Potential energy and proton acceleratoin

[physics100000]

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?

 

[nickjer]

In EPE = qV, the 'q' is the charge being acted on by the spherical charge distribution, since you set V to be the potential caused by this distribution. So in this case it would be the charge of the proton.

 

[physics100000]

So when I plugged in 1 *10^-19 C [charge of proton] I ended up with 9.2*10^-6 m which was not even an option on the multiple choice. I am confused, please help. If this helps the answers are:
1.9* 10^6m
3.1*10^-13m
5.5*10^-7 m
9.2 *10^-7 m

 

[nickjer]

The charge of the proton is really 1.6*10^-19 C. But it looks like you made a mistake somewhere and lost a factor of 10. You actually had the right answer initially, but did the math wrong.

So please write out all the steps and I can point out where you went wrong.

 

[physics100000]

so here's what I did:

Kf+Vf = Ki + Vi
0+qv= 1/2mv^2+ 0
qv= .5mv^2
(1.6*10^-19)(v)= (.5)(1.67*10^-27)(3*10^6m/s)
v= .01565

V= kQ/R
.01565= (9*10^9)(1.6*10^-19)/(R)
R= 9.2* 10 ^-8 m

but then I did this:
leaving q = 1.6*10^-18 and changing Q= 1.6 *10^-19 [ proton charge] and then solving the same way to get a final answer of
9.2 *10^-7m.

so which way is correct?
By the way thanks for helping.

 

[nickjer]

Changing the order of q and Q should not affect your answer. You are making a mistake when you calculate R in this step:

.01565= (9*10^9)(1.6*10^-19)/(R)

Originally you have q = 1.6*10-19 C, and here in this step you are also setting Q = 1.6*10^-19 C. You should have set Q = 1.6*10^-18 C.

 

[physics100000]

Changing the order of q and Q should not affect your answer. You are making a mistake when you calculate R in this step:

.01565= (9*10^9)(1.6*10^-19)/(R)

Originally you have q = 1.6*10-19 C, and here in this step you are also setting Q = 1.6*10^-19 C. You should have set Q = 1.6*10^-18 C.

well i found the right answer and it was nowhere near what I got.

by the way i did originally have Q = 1.6*10^-18 C but you told me to change it.

 

[nickjer]

Well, you can interchange q and Q. But only one of them can be 1.6*10^-19 C. Unfortunately, you set them both to that.