2 = 2.25E5?
 
pat666 said:
I rechecked it and got the same F=9E9*5E-6*5E-6/.001^2 = 2.25E5?

So you happen to know the correct answer?
 
no its for one of those stupid online tests
 
pat666 said:
no its for one of those stupid online tests

I ask as my calculation gives around 300 m/s.
 
  • #10
really that's significantly different to my answer - exact same procedure as me??
 
  • #11
I just tried it again from scratch and i keep getting 402m/s can you tell me how you did it please
 
  • #12
pat666 said:
really that's significantly different to my answer - exact same procedure as me??
Yes. Post what you did.
 
  • #13
ok for the Force i got 2.25*10^5 N (F=KQ^2/.001^2)
then a = 2.25*10^5/.025 = 9E6m/s^2
v^2=0+2*9E6*0.009
 
  • #14
v^2=u^2+2as only applies if the acceleration is constant. It's not. Use potential energy and energy conservation.
 
  • #15
yeah I figured out to use E=K+U
 

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