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Finding the surface charge density of plates in order to stop a proton!

  • Thread starter limonysal
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Homework Statement


You have a summer intern position at a laboratory that uses high speed proton beam. The protons exit the machine at a speed of 2.0*10^6 m/s [...] You decide to slow the protons to an acceptable speed (2.0*10^5 m/s), then let them hit a target. you take two metal plates, space them 2.0 cm apart, then drill a small hole through the center of one plate to let the proton beam enter. The opposite plate is the target in which the protons will embed themselves.

a. What are the minimum surface charge densities you need to place on each plate? Which plate, positive or negative, faces the incoming proton beam?

b. What happens if you charge the plates to +/- 1.0*10^-5 C.m^2? Does your device still work?


Homework Equations


W=F*d
E(electric field vector)= [tex]\eta[/tex]/[tex]\epsilon[/tex]0, from pos to neg)
F=Eq
vf^2=vi^2+2a[tex]\Delta[/tex]s
W=[tex]\Delta[/tex]U=[tex]\Delta[/tex]KE=0.5m(vf^2-vi^2)


The Attempt at a Solution


I'm not sure if I have the signs right with Work and Energy. Eventually I combined the equations to get:
[tex]\eta[/tex]=[m*(vf^2-vi^2)*[tex]\epsilon[/tex]0]/2qd

and d=0.02m (the distance between the plates)

I actually know that the answer iw [tex]\eta[/tex]=-9.1*10^-6 C/m^2. I know that the plate with the hole has to be the negative one, other wise the proton would speed up.

I tried using 1.60*10^-19 C for q, although I'm not sure what it should be. So that's one point I'm still confused on. I'm fairly sure that I'm on the right track since our professor "reminded" us that W=Fd...But since there isn't a specific charge mentioned, I'm not sure what to use for q.

Any insight? Maybe find an equation that does not depend on q?
 

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