# Homework Help: Speed of positron

1. Mar 26, 2016

### jtb93

1. The problem statement, all variables and given/known data
A positron is a particle with the same mass as an electron, but with a positive charge. If a positron begins with a speed of 1/10th the speed of light, and then moves from a high voltage region to a low voltage region, what happens to its speed? Ignore any gravitational effects.

2. Relevant equations
V=IR
possibly PE=qdeltaV?

3. The attempt at a solution
I approached this problem using V=IR. By plugging values into this equation, current increases when voltage difference increases. Hence the positron speeds up. This answer matches what's in the back of the book, but I'm not sure if I approached the problem the intended way.

2. Mar 26, 2016

### Drakkith

Staff Emeritus
Well, the equation V=IR is known as Ohm's law and is usually used for electric circuits. If this had been an electron instead of a positron you would have gotten the wrong answer.

So, you also have PE=QΔV (U=QΔV). How might this equation relate to the problem?

3. Mar 26, 2016

### jtb93

The charge (Q) is going to stay the same throughout the problem but the voltage difference is going to change from high to low?

4. Mar 26, 2016

### Drakkith

Staff Emeritus
So if the voltage changes from high to low, what is the sign of ΔV?

5. Mar 26, 2016

### jtb93

Positive.

6. Mar 26, 2016

### Drakkith

Staff Emeritus
ΔV=Vf-Vi, right? If the initial voltage is higher than the final voltage, what is the sign of ΔV?

7. Mar 26, 2016

### jtb93

Nevermind, I understand how you did that now. So ΔV is negative.

8. Mar 26, 2016

### Drakkith

Staff Emeritus
Indeed. So if ΔV is negative, that means ΔU is negative. What does a negative change in potential energy mean here?

9. Mar 26, 2016

### Drakkith

Staff Emeritus
Sorry, I forgot to point out earlier that the voltage changes from high to low, not the voltage difference. I'm pointing this out to help avoid confusion for others that might read this thread.

10. Mar 26, 2016

### jtb93

More work has been done so the speed increased?

11. Mar 26, 2016

### Drakkith

Staff Emeritus
And how is that energy being used in this problem? Potential energy is changing to what form?

12. Mar 26, 2016

### jtb93

Kinetic energy?

13. Mar 26, 2016

### Drakkith

Staff Emeritus
Does that agree with your answer of an increase in the velocity of the particle?

14. Mar 26, 2016

### jtb93

Yes, the kinetic energy is increasing so the positron speeds up.

15. Mar 26, 2016

### Drakkith

Staff Emeritus
Exactly!

16. Mar 26, 2016

### jtb93

Took more step than expected but it finally makes sense, haha. Thanks!

17. Mar 26, 2016

### Drakkith

Staff Emeritus
Note that since ΔU=QΔV, an electron in this electric field would experience a change in potential energy of: ΔU=(-Q)ΔV, and so ΔU would be positive. The electron would slow down.

18. Mar 26, 2016

### Drakkith

Staff Emeritus
Indeed. And they get longer. I just did a homework problem that made me go from voltage → potential energy → kinetic energy → velocity → centripetal acceleration → centripetal force → magnetic force → magnetic field.

(And helping you here actually made me catch a mistake I had made)

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