# Homework Help: Time it takes for two charges to collide

1. Dec 5, 2012

### jpark31

1. The problem statement, all variables and given/known data

Two charges with e+ and e- charge are separated a distance "a" apart. Assuming there's no gravity, how long does it take for the charges to collide?

2. Relevant equations

My professor said to use conservation. I'm guessing he means Wnc = (KE + PE) - (KEo + PEo)

3. The attempt at a solution

I have no clue where to start. Can anyone help me out?

2. Dec 6, 2012

### ehild

Conservation of what? What is Wnc? What else is conserved when both particles are free?

Have you studied calculus?

ehild

Last edited: Dec 6, 2012
3. Dec 6, 2012

### jpark31

Yes I have studied calculus. I think he meant Conservation of energy?
Mechanical Energy= (Kinetic+Potential)-(Kinetic_0+Potential_0).

4. Dec 6, 2012

### ehild

Yes, you can use conservation of energy. But your equation is wrong. Mechanical energy is the sum of kinetic energy and potential energy.

ehild

5. Dec 6, 2012

### jpark31

Ahh I see. So the conservation of energy isn't the sum of kinetic and potential? E=U+K=1/2kx^2 + 1/2mv^2?

6. Dec 6, 2012

### ehild

Conservation of energy is a law, that the mechanical energy is conserved during the motion.

What is 1/2 kx^2? Is it related to this problem? What is the potential energy of a system of those charged particles?

7. Dec 6, 2012

### rcgldr

What is the mass of each particle, or are you suppose to solve this assuming mass is a variable? Are the effects of an accelerating charge supposed to be taken into account?

Last edited: Dec 6, 2012
8. Dec 6, 2012

### jpark31

Potential energy = mgh. But gravity is assumed to be 0, according to my professor. So potential is 0?
One charge is an electron and the other is a proton, so e-=9.109-10^-31 kg and e+=1.673*10^-27 kg.
Yes, the effects of an accelerating charge are supposed to be taken into account! Sorry if I wasn't very clear on the problem, but I appreciate all the help I am getting! Thank you!

9. Dec 6, 2012

### ehild

Gravity is ignored between the particles as it is much-much weaker then the electric force between them. What force acts between two charges?
What is the electric potential? What is the electric potential energy of two charged particles?

ehild

10. Dec 6, 2012

### jpark31

F=kq1q2/r^2 so, F=[(9*10^9)(1.602*10^-19)(-1.602*10^-19)]/a^2=(-2.31*10^-28)/a^2
Electrostatic Potential is V(r)=k int(a to b) dq/r
So the limits of integration would be 0 to a right?
Then V=kq/r^2=(9*10^9)(1.602*10^-19)/a^2=1.4418*10^-9/a^2

11. Dec 6, 2012

### ehild

No, you get the electrostatic potential difference by integrating the negative electric field strength with respect to the distance. In the field of a point charge q the potential difference is

$$V(b)-V(a)=-\int _a^b{k\frac{q}{r^2}dr}$$

We choose the potential zero in infinity. You can find the potential V(r) at any distance r from the proton, and the potential energy of the electron is PE=-eV(r). From conservation of energy, you get the speed of the electron at r. The integral of the speed with respect to time will give the distance "a". That is how the time and the distance are related, from where you can get the time.

ehild

12. Dec 9, 2012

### jpark31

Wait so the limits of integration are zero to infinity? Is that what you meant when you said "potential zero in infinity"? I'm sorry but I've been searching and the only formula for conservation of energy that I see is the summation and differences in kinetic and potential energy.

13. Dec 9, 2012

### rcgldr

It doesn't matter, you need to create a formula that relates velocity to distance (call this r), assuming the initial velocity is zero at distance a, based on conservation of energy (PE + KE = constant). My guess is that you're supposed to assume that velocity = dr/dt (derivative of distance beteen charges versus time). See if you can turn this information into an integral that you can solve.

14. Dec 9, 2012

### ehild

You can chose the potential arbitrary at a certain point. But it is convenient to choose the potential in the field of a point charge zero at infinity, and considering the potential at distance r from a point charge q equal to

V(r)=kq/r.

Don't you remember it? Then you can write conservation of energy for the electron as

KE +(-e)V(r)=constant, and find v as function of r.

ehild

Last edited: Dec 9, 2012
15. Dec 9, 2012

### jpark31

Is the constant used in the conservation of energy just a arbitrary number "C"?
Okay, so here's my latest attempt with your guys' help. V(r)=kq/r= 9*10^9(1.6*10^-19)/r=1.44*10^-9/r
1/2(mv^2)+(1.6*10^-19)(1.44*10^-9/r)= C
4.5545*10^-31v^2+2.304*10^-28/r=C
v=sqr root[(C-2.304*10^-28/r)/(4.5545*10^-31)]
And then from here I integrate Vdt=dr?

16. Dec 10, 2012

### ehild

Take into mind that the speed of the electron is zero at r=a, it has got only potential energy there. And the charge of the electron is negative. So 1/2mv2-ke2/r=-ke2/a.

Do not plug in the data yet, solve the problem symbolically.

So you have the v(t) function as

$$v(t)=dr/dt=-\sqrt{\frac{2ke^2}{m}(\frac{1}{r}-\frac{1}{a})}$$

it is negative as the electron moves towards the proton, decreasing the distance r.
The equation can be written as $$dt/dr=-\frac{1}{\sqrt{\frac{2ke^2}{m}(\frac{1}{r}-\frac{1}{a})}}$$

Integrate it from r=a to r=0.

ehild

17. Dec 10, 2012

### jpark31

THANK YOU SO MUCH! Starting from scratch I didn't know anything and so you have been very patient with me, thank you for that. After integrating and plugging in the variables, I came out with approximately 6.984*10^-2*a^3/2 seconds.

18. Dec 10, 2012

### ehild

How could you integrate it so fast? It is correct.

Now, there is the question, if the speed of the electron approximates the speed of light. Then you need to calculate with increasing mass... but I think the problem does not require that.

ehild