Determine the velocity of a particle in mass spectrometry

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To determine the velocity of a particle in mass spectrometry, the relationship between potential energy (Ep) and kinetic energy (Ec) must be correctly applied. The equation Ep = 0.5 (sigma(i) qiVi) is valid for multiple charges, while for a single particle in an external field, Ep simplifies to U = qV. This distinction is crucial to avoid misapplication of the formulas. Clarification is needed on the origin of the equation Ec(f) = Ec(i) - Ec(f), as it appears to be incorrectly stated. Understanding the context and correct application of these equations is essential for accurate calculations.
duchuy
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Homework Statement
V=3000V, B= 0,035 T, m = m = 2.10^(-26) kg, q = e
Relevant Equations
x
Hi, I am trying to determine the velocity of the particle with the mass m coming out of the acclerator.
I tried writing :
Ep(i) + Ec(i) = Ep(f) + Ec(f)
Ec(f) = Ec(i) - Ec(f)
But at this step, I'm no longer sure how to express Ep with V because :
In my textbook, it's written :
Ep = 0,5 (sigma(i) qiVi)
But I am not sure if I am supposed to put that 0,5 in this case or not, which would imply Ep = qV only.

Ep = potential energy
Ec = Kinetic energy
Sorry in advance if these aren't notations that are used in English.

Thank you so much for your help!
 
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duchuy said:
Homework Statement:: V=3000V, B= 0,035 T, m = m = 2.10^(-26) kg, q = e
Relevant Equations:: x

Ep = 0,5 (sigma(i) qiVi)
This is true for a collection of charges where the potentials are due to the same charges you are considering. This is to avoid double counting because in actuality the potentials are between pairs of charges. The text you are reading should explain this.

If you are considering a single particle in an external field, then potential energy is U = qV. This is an example of the need to understand formulae that you find and to make sure they are applicable.
 
Oh okay thank you sir!
 
A diagram would help in understanding what the symbols and the formulae mean.
I am not sure where the second formula
Ec(f) = Ec(i) - Ec(f), comes from.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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