Why negative charge moves from low to high potential?

[MIA6]

For electric potential energy, the figure on my book is two parallel plates, one is positive, the other one is negative. Then a tiny positive point charge q placed near the positive plate, and then it says positive charge q has its greatest PE near positive plate. If it’s negative charge, its PE is greatest near negative plate. I understand that. But when it comes to Electric Potential, positive plate is at a higher potential than the negative plate. Why? How can we tell? Why negative charge moves from low to high potential? In my opinion, V=PE/q, so it should be the same as PE because negative charge has high PE near negative plate, so accordingly, negative plate should be the high potential when it comes to negative charge; positive plate should be the high potential when it comes to positive plate.
Or it just doesn't depend on what charge it is, people just assign the positive one be the high potential. Thanks a lot for your help.

 

[atyy]

Potential is defined as work done on/against (I never remember which) a positive test charge (a charge so small it doesn't cause the surrounding charges to move). So it would predict that if you place a negative charge in the same situation, it should move in the opposite direction as a positive charge, as we imagine to be true.

 

[imoses44]

A negative charge does indeed move from a region of high potential to one of low potential. Electrons move from the negative terminal to the positive - the opposite of current flow.

 

[MIA6]

Potential is defined as work done on/against (I never remember which) a positive test charge (a charge so small it doesn't cause the surrounding charges to move). So it would predict that if you place a negative charge in the same situation, it should move in the opposite direction as a positive charge, as we imagine to be true.

On my book, they assumed the positive test charge move from the positive plate to the negative plate, since PE=-W=-qEd, and V=PE/q=-W/q=-Ed. Then it tells us that the potential of the positive plate is greater than the potential of the negative plate, by the amount Ed. (because of that negative sign). But then I would say how about putting a negative charge near the negative plate, then PE is still equal to -W, so V will equal to -Ed as well, then I can also say the potential of negative plate is greater than the potential of the positive one. See, it goes both ways. like you said, if people experimented the positive charge first, then they may just make the conclusion that positive plate is always on the high potential, negative is on the low, so then each time, no matter what positive or negative the charge is, they just automatically say positive moves from high to low, negative move from low to high. (but according to my proof, i can assign negative plate is at high potential, positive at low) So my point is why positive plate is high potential all the time? It doesn't depend on the charge in any problem?

 

[atyy]

On my book, they assumed the positive test charge move from the positive plate to the negative plate, since PE=-W=-qEd, and V=PE/q=-W/q=-Ed. Then it tells us that the potential of the positive plate is greater than the potential of the negative plate, by the amount Ed. (because of that negative sign). But then I would say how about putting a negative charge near the negative plate, then PE is still equal to -W, so V will equal to -Ed as well, then I can also say the potential of negative plate is greater than the potential of the positive one. See, it goes both ways. like you said, if people experimented the positive charge first, then they may just make the conclusion that positive plate is always on the high potential, negative is on the low, so then each time, no matter what positive or negative the charge is, they just automatically say positive moves from high to low, negative move from low to high. (but according to my proof, i can assign negative plate is at high potential, positive at low) So my point is why positive plate is high potential all the time? It doesn't depend on the charge in any problem?

V is the potential difference between the positive and negative plates.

Let the potential at the positive plate be Vp.
Let the potential at the negative plate be Vn.

We know that V=Vp-Vn=Ed.

We can set the zero of potential anywhere we want, because only the potential difference matters. So for convenience in this case, we set the zero of potential in the middle of the plates, which means: Vp=Ed/2, and Vn=-Ed/2.

So to move a positive charge from the negative to the positive plate requires W=-PE=-q(Vp-Vn)=-q(Ed/2-(-Ed/2))=-qEd.

To move a negative charge from the positive to the negative plate requires W=-PE=-q(Vn-Vp)=-q((-Ed/2-Ed/2))=qEd ...

but for a negative charge, q is a negative number, so qEd is still a negative number.

 

[MIA6]

V is the potential difference between the positive and negative plates.

Let the potential at the positive plate be Vp.
Let the potential at the negative plate be Vn.

We know that V=Vp-Vn=Ed.

We can set the zero of potential anywhere we want, because only the potential difference matters. So for convenience in this case, we set the zero of potential in the middle of the plates, which means: Vp=Ed/2, and Vn=-Ed/2.

So to move a positive charge from the negative to the positive plate requires W=-PE=-q(Vp-Vn)=-q(Ed/2-(-Ed/2))=-qEd.

To move a negative charge from the positive to the negative plate requires W=-PE=-q(Vn-Vp)=-q((-Ed/2-Ed/2))=qEd ...

but for a negative charge, q is a negative number, so qEd is still a negative number.

your proof seems to work very well, but just few questions here. why Vp=Ed/2, but Vn=-Ed/2, (why Vn is negative sign?) and /2 here means in the middle of the plates?

 

[atyy]

your proof seems to work very well, but just few questions here. why Vp=Ed/2, but Vn=-Ed/2, (why Vn is negative sign?) and /2 here means in the middle of the plates?

In a problem using the concept of potential, I am allowed to set the potential at one (and only one) point in space to be zero, and calculate all other potentials relative to it. This is because we only use *potential differences* to predict what will happen.

Setting #1: For example, I could have set the potential at the negative plate to be zero, in which case Vn=0, Vp=Ed. If I had done that, then the potential exactly half way between the plates would have been Ed/2.

Setting #2: Since I had the freedom to choose one location to have zero potential, I chose the potential exactly half-way between the plates to be 0. In which case Vn=-Ed/2, Vp=Ed/2.

In both cases, Vp-Vn=Ed. Your question was easier to answer using Setting #2, so I chose that. But the answer should be the same if you use Setting #1. Try it.

 

[MIA6]

Setting #1: For example, I could have set the potential at the negative plate to be zero, in which case Vn=0, Vp=Ed. If I had done that, then the potential exactly half way between the plates would have been Ed/2.

I actually found setting 1 easier because I don't get why you chose the potential in the middle of the plates, but I get it now why positive is always at high potential, and negative is at low potential.

 

[MIA6]

btw, If a negative charge is near positive plate, then we have to do work in order to move it to the negative plate, right? so in this case, it's moving against electric field? what If a negative charge is near negative plate, then do we still need to do work in order to make it go to positive? i think it will go by itself directly because of the attraction. But in which case does the negative charge have the maximum PE? the second case? Is there PE in the first case?