Physics 12 Electric Energy Question

[Alameen Damer]

Homework Statement

A proton starts at rest and travels through a potential difference of 8000V. What's it's final speed.

Homework Equations

ΔEe+ΔEk=0
ΔEk=-ΔEe
Ee=qV

The Attempt at a Solution

I know how to find the correct magnitude, but I am getting the wrong sign.

Ek=-ΔEe
1/2mv^2=-qΔV
v=root [-(2qΔV)/m]

Note mass of proton=1.67 x 10^-27 kg

v=root[-2(1.6.x10^-19)(8000))/1.67 x 10^-27]

I get an error because obviously there isn't a root for a negative value of [-2(1.6.x10^-19)(8000))/1.67 x 10^-27].
I want to know where I am going wrong in terms of positive/negative.

 

[haruspex]

The question doesn't state which way the potential difference is, it only gives you the magnitude.
But you are given that it starts at rest and only moves because of the field, so which way will it move, to a higher potential or to a lower one?

 

[Alameen Damer]

Won't it be lower because it's losing potential electric energy and getting kinetic energy?

 

[haruspex]

Won't it be lower because it's losing potential electric energy and getting kinetic energy?

An electron has a negative charge. If it moves to a lower potential does it gain or lose PE?

 

[Alameen Damer]

Lose

 

[haruspex]

Lose

Try a different question. If a positively charged particle moves to a lower electrical potential, does it gain or lose PE?

 

[Alameen Damer]

Oh, does the elctron gain due to the negative charge? If so then the positive particle would lose PE.

 

[haruspex]

Oh, does the elctron gain due to the negative charge? If so then the positive particle would lose PE.

An electron moving to a lower electrical potential would gain PE: $q \Delta V > 0$. So which way is this electron moving?

 

[Alameen Damer]

The proton would be moving against the electric field I believe.

However, this question is talking about a proton.

 

[haruspex]

The proton would be moving against the electric field I believe.

However, this question is talking about a proton.

Oh, I'm sorry - don't know why I read it as electron.
So, restart.
The proton will move to a lower potential, so $\Delta V < 0$, $\Delta PE = q \Delta V < 0$. Now, is $\Delta KE$ equal to $\Delta PE$ or to $-\Delta PE$?

 

[Alameen Damer]

KE will be equal to -PE

 

[haruspex]

KE will be equal to -PE

Right. So does that get rid of your minus sign?

 

[Alameen Damer]

Yes, but how come when solving algebraically there is a minus sign. I want to understand it algebraically.

 

[haruspex]

Yes, but how come when solving algebraically there is a minus sign. I want to understand it algebraically.

I thought I explained it algebraically.
The proton will move to a lower potential, so $\Delta V < 0$, $\Delta PE = q \Delta V < 0$, $\Delta KE = -\Delta PE= -q \Delta V > 0$.
If it were an electron, both the sign of the charge and the sign of the potential difference would change, so the change in KE would still be positive.