Trying to understand the electric potential and potential energy

Question

I am trying to understand some facts on electrical potential and potential energy. It is quite confusing in the text to say that the zero potential could be freely chosen for convenience. In that case, how do you interpret the associated energy and the sign.

Let me start with the gravitational energy as analogy. Let's said I am standing on a high platform H h meter above the ground. If I am free to chose where the zero potential is, I consider the following two cases:

1) Assuming the ground has zero potential energy, the potential energy that I have is $mg(H-0)=mgH$. This makes sense to me since I have positive energy since I am standing in high place and I have potential to fall if I jump so my positive energy will convert into kinetic energy.

2) But what happens if I choose the zero potential at where I stand so the ground is at $-H$, so my energy is $mg(-H-0) = -mgH$, so is that negative really mean? I can only have one specific energy because I am standing at one place. So how does this negative potential energy related or equal to the positive one at 1)?

Now come to my question on electrical potential and energy, it is even more confusing because we have two different kind of charge (positive and negative). Let's consider the following case

Two metal terminals are connected to a battery (voltage is $V$). If I have an electron between the terminals, what is the potential energy the electron experience? As I learn from the text, the electrical potential energy should be $qV$, where $q$ is the charge of the particle, $V$ is the potential difference. Again, if we are free to choose the zero potential point, it may end up with two potential energy

1) By choosing the negative terminal as zero potential point so the potential energy for the electron should be $-e(V-0)=-eV$, which is negative. This makes sense to me because a negative charge will move away from the the negative terminal instead of the positive one, so it has negative potential energy (it won't move by itself towards the negative gate).

2) By choosing the positive terminal as zero potential point so the potential energy for the electron should be $-e(0-V)= eV$. It is hard to understand what is this positive potential really mean.

eshaya · Answer

But your height on the hill is Rearth + H, where Rearth is the radius of the Earth and, so would U = mg(Rearth + H) (never mind the integral needed for now)? But wait, the Earth is in the potential well of the Sun so shouldn't U = mg(sun)(Rorbit - H)? Luckily, you don't need to know an absolute potential which would require knowing the distances, masses, and charges to everything in the universe. You can pick any point as a reference point and work with differences from that point. Forces depend on the gradients (instantaneous change with distance) in the potential and work depends on changes in U from one point to another. Electrons move up the local potential gradient.

Answered by eshaya on December 3, 2020

nasu · Answer

I think you confused yourself with your example.

The zero of potential energy is at ground level. So you have mgH at initial position and zero at ground level. 
So your initial energy is +mgH above ground energy.
The zero of the potential is at height H. So your initial energy is zero and the energy at ground level is -mgH. 
So your initial energy is +mgH above ground energy.

See? Compare the last sentences of the two cases. There is no difference.
In both cases the PE(initial) - PE(final) is the same. In your calculation you switched the terms for case 2 so of course you got negative sign.

Shreyansh Pathak · Answer

I would like to give a general answer for both gravitational and electric systems considered here.Please read the definition below:-

The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system.$$Delta U_{system}=-int vec{F_{cons}}.dvec{s}$$

For example:- Take a ball and earth as your system then gravitational force will be an internal conservative force for the system.Now release the ball from a certain height $h$ then the work done by the gravitational force is $mgh$ so,corresponding change in gravitational potential energy will be given as $-mgh$.
Now consider the following points:-

Change in potential energy of the system is only defined for multi particle system.

Absolute potential energy is undefined.

As absolute potential energy is undefined and we can just calculate change in potential energy of a system then if we say that change in gravitational potential energy is $-mgh$ when we release a ball from certain height $h$ then the change will remain the same whether you choose your reference point as $infty$ or $0$ or $2h$ etc.

Sarah Messer · Answer

In all cases, you should have - = . Your confusion stems from swapping variables / signs halfway through the calculation. Gravity case 1: initial energy = mg(H-0) = mgH final energy = reference energy (ground level) = mg0 = 0 potential energy relative to ground = mgH - 0 = mgH Gravity case 2: initial energy = reference energy (platform level) = mg0 = 0 final energy = mg(-H) = -mgH, because the ground is H below the platform. potential energy relative to ground = 0 - (-mgH) = mgH Circuit case 1: initial energy = electron on positive terminal = -eV final energy = electron on negative terminal = reference energy = 0 potential energy = -eV - 0 = -eV Circuit case 2: initial energy = electron on positive terminal = reference energy = 0 final energy = electron on negative terminal = eV potential energy = 0 - eV = -eV

Trying to understand the electric potential and potential energy

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