Relationship of particle charge to speed?

  • Thread starter temptasian
  • Start date
  • #1

Homework Statement



The four points shown in the attached picture are near a positively charged rod (shaded circle). Points W and Y are equidistant from the rod, as are points X and Z. A charged particle with mass 3.0 x 10^-8 kg is released from rest at point W and later observed to pass point X.
Suppose the magnitude of the charge is 2 microcoloumbs and that the speed of the particle is 40 m/s as it passes point X. Suppose that a second particle with the same mass as the first but NINE times the CHARGE were released from rest at point W. Would the speed of the second particle as it passes point X be greater than, less than, or equal to the speed of the first particle (with charge of 2 microcoulombs) as it passes point X?


Homework Equations


From KE = 1/2 (mv^2), I know that charge is not involved here, and since mass is the same, I believe the speed of the second particle is equal to the speed of the first particle.
However, I'm not sure if I should be using this equation to make a relationship between speed and charge. Other possible equations to use may be: U(potential energy) = qV, and W = -qEd = ΔU
Am I on the right track?

The Attempt at a Solution

 

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Answers and Replies

  • #2
BruceW
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yes, you need to use the 'other possible equations'. You've got ΔU=qΔV This is the change in potential energy, so what will be the change in kinetic energy? Also, is ΔV the same or different in the two situations?
 
  • #3
ΔV is constant for both situations. So if I use ΔU = qΔV, when charge increases, ΔU has to increase also, which means kinetic energy has decreased. Since KE has decreased, and we know that KE = 1/2 mv^2, the speed of the second particle will be less than the speed of the second particle. I hope I'm correct here?
 
  • #4
BruceW
Homework Helper
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Yes, the ΔV is the same in both situations.
temptasian said:
So if I use ΔU = qΔV, when charge increases, ΔU has to increase also, which means kinetic energy has decreased. Since KE has decreased, and we know that KE = 1/2 mv^2, the speed of the second particle will be less than the speed of the second particle.
But this isn't right. Think about the sign of ΔU. The particle is being pushed away from the rod, so how does the potential change when the particle moves away from the rod?
 

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