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Homework Statement
A proton is accelerated to 3*10^6 m/s and heads towards a spherical charge distribution having a charge of 1.6*10^-18 C. Approximately how close to the distribution does the proton travel before stopping? (Assume inital potential energy of proton is zero).
Vi= 0
Q= 1.6*10^-18 C
a= 3*10^6 m/s
Homework Equations
energy conservation:
Kf+Vf = Ki + Vi
V= kQ/R
The Attempt at a Solution
Kf+Vf = Ki + Vi
0+qv= 1/2mv^2+ 0
qv= .5mv^2
(1.6*10^-18)(v)= (.5)(1.67*10^-27)(3*10^6m/s)
v= .001565
V= kQ/R
.001565= (9*10^9)(1.6*10^-18)/(R)
R= 9.2 * 10^-7m
Is this correct?