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

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Homework Help Overview

The problem involves two identical charged particles, one fixed and the other released from a distance of 1.0 mm, with the goal of finding the speed of the moving charge when it is 1.0 cm from the fixed charge. The subject area pertains to electrostatics and kinematics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating the force between the charges using Coulomb's law and then applying Newton's second law to find acceleration. Some express confusion over the results leading to imaginary solutions, while others question the accuracy of force calculations and the method used to determine speed.

Discussion Status

There is ongoing exploration of different approaches, including the use of energy conservation principles. Some participants have provided calculations and results, but discrepancies in answers suggest a lack of consensus on the correct method or values.

Contextual Notes

Participants note that the problem is part of an online test, which may impose additional constraints or expectations on the solution process. There is also mention of the initial velocity being zero when the charge is released.

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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