# Motion of a charged particle in an electric field

1. Apr 30, 2013

### Nemo's

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

Each of the nuclei below are accelerated from rest through the same potential difference.Which one completes the acceleration with the lowest speed?
1H1 4He2 7Li3
9Be4
2. Relevant equations

F=ma=QE

3. The attempt at a solution
v∝a∝Q/m so for lowest v Q/m lowest is 1 so answer is 1H1
1. The problem statement, all variables and given/known data

The following particles are each accelerated from rest through the same potential difference. Which one completes acceleration with the greatest momentum?
α β proron neutron

2. Relevant equations
F=ma=QE

3. The attempt at a solution
mv∝QE so biggest mass and biggest charge give biggest momentum so the answer is α

2. Apr 30, 2013

### Staff: Mentor

1H1 has the largest Q/m.
While a larger Q/m leads to a larger velocity, those are not proportional to each other.

mv∝QE is not true.

Last edited: Apr 30, 2013
3. Apr 30, 2013

### technician

The energy gained by each particle = q x V
They all pass through the same V so the energy gained for each in order is 1, 2, 3 and 4 units.
This is KE so for each use 0.5mv^2 with the appropriate masses to find the values of velocity, v.

4. Apr 30, 2013

### Nemo's

Yes that's right I don't know why I reversed it. My bad. So it must be the Lithium.

Why's that?

5. Apr 30, 2013

### Nemo's

Thank you so much the energy approach is a lot
easier

6. Apr 30, 2013

### Staff: Mentor

Why should the equation be true?
Energy and momentum are not the same.

7. May 1, 2013

### Nemo's

O.K so force (EQ) is directly proportional to the rate of change of momentum(Δmv/t) This one is true right?

8. May 1, 2013

### Staff: Mentor

$F=\frac{dp}{dt} = m \frac{dv}{dt} = ma$ (for constant m), indeed. As time is not given, I don't think this helps here.

9. May 3, 2013

### Nemo's

I see. So using proportionality here can't solve this problem.
O.K so by using the energy method I can get the speed of each particle then I could multiply each speed by the corresponding mass to get the momentum?
The problem is it's a multiple choice question and I shouldn't be wasting so much time on a single question in an exam. Isn't there a faster way ?

10. May 3, 2013

### Staff: Mentor

You can do this, or just compare the different particles ("larger mass leads to whatever, that leads to ..., ...")

11. May 4, 2013

### technician

The charge of the particles is1, 2, 3 4....so the energy gained is also 1, 2 3 and 4 (relatively!!)
The mass of the particles is 1, 4, 7 and 9.
so using energy =1/2 m v2 gives
1 = 1/2 x 1 x v2
2 = 1/2 x 4 x v2
3 = 1/2 x 7 x v2
4 = 1/2 x 9 x v2

if you sort this out you should see that for 1 v2 = 2
for 2 v2 = 1
for 3 v2 = 6/7
for 4 v2 = 8/9

12. May 4, 2013

### Nemo's

O.K Thank you so much :D