Speed of a Positron: How Does Voltage Affect It?

[jtb93]

Homework Statement

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.

Homework Equations

V=IR
possibly PE=qdeltaV?

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.

 

[Drakkith]

This answer matches what's in the back of the book, but I'm not sure if I approached the problem the intended way.

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?

 

[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?

 

[Drakkith]

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

 

[jtb93]

Positive.

 

[Drakkith]

ΔV=V_f-V_i, right? If the initial voltage is higher than the final voltage, what is the sign of ΔV?

 

[jtb93]

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

 

[Drakkith]

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

 

[Drakkith]

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

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.

 

[jtb93]

More work has been done so the speed increased?

 

[Drakkith]

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

 

[jtb93]

Kinetic energy?

 

[Drakkith]

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

 

[jtb93]

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

 

[Drakkith]

Exactly!

 

[jtb93]

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

 

[Drakkith]

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.

 

[Drakkith]

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

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)

 

[Tom Atkinson]

When you say in the problem that the positron moves from a high voltage region to a low voltage region, how do you determine the charge and how did it get that way? Is the high voltage region of space charged by lack of electron (positive) or many electrons (negative)? Or did you mean it is high voltage because of many positrons (positive) making them repel and speed up... or less likely is it charged by a gap or lack of electrons (positive) causing the positron to accelerate to a more electron rich environment?

 

[Drakkith]

When you say in the problem that the positron moves from a high voltage region to a low voltage region, how do you determine the charge and how did it get that way? Is the high voltage region of space charged by lack of electron (positive) or many electrons (negative)? Or did you mean it is high voltage because of many positrons (positive) making them repel and speed up... or less likely is it charged by a gap or lack of electrons (positive) causing the positron to accelerate to a more electron rich environment?

The cause of the electric field is irrelevant to the problem. What matters is that the electric field is not uniform, so there is a voltage difference between different parts of the field. And 'charge' doesn't apply to an electric field, it only applies to particles.