Electric Potential -- Is my understanding correct?

[guyvsdcsniper]

I have been having a hard time understanding Electric Potential and believe I finally have a grasp on what is trying to say. I wanted to right out my understanding here and hopefully have someone confirm what I am saying is somewhat accurate as I feel like when you write stuff out you tend to get a better understanding of the topic. I would appreciate any input.

From what I understand, Electric Potential does not depend on a test charge. It is always there if we have a source charge. What it does is tell us how much Energy per Charge we will have at a certain distance from the source charge.

So by V = kQ/r, if we know the charge of the source charge and when we know what distance we want to evaluate, we plug that in for "r" and that will tell us the Energy per charge at that point. By knowing that, we can then arrange the equation V = U/q into U= qV, and when we know q, a test charge, we will get potential energy. It is a very simple way of finding the potential energy a distance from the source charge.

Just by looking at the equation for V, we can tell that when the distance from the source charge is small, we will have a larger Electric Potential (Voltage). It shouldn't matter if the source charge is negative or positive, If you are close to the source charge there is higher Voltage.

It is also my understanding that the potential difference just tells us the difference in potential between two points away from the source charge. So if a source charge Q has two points away from it, let's say a = 1m and b = 2m, if we do b-a, this would tell us we be going up in voltage since the difference is positive and if we do a-b we would get a negative value, telling us voltage is decreasing. Lastly, this is the part I have been struggling with in class is Electric Potential in a Uniform Field. If we have a uniform field in a capacitor, going from positive to negative, the Electric field vector goes from positive to negative (we can say left to right in this example). We also have a direction vector as well for particles moving through this uniform field. If we have let's say a positive charge going in the direction of the electric field it falls to lower PE, and since it would be "stuck" and attracted to the negative plate, it would fall to lower potential, considering it is now a farther distance from the positive charge.

Im a little lost on a negative charge in that situation. It moves against the field, getting attracted to the positive plate, So its Electric Potential would be high? how does the fact that it travels opposite to the electric field direction factor into its voltage? is it considered a negative voltage since you have a -x vector and pos x vector?

 

[hutchphd]

(Voltage). It shouldn't matter if the source charge is negative or positive, If you are close to the source charge there is higher Voltage.

"Higher" is not well defined. Nearing a negative charge the potential will be a bigger (but more negative) number. Most people would call that "lower". I don't know whether your physics or English is suspect.

 

[guyvsdcsniper]

"Higher" is not well defined. Nearing a negative charge the potential will be a bigger (but more negative) number. Most people would call that "lower". I don't know whether your physics or English is suspect.

I believe my physics was suspect but I think your response has helped me realize why I was wrong.

Near a positive source charge, the potential will be higher, labelled as Higher Potential.
Near a negative source charge, the potential will be bigger, but given that it is a negative value, we call that a lower potential.

Is that correct or am I still off?

 

[hutchphd]

I think you are correct. The hypothetical "test charge" is always assumed positive.

 

[guyvsdcsniper]

I think you are correct. The hypothetical "test charge" is always assumed positive.

Are you talking about the "test charge" that we use to probe electric fields?

 

[vanhees71]

I'm a bit lost about what your level of understanding is. Since the thread is labelled I, I assume you know vector calculus. Then you should know that a (scalar) potential $\Phi$ of a vector field $\vec{E}$ means that
$$\vec{E}=-\vec{\nabla} V,$$
and that such a potential exists (at least locally in simply connected regions of space) you must have
$$\vec{\nabla} \times \vec{E}=0.$$
A "voltage" is simply a potential difference with a definite choice of a reference point.

Since the electric field by definition does not depend on test charges which can be used to measure it, of course also the electrostatic potential does not depend on any test charges.

 

[sophiecentaur]

Near a positive source charge, the potential will be higher, labelled as Higher Potential.
Near a negative source charge, the potential will be bigger, but given that it is a negative value, we call that a lower potential.

Potential is defined in terms of work done on a positive charge. Too move a + charge to a + charge (positive plate) involves positive work against a repulsive force. So the potential is higher.
Moving it to a negative plate involves doing negative work in the direction of a lower potential.
We don't just "call" it a lower or higher potential; the sign is there in the original definition.
It's funny, really but I tend to favour explanations in terms of Energy, rather than Forces. However, here, the idea of repulsion or attraction is more intuitive to me. But take your pick.