Electrostatics and Velocity problem

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kokocipher

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  1. Pre-Medical
3) What voltage is required to accelerate protons to a speed of 10^4 m/s ?
(e =1.6E-19C, proton's mass =1.67E-27kg)
(k=9E9 Nm/C^2)

This Q. is from Kaplan's MCAT physics review notes (p.419)

It's in the High-yield problems category under "similar questions," so it did not offer an answer. I worked out an answer but could not find a way to confirm it. (I did it by using conservation of energy.) I want to see if anyone else could work the problem and can arrive an answer that matches mine, so I can finally get it off of my head 😉.
 
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PEnonconstant E field = kq1q2/r^2=vq
Peconstant E field = q*E*d.x=vq

vq = .5mv^2
 
3) What voltage is required to accelerate protons to a speed of 10^4 m/s ?
(e =1.6E-19C, proton's mass =1.67E-27kg)
(k=9E9 Nm/C^2)

This Q. is from Kaplan's MCAT physics review notes (p.419)

It's in the High-yield problems category under "similar questions," so it did not offer an answer. I worked out an answer but could not find a way to confirm it. (I did it by using conservation of energy.) I want to see if anyone else could work the problem and can arrive an answer that matches mine, so I can finally get it off of my head 😉.

V=J/C*C=Joules The KE=.5(m)v(^2). So, 8.33e-20=.5(1.67e-27)(1e-4)^2
J/C=Volts so 8.33e-20/1.6e-19= 5.2e-1 Volts is what I get. I did this in my head so that should be pretty close.
 
V=J/C*C=Joules The KE=.5(m)v(^2). So, 8.33e-20=.5(1.67e-27)(1e-4)^2
J/C=Volts so 8.33e-20/1.6e-19= 5.2e-1 Volts is what I get. I did this in my head so that should be pretty close.

0.5 Volt is my answer as well, thanks.
 
V=J/C*C=Joules The KE=.5(m)v(^2). So, 8.33e-20=.5(1.67e-27)(1e-4)^2
J/C=Volts so 8.33e-20/1.6e-19= 5.2e-1 Volts is what I get. I did this in my head so that should be pretty close.

Very nicely done sir.
 
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