# Homework Help: Electrostatics (Distance of closest approach)

1. Jul 17, 2010

### Mandeep Deka

1. A proton an an alpha-particle are projected with a velocity v=(Ke^2/mL)^0.5 each, when they are far away from each other. If their initial velocity vectors are anti parallel to each other, with a distance "L" between them, find the distance of closest approach.(mass of proton = m, charge of proton= +e, mass of alpha-particle= 4m, charge of alpha particle= +2e)
Ans: L[5+(89)^0.5]/8

2. F=kqQ/r^2, U=kqQ/r

3. Conservation of momentum and conservation of energy are the two prime theorems that are applicable in this case, as there is no external agent doing work, but they still are insufficient to eliminate the number of variables. As both the particles experience force (changing with time), towards each other they deflect from each other and at some point they are closest to each other... this is all i could sort out...

Plz help!!!!!

2. Jul 17, 2010

### Staff: Mentor

Why do you think they are insufficient? Hint: What will be the speed of the particles when they reach the distance of closest approach?

FYI, this answer doesn't quite make sense. Better double check it.

Last edited: Jul 17, 2010
3. Jul 18, 2010

### graphene

Could you put down your solution here? How did you obtain that value for L?

4. Jul 18, 2010

### Swap

you don't even need to bother about the momentum. The required eq.(in words)
Kinetic energy of both the particles= Potential energy of the system at the closest point of approach. ( Remember the velocity at the closest point of approach is zero)

5. Jul 18, 2010

### Staff: Mentor

That is incorrect.

6. Jul 18, 2010

### Swap

why?

7. Jul 18, 2010

### Staff: Mentor

Consider the total momentum of the particles.

8. Jul 18, 2010

### Swap

oh yes. At the time of minimum approach the velocity of proton will be 0 while the velocity of the alpha particle will be 3V right.??

9. Jul 18, 2010

### Staff: Mentor

No. If that were true, they'd still be approaching each other.

10. Jul 18, 2010

### Swap

no not really. once the protons velocity has changed direction the proton will move away and thus distance between them will start to increase. We can consider it as two carts of different masses coming towards each other with a spring attached to them.
If not so, it is really not possible to solve this since u r left with 3 variables.

11. Jul 18, 2010

### Staff: Mentor

What you are missing the relative velocity of the particles at the point of closest approach. (It is the same as the two carts with spring problem!)

12. Jul 18, 2010

### Swap

can you explain more. I dun understand what do u mean by the "relative velocity of the particles at the point of closest approach".

13. Jul 18, 2010

### Staff: Mentor

Think about it. You made the statement that at the distance of closest approach, one particle would have speed 0 while the other had speed 3V. If that were so, the second would still be approaching the first particle. They would still be getting closer.

It's exactly the same with the two carts and spring problem. How must the speeds of the two carts relate when the spring is maximally compressed?

14. Jul 18, 2010

### Swap

oh ok....I got it. That means at the point of minimum separation the velocity of both of them will be equal is it?

15. Jul 18, 2010

### Staff: Mentor

Exactly!

16. Jul 18, 2010

### Swap

Thanks a lot to make my concept clear. =p

17. Jul 19, 2010

### Mandeep Deka

so guys is there any way to get that weird looking answer.....
since the difficulty in doing the qs has been established, is there any way out of it.... coz i havnt found any n nor has my teacher been so far!!!!!

so far the qs is concerned it is perfectly correct... and the ans too is!!
if anyone can find the answer to that question...plz do inform!!!!!

(i would specially request any student from kota institutes, like bansal, vibrant etc, to get that solution coz as far as i know, the qs has its origin in some institute of kota only!!)

18. Jul 19, 2010

### Staff: Mentor

What makes you think the answer is correct? Where's your attempted solution? It seems like a fairly straightforward problem to me.

19. Jul 20, 2010

### Swap

Agree with Doc al. It is a straight forward question. Moreover I got a very nice answer. It is (5/9)L. ( I dont really trust the coaching institutes u r talking abt, I have sometimes find mistake in their solutions to IIT JEE question paper.)

20. Jul 20, 2010

### Mandeep Deka

even, i had thought that the answer is wrong, but i just wanted to know if anyone had some solution or if there was anyway that answer came up...
thanks for helping...

by the way there was another problem of this sort...

Two particles of equal charge and mass (q,m) are infinitely far from each other. At some instant of time one particle is projected towards other with a velocity 'v'. Find the distance of closest approach.