[Physics] Higher To Lower Electric Potential

electric-fieldselectricityelectromagnetismpotentialpotential energy

The question I am working on is:

"An electron moving parallel to the x axis has an initial speed of $4.65 \cdot 10^6~m/s$ at the origin. Its speed is reduced to $1.27 \cdot 10^5 ~m/s$ at the point x = 2.00 cm.

(a) Calculate the electric potential difference between the origin and that point.

(b) Which point is at the higher potential?"

I am having difficulty visualizing this situation. Obviously, because of the reduction in speed, the kinetic energy is reducing; but, because of conservation of energy, the reduction of kinetic energy corresponds to an increase in potential energy.

Is something giving the electron an initial "push," and a positive source charge distribution is pulling the electron as it passes by, thereby reducing the speed of the electron?

Also, what confuses me is this paragraph:

We assume that the electron is moving in a vacuum. Its speed is reduced because it is >moving to a higher electric potential energy. Since its charge is negative, this means >that it is moving to a more negative potential; that is, its starting position is at a higher potential (a higher voltage) than its final position.

What do they mean by "more negative potential," what does a negative potential physically mean? I realize that at our initial position, we arbitrarily choose the initial electric potential the value zero, and so if we chose another point in the electric field, relative to our initial point our electric potential will either be positive or negative–assuming that our second point is not in the equipotential region that the first point is. I suppose I am also having trouble understanding what situation a negative electric potential corresponds to, and what situation a positive electric potential corresponds to.

I'd appreciate the help! Thanks in advance!

EDIT: Also, is it true that if a charge has negative electric potential energy, then it is not necessary that it has lost that energy. In other words, negative electric potential energy can mean a gain in that sort of energy?

I've heard electric potential being described as being "the measure of potential energy per unit charge." From this, I have developed the understanding that electric potential is a scalar that is assigned a different value at every point in an electric field.It almost seems like a conversion factor. It's how much potential energy a charged particle could possession if it were there. It's a potential of a potential. Again, sorry for all of the comments. I hope you can help me

Another question, if I push a negative test charge near a negative source charge, the test charge will be gaining potential energy, and that potential energy will be positive, right? Alternatively, if I move a negative test charge near a positive source charge, making sure it doesn't accelerate towards it, it will loss potential energy, showing up as a negative potential energy, right? And the mathematics will show that these are true? I guess the root of my problem is seeing a generalized situation, and the mathematics that accompany it.

Actually, I have one more question. I was able to answer the question I originally posed, except for part (b). I thought, because the electric potential energy was increasing at that point, that the electric potential at that point would be higher. After all, electric potential is defined one way as, "the amount of electric potential energy that a unitary point charge at that location would have." So, if the electron has more PE at the final position, wouldn't that mean the electric potential was greater at that final position too?

Best Answer

First the math:

Given the electric potential $V(\mathbf a)$ at a point $\mathbf a$ in space and the potential $V(\mathbf b)$ at another point $\mathbf b$, the electric potential energy of a charge moving from point $\mathbf a$ to point $\mathbf b$ will change by $$ U(\mathbf b) - U(\mathbf a) = q[V(\mathbf b) - V(\mathbf a)] $$ If the speed of an electron decreases in moving from a point $\mathbf a$ to a point $\mathbf b$, then that must mean that its potential energy has increased: $U(\mathbf b) - U(\mathbf a) >0$, so that since the electron has charge $-e$, we get $$ (-e)[V(\mathbf b) - V(\mathbf a)]>0 $$ which, upon dividing both sides by $-e$ (and noting to change the direction of the inequality when we do so) implies $$ V(\mathbf b) < V(\mathbf a) $$ So we see that the starting position $\mathbf a$ is at a higher potential than the final position $\mathbf b$ as the second paragraph says.

The intuition here is that there are some other charges around (they could be positive or negative depending on where we place them) that setup the potential "landscape" V in such a way that the potential at the origin, where the charge began, is higher than the potential at the final point, and a negative charge that rolls "down this potential hill" will be going slower when it gets to the bottom of the hill.

Hope that wasn't confusing!

Cheers!

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[Physics] Why the electrons flow from higher potential to lower potential

You have a slight misconception that is that higher potential i.e. Positive side to lower potential i.e. Negative side. Both sides have the same electric potential just like two masses share the same gravitational potential between each other. The direction of the current is determined by which charge is less inert (the electrons). Here the gravity analogy can be used again. If you fall towards earth you are moving while the earth stays more or less put because it is much more inert. It does however have the same potential as you do. So in a circuit, the positive side is the earth and one electron is analogous to you falling.

Electromagnetism – Does Electric Current Flow from Higher Potential to Lower Potential?

Electric fields exert forces on charges. The charges then respond to these forces according to Newton’s laws.

Suppose you have two objects with charges $q_1\neq q_2$ and masses $m_1\neq m_2$, initially at rest. Coulomb’s Law tells you the magnitude and direction of the force on each charge. In order to conserve momentum, the less-massive charge will do more of the moving. This is why, for instance, we can approximate the nucleus as “stationary” when we discuss atoms.

If you connect two charges with a macroscopic conductor, you can generally consider the motion of the charge carriers without worrying about the back-reaction of neutral bulk of the conductor — just like, when you blow out a candle, you don’t generally have to treat your body like a rocket engine which recoils away from the candle.

In metallic conductors the current is carried by electrons which flow towards high potentials — that is, antiparallel to the electric field. But in electrolytes, it can be that the positive charges have more mobility than the negative charges.

If I understand your specific scenario, you have (collapsing to one dimension)

(Q)              (q)         (2q)
 O                A           B

where the charge $Q$ at the origin is much larger the charges $q,2q$ at $A$ and $B$. You weren’t specific about the ratio of charges, here $q_B = 2q_A$, so I picked one arbitarily; you also weren’t specific about the ratio of distances, except that $r_B > r_A$.

If you connect your two “test” charges with a conductor, you’ll have two charged objects instead of three:

(Q)              (      3q     )
 O                A———————————B

The charges on the surface of the wire will distribute themselves so that the electric field on the inside of the wire is zero. Exactly how this is done can be quite complicated, especially if the wire isn’t straight: electric charges have a habit of accumulating on the high-curvature parts of surfaces, like corners or like the pointy ends of lightning rods. In general the combined positively-charged object $A—B$ will be more negative at its $A$ end and more positive at its $B$ end.

You set up $B$ to be more positive than $A$ in your initial condition; therefore the question of whether charge flows from $A$ to $B$ or vice-versa is quantitative, rather than qualitative.

Best Answer

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