postion vs time graph for moving charged particle?

Sep 2016
1
0
Question
{Consider the system shown to the right. Particles 1 and 2 are fixed in place while particle three is placed at the location shown and released.



Let’s look at the mechanics of q3, which has a mass of 3.00 X 10-5 kg.


What will be the net force on q3?

What will be q3’s acceleration?

Can you use the acceleration equation [x(t)=1/2at^2+v0t+x0] to predict where q3 will be after 0.50 seconds? Explain.
Project: You al make a spreadsheet that graphs q3’s position vs. time. Here are the requirements:

Spreadsheet accurately models the movement of particle 3.

You can adjust the magnitude and sign of each of the charges.

You can adjust the initial position and mass of q3 (including to the left or right of both charges or right in the middle).}



ok now aside from the excel spreadsheet which I dont know how to do since they want an XvsT graph I am a little bit confused about how to tackle this problem . I take AP physics E&M which i am fairly good at but i always hated mechanics so that is why I might sound like i am a little more lost than i should be. I apologize. at my school juniors take pre-calc I take advanced pre-calc, since the school year is still young I dont really know that much calculus so maybe my derivatives were wrong nevertheless this is what I was thinking.

1. q1*q3*k/(r^2)= F1_3(this is the electric force between 1 and 3)

2. q2*q3*k/r^2)= F1_2(force between particle 1 and 2)

3. from there i can find the net force between the two charges with f1_3+f1_2=fnet

4. with the net force i would then find the acceleration using a=fnet/m (with m being the mass. (now everything above i was okay with doing but now heres where i get confused)

5. take the acceleration just found and find velocity. v=-(at) being time intervals (i got that equation aver deriving it from the equation in step 6 and the initial equation was x(.05)=at+v

6.take that velocity and previous acceleration and find new position: x=1/2at^2+v*t+x

7. the value for x becomes the new position for particle 3 and now i go back to to the top in order to calculate electric force then acceleration etc rinse and repeat.

the equations that i used derived them from suvat equation x=1/2at^2+v*t+x to find the values i needed. My position value for .00-.05 value didn't make sense because it equated to about 2000 meters I am not understanding where i went wrong. my process went like this:

F1-3=-899 X componet
F2_3= -4495 X componet
Fnet3- 5394 N
A=F/m or -5394/3E-5= -1.798E8
from here this what i did since the problem wants a x vs t graph i decided to do the t in intervals
x(.05)=a
x(.05)=at+v (here is where i get my velocity which is a whopping 8990000.0
x(.05)=.5*-1.798E8*.05^2+ 8990000.0 *.05+.01
this value of x equates to 224750 but how can the particle move to such a high value i know that something is off i just cant find it.

but if that value was to make sense i would take that x and put that as the new position of q3 and then do the distances between the charges again find fnet again new acceleration etc. and then repeat the process of equations till I have done as many time intervals as necessary
 

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Last edited:
Dec 2013
19
8
Encinitas, CA
Your basic plan sounds correct.

I'm seeing that you have to use a very small $\Delta t$

$1 \mu s$ seems to be sufficient.

The motion is stabilized by about $800 \mu s$ with the parameters shown in the picture

You should see a damped sinusoid.