gre Posted January 4, 2009 Share Posted January 4, 2009 How is it known the proton and electron have equal charge magnitudes? Link to comment Share on other sites More sharing options...
Klaynos Posted January 4, 2009 Share Posted January 4, 2009 Experimentation. Take a hydrogen atom, measure it's charge. Remove an electron, they are now both charged, the only way this is possible is if they are equal. Link to comment Share on other sites More sharing options...
gre Posted January 4, 2009 Author Share Posted January 4, 2009 Well, how is an electron, proton, or hydrogen atom's charge measured? Link to comment Share on other sites More sharing options...
Klaynos Posted January 4, 2009 Share Posted January 4, 2009 Well you can tell if something is charged, and whether it is positive or negative charge by passing it threw a magnetic field and seeing what its path is. Charged particles will curve. So if you detect them afterwards you will get distinct regions based on mass and charge. You can also measure the charge of an electron independently of mass using an oil drop experiment: http://www.juliantrubin.com/bigten/millikanoildrop.html Link to comment Share on other sites More sharing options...
ydoaPs Posted January 4, 2009 Share Posted January 4, 2009 Well you can tell if something is charged, and whether it is positive or negative charge by passing it threw a magnetic field and seeing what its path is. Charged particles will curve. So if you detect them afterwards you will get distinct regions based on mass and charge. [math]F=q(B{\times}V)[/math] where F is force, q is charge, B is magnetic field, and V is velocity. Right? Link to comment Share on other sites More sharing options...
swansont Posted January 4, 2009 Share Posted January 4, 2009 [math]F=q(B{\times}V)[/math] where F is force, q is charge, B is magnetic field, and V is velocity. Right? All except a sign. It's v X B Link to comment Share on other sites More sharing options...
gre Posted January 4, 2009 Author Share Posted January 4, 2009 When you apply F = q(v x B) to say the hydrogen atom.. What is the V, Since electrons don't actually fly around the proton? Link to comment Share on other sites More sharing options...
Klaynos Posted January 4, 2009 Share Posted January 4, 2009 (edited) When you apply F = q(v x B) to say the hydrogen atom.. What is the V, Since electrons don't actually fly around the proton? You apply it to the bound system as a whole so the whole atom moving through the field. Edited January 4, 2009 by Klaynos Link to comment Share on other sites More sharing options...
gre Posted January 4, 2009 Author Share Posted January 4, 2009 You apply it to the bound system as a whole so the whole atom moving threw the field. I got a little off topic. But how does "V" in the Lorentz force apply to the hydrogen atom.... ? Well you can tell if something is charged, and whether it is positive or negative charge by passing it threw a magnetic field and seeing what its path is. Charged particles will curve. So if you detect them afterwards you will get distinct regions based on mass and charge. http://www.juliantrubin.com/bigten/millikanoildrop.html Don't magnetic fields accelerate or decelerate particles as they pass through? Link to comment Share on other sites More sharing options...
Klaynos Posted January 4, 2009 Share Posted January 4, 2009 I got a little off topic. But how does "V" in the Lorentz force apply to the hydrogen atom.... ? The v is the velocity of the atom relative to the field. Don't magnetic fields accelerate or decelerate particles as they pass through? Yes, that is why the particle curves, the acceleration is perpendicular to the direction of movement ([math]\bold{ v \times B}[/math]). Link to comment Share on other sites More sharing options...
gre Posted January 4, 2009 Author Share Posted January 4, 2009 The v is the velocity of the atom relative to the field. How is the movement (velocity) of the electron described? Yes, that is why the particle curves, the acceleration is perpendicular to the direction of movement ([math]\bold{ v \times B}[/math]). Thanks. Link to comment Share on other sites More sharing options...
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