Comparison of voltage to gravitational potential

Question

I'm on my way understanding what Voltage is, and came across this great video explaining the concept of electrical potential using an analogy to gravitational potential. I'll write what I understood from it and what I didn't.

Gravitation

With gravitation, we associate an object with a gravitational potential energy, measured in Joules. That it possesses when placed in some point in space. This is how much work can the object do on its way down to the lowest position possible, or conversely, how much work needed to lift it up to where it is now. This value is a function of the height, the mass of the object, and the gravitational strength.

We associate gravitational potential to a position in space, measured in Joules per kg, which tells us how much potential energy will a 1kg hold when it will be placed in that position. This value is a function of the strength of the gravitational field, and the height of that position from another position to which an object can fall to.

When we place an object in some height, say 2 meters, and let it fall to a height of 0.5 meter, each kg of that object is losing 1.5 * 9.8 Joules of potential gravitational energy when it gets to the lower position. The object will fall because nature tends to lower potential energy.

So we can say that two points in space are associated with gravitational potential, which is how much 1kg of mass will do when it falls from the higher to the lower.

Electricity

Now let’s talk about electricity. Instead of mass we talk about charge, and instead of gravitation we talk about electric field.

A charge, when placed in an electric field, is associated with electrical potential energy, measured in Joules, that it possesses when held in that place. This is how much work can that charge do when released and repealed by the electric field, or conversely, how much work needed to be done to get it to this place. This value is a function of the strength of the electric field, the size of the charge (number of coulombs), and the distance from the charge creating the field.

We associate electrical potential to a position in space, measured in Joules per Coulomb. This is how much potential energy 1 Coulomb of charge will hold when placed in that position. This value is a function of the strength of the electrical field and the distance from the charge creating it. As with gravity, when we place a charge in an electrical field - a place with some electrical potential associated with it - and let it repealed to a point that has less electrical potential, each coulomb of charge will lose the difference of electric potential between the two points. If the difference in electrical potential between the two points is 9 Joules/Coulomb, each coulomb will lose 9 Joules of electrical potential energy when moved from the higher potential point to the lower.

Voltage

Voltage is the difference in electric potential between two points. A battery's voltage for example is the difference in electrical potential between its two poles. This difference is basically the amount of Joules of energy 1 Coulomb has when placed in the positive pole more than it has when placed in the negative pole. A coulomb placed in the positive side of a 9 Volt battery can do 9 Joules of work more than a coulomb in the negative side. The prof. in the video compares it to a ball placed 2 meters above the ground, that we hold above a table placed 0.5 meter above the ground. There's a difference of gravitational potential between the two points: the top has 2 * 9.8 Joules/kg and the bottom has 0.5 * 9.8 Joules/kg. When the ball is released, each kg will lose 15 Joules/kg of gravitational PE.

What I don't understand

My question is as follows: Consider the analogy to the ball placed above a table. Eventually, the bottom point (i.e. the table) has zero gravitational potential energy. The professor present it as a point with some gravitational potential, but as far as we concern it has no potential since once the ball reaches that point, it will have no potential energy since it can't fall anymore, so when we talk about gravitational potential, we are really talking about a potential of one point, which is some point that from there the object can fall, and that potential is a function of the height from that bottom point. We can express that height as the subtraction of the distances of the two points from a third point, but what is the use of this?

I guess my question comes down to this: We say that Voltage is the difference in electrical potential between two points, and we define electrical potential of a position as the amount of Joules of work that can be done by 1 Coulomb of charge placed in that position. From this I understand that for a given position, there is some potential, regardless to any other point, and voltage is two points with a difference in their electrical potential. Put another way: the professor in the video explains that: When we say that this battery is of 9 Volt we're actually saying that one Coulomb placed at the positive side can do 9 Joules of work more than a Coulomb placed at the negative side. From that I understand that there is some amount of work that a Coulomb at the positive side can do, and another amount of work that a Coulomb at the negative side can do, and the difference between them is the voltage of the battery.

But I understand that it's not the case... Can someone clear this for me?

Answer 1

The electrostatic force and the gravitational force are both conservative forces, meaning that the work done by the force (change in kinetic energy) is independent of the path taken and only depends on the starting and ending positions. The negative of the work done by a conservative force is defined as a difference in potential energy.

The electric field $\vec E$ is defined as the force per unit charge, and voltage is the potential energy per unit charge (joule per coulomb) for an electrostatic field. $V = -\int_{r_a}^{r_b} \vec E \cdot d \vec r$ where is $V$ is the difference in voltage from position $r_a$ to $r_b$.

Similarly, $PE_g = -\int_{r_a}^{r_b} \vec F_g \cdot d \vec r$ where is $PE_g$ is the difference in gravitational potential energy from position $r_a$ to $r_b$, and $\vec F_g$ is the force of gravity.

With respect to your question, a 9V battery has a difference in voltage of 9 V across its terminals. The voltage of either terminal with respect to say the earth is undefined unless one of the terminals is connected to earth in which case that terminal has the same voltage as earth which is typically taken as $0$ volts. No matter the absolute potential of either terminal with respect to earth, the difference in voltage between the battery terminals is 9 V. A charge $q$ coulombs (positive) experiences a change of voltage of -9 V ( change in potential energy of -9*q joules) when moving from the positive to the negative terminal, typically through an electrical circuit attached to the battery. For a purely resistive circuit the decrease in potential energy is equal to the change in internal energy (heating) of the resistance. (In reality negatively charged particles, electrons, move from the negative terminal to the positive terminal but the effect is the same as described for a positive charge.)

Answer 2

It depends on what the potential of the person is, defined with respect to the battery, since current flows according to potential differences (as you've observed, this difference is physically meaningful, not the absolute voltages). If the person is at $0V$ with respect to how you've defined the voltages of the battery terminals, then the terminal with $+9V$ will try to equalize the voltage difference by producing a current to the person. If the person is at $9V$, however, the opposite terminal will produce the current.

Notice that in the latter case, we can shift all three voltages by $-9V$, and we see that the scenario is equivalent to the one in which the person is at $0V$, and the terminal of the battery with $-9V$ will produce a current. This constant shift across the board can be done in any case, since it will not affect the differences between voltages: $$(V_1 + V_0) - (V_2 + V_0)= V_1 - V_2,$$ and this voltage difference is again, the only physical quantity that induces electrical current here.