# Electrostatics and Velocity problem

#### kokocipher

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 .

Last edited:

#### IntelInside

10+ Year Member
7+ Year Member
PEnonconstant E field = kq1q2/r^2=vq
Peconstant E field = q*E*d.x=vq

vq = .5mv^2

#### PhilIvey

##### Poker Aficionado
10+ Year Member
5+ Year Member
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.

OP
K

#### kokocipher

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.

#### mipp0

10+ Year Member
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.