# A particle's travels

1. Jul 24, 2014

### Myr73

A charge of -4.00uc is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 X 10^-3kg and charge -3.00uc is fired with an initial speed of 15.0m/s directly toward the fixed charge. How far does the particle travel before it stops and begins to return back?

q1= -4X 10^-6C d= 0.55m m=2.5 X 10^-3 kg q2= -3 x 10^-6 C
V0= 15 m/s x=? x= 0.55-r {r being the distance in between charge 2's maximum distance traveled(x) and charge 1}

2. Relevant equations

KE=PE Ke= 0.5mv^2 and then PE either is =q2V or = kQ1Q2/r

3. The attempt at a solution

So I would either do a) 0.5mv^2=q1V --> V= 0.5(2.5X 10^-3)(15^2)/(-4 x 10^-6)
and then --> r= V/kQ --> x= 0.55-r

or I'm thinking more likely b) 0.5mv^2=kQ1Q2/r --> r=kQ1Q2/0.5mv^2
--> x=0.55-r

I think it would be b) but I am unsure and would I just use the mass given of 2.5 X 10^-3kg? the charges wouldnt have a mass I guess -

2. Jul 24, 2014

### BiGyElLoWhAt

That's really good right there.

You can use $q_{2}V$, but you need a potential function for the electric field, and in my opinion, that's overkill for this problem.

Assuming you solved and plugged in the numbers correctly, you should get the right answer. That being said, I'm too lazy to sift through all those numbers. Next time solve it algebraically with symbals first, then go through and solve for your desired variable, and finally plug in the numbers.

3. Jul 24, 2014

### Myr73

sweet-

4. Jul 28, 2014

### Myr73

Ok, i thought I had it all right but the answer seems way wrong-->
KE=PE Ke= 0.5mv^2 and then PE = kQ1Q2/r m=2.50X10^3kg Q1= -4x10^-6 Q2= -3X10^-6 v=15.0m/s
0.5mv^2=kQ1Q2/r --> r=kQ1Q2/0.5mv^2=0.107856/281 250= 3.85 X 10^-7m
--> x=0.55-r= 0.55m-3.85 X 10^-7m=0.549m,
witch seems way to close the original distance 0.55m- Is that possible?

5. Jul 29, 2014

### BiGyElLoWhAt

Hmm... I have the same equation you have: $r = 2K\frac{q_1q_2}{mv^2}$, but when I plug in the values I get something slightly smaller for r. Double check your math for me, would ya?

6. Jul 30, 2014

### Myr73

alright-- so r=2k q1q2/mv^2 r= 2(8.988 x10^9) { (4 X10^-6)(3 x 10^-6)/( 2.5X10^3 x 15^2)

r= 1.798 X10^10 {(1.2 x 10^-11)/ 562 500} =1.798 X 10^10 {2.13 x 10^-17} = 3.84 x 10^-7

Witch for x would still give me 0.549m --> x=0.55-3.84 X 10^-7=0.5499

7. Jul 30, 2014

### BiGyElLoWhAt

Alright well lets see...

$r = 2K\frac{q_1q_2}{mv^2}$
$r = 2*(8.988*10^9)\frac{(4*10^{-6})*(3*10^{-6})}{(2.5*10^{-3})(15^2)}$

group all the 10's together...

$\frac{2*4*3*8.988}{2.5*15^2} \frac{10^9*10^{-6}*10^{-6}}{10^{-3}}$

multiply all the 10's out and you get
$\frac{2*4*3*8.988}{2.5*15^2} * 10^0$
Multiply that fraction out and you should get .383488, take 2 sig figs (that's what you have in the problem) and that gives you r =.38m

now find the distance the particle travels.

8. Jul 30, 2014

### Myr73

Thanks ,it makes sense, all good now :)

9. Jul 30, 2014

### BiGyElLoWhAt

cool no problem