Potential difference: positive or negative?

[nmsurobert]

I am working out an example problem from one of my textbooks and I am a bit confused on why a value is negative. The problem asks: Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V.

This is a conservation of energy problem. Ultimately you would use:
v = √(2qV/m)
v = √(2(-1.6e-19)(-100V)/(9.1e-31))
v = 5.6e6

We are also provided with a picture of an electron moving between two plates from A (the negative plate) to B (the positive plate).

Why is potential difference -100V and not 100V? I also understand that +100V would result in an imaginary number, but I am having a hard time keep up with sign conventions the book is using. In a different example regarding a battery, the book says "...since the electrons are going from the negative terminal to the positive, we see that ΔV = +12.0 V ."

Any input would be helpful. Thanks.

 

[RPinPA]

That is a little confusing.

There doesn't seem to be a good reason for putting that negative sign in there in that position, other than the fact that there must be one somewhere. By conservation of energy, the total change in energy must be 0, so (using $T$ for kinetic energy) $\Delta T + \Delta U = 0$ or $\Delta T = -\Delta U$. Kinetic energy increases if you go downward in potential energy.

I would rather just reason it out this way: Since the electron is accelerating, it is decreasing in potential energy. The magnitude of the change in potential energy is $|qV|$ and that's equal to the gain in kinetic energy $(1/2)mv^2$. Now everything is positive and I don't worry about sign conventions.

And then you can note that a positive $\Delta U = q \Delta V$ implies a negative $\Delta V$ for an electron.

 

[Herman Trivilino]

$\Delta T + \Delta U = 0$ or $\Delta T = -\Delta U$.

That's correct. Therefore the equation the OP is using would instead be $v=\sqrt{-2qV/m}$ where $V$ equals +100 volts.

 

[RPinPA]

That's correct. Therefore the equation the OP is using would instead be $v=\sqrt{-2qV/m}$ where $V$ equals +100 volts.

Right. In case I wasn't clear, that's what I meant by "the minus sign has to be somewhere", and I found it strange to attach it to the voltage difference. Perhaps somewhere in this textbook the author has written this equation as $q(-V)$.

 

[gleem]

WRT charge sign convention. Negative charges always go to the more positive electrode while positive charges go to the more negative electrode. In either case the charge goes from a higher potential to a lower potential relative to its charge. It is unfortunate that Benjamin Franklin chose the positive charge as the charge of the electrical current instead of the negative charge.

I would rather just reason it out this way: Since the electron is accelerating, it is decreasing in potential energy. The magnitude of the change in potential energy is |qV||qV||qV| and that's equal to the gain in kinetic energy (1/2)mv2(1/2)mv2(1/2)mv^2. Now everything is positive and I don't worry about sign conventions.

and I second this. As an exercise think about a positron.

 

[nmsurobert]

That's correct. Therefore the equation the OP is using would instead be $v=\sqrt{-2qV/m}$ where $V$ equals +100 volts.

The negative sign there make sense. However, q is still negative so you would end up with an imaginary number.

Here is the chapter in the textbook I'm referring to. The example in question is Example 3.
https://courses.lumenlearning.com/p...ectric-potential-energy-potential-difference/

 

[jtbell]

Therefore the equation the OP is using would instead be v=√−2qV/mv=\sqrt{-2qV/m} where VV equals +100 volts.

The negative sign there make sense. However, q is still negative so you would end up with an imaginary number.

How so? Doesn't -2 times a negative q give you a positive number?

 

[nmsurobert]

How so? Doesn't -2 times a negative q give you a positive number?

you're right. I'm sorry. I stayed up a little late trying to figure this out lol.

so... is this a mistake in the text? should V be positive? I completely understand where the negative sign comes from the way that you explained it. the example problem is written exactly as I wrote it in the first post.

 

[jtbell]

I've been there. Combinations of explicit and hidden minus signs can really mess with your mind.

 

[Herman Trivilino]

so... is this a mistake in the text? should V be positive?

Yes. And yes.

 

[Herman Trivilino]

Negative charges always go to the more positive electrode while positive charges go to the more negative electrode. In either case the charge goes from a higher potential to a lower potential relative to its charge.

A negatively charged particle that is otherwise free will experience a force in the direction of increasing potential.

What you're saying is true of the potential energy, but not the potential.

 

[gleem]

Yes. Blame it on Franklin.

 

[Herman Trivilino]

Yes. Blame it on Franklin.

No. Electric charge comes in two flavors regardless of the name chosen for those flavors by Franklin or anyone else. The blame lies in the fact that the electric force can be either attractive or repulsive.

 

[Herman Trivilino]

Here is the chapter in the textbook I'm referring to. The example in question is Example 3.

Okay, so the way the author is working that problem is correct because he's assuming (without saying so) that the initial potential is -100 volts and the final potential is zero volts. That's okay to do but not pointing it out explicitly is, in my opinion, sloppy and therefore confusing.

Note he writes $qV=\frac{mv^2}{2}$. Since the right hand side of that equation can never be negative, the left hand side can't be either. Since $q$ is negative $V$ must also be negative.

 

[nmsurobert]

Okay, so the way the author is working that problem is correct because he's assuming (without saying so) that the initial potential is -100 volts and the final potential is zero volts. That's okay to do but not pointing it out explicitly is, in my opinion, sloppy and therefore confusing.

Note he writes $qV=\frac{mv^2}{2}$. Since the right hand side of that equation can never be negative, the left hand side can't be either. Since $q$ is negative $V$ must also be negative.

ahhhh ok. I tried to reason it out like that before I made the post, but the previous example doesn't make that same assumption.

thanks for all the replies. I worked myself to sleep, had a dream about it, and continued to think about it this morning lol

 

[RPinPA]

Not to add to the confusion about signs, but $qV$ doesn't always have to be positive, any more than projectiles always have to go down. Particles can go upward in potential energy. But it takes work to do so, and so that will cause the kinetic energy to decrease. You could represent that with a variety of notations, but however you do it and wherever you put your negative signs, you should find that increasing PE by 1 Joule causes a decrease of KE of 1 J.

So you could for instance write it this way: $- q \Delta V = m (\Delta v^2)/2$. In this equation both sides might be positive or negative. If the change in PE $q \Delta V$ is positive (it goes up in PE) then $v^2$ must decrease.