Electric Potential Energy and Work

[rainstom07]

Hello.

I need help clearing up who's gaining potential energy and who's losing potential energy from an particular object perspective. And, from that object perspective, whether the object doing positive work or negative work.

$$W = \int \vec F \cdot d\vec x$$
$$\Delta U = -W$$

Question 1:
From the left-hand perspective, the left-hand moved in the positive x-direction and it applied a force in the positive x-direction therefore... it did positive work on the right-hand. Is this correct?

Question 2: continuing from question 1
If the left-hand did positive work on the right-hand, does this mean that the left-hand has lost potential energy?

Question 3: continuing from question 2
If the left-hand has really lost potential energy, how come when the left-hand "lets go" (stop applying a force in the positive x-direction), the left-hand want to naturally accelerates in the negative x-direction. Doesn't accelerating mean that the left-hand is converting it's potential energy into kinetic energy? Where does it get that energy from?

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Question 4:
From the right-hand perspective, the left-hand is moving in the positive x-direction, but it's also moving against the electric field from the right-hand so... the work done by the right-hand on the left-hand is negative.

Question 5: continuing from question 4
The left-hand gains potential energy while the right-hand loses potential energy. Is this correct? Why/why not?

Also... what hell does it mean for the right-hand to lose/gain potential energy?

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Thanks in advance!

 

[kuruman]

A single object does not "have" potential energy and therefore it can neither gain nor lose potential energy. Potential energy is associated with at least two objects. In the absence of non-conservative forces, if the potential energy associated with the objects decreases, the kinetic energy of both objects will increase by exactly the same amount. It is a mistake to think otherwise, but the confusion originates in introductory physics instruction. Here is an example.

Question
A man tosses a rock straight up in the air. Find the rock's change in potential energy when it rises to height $h$ above the surface of the Earth.

What you may see in a textbook or class lecture
Since the rock is rising, it gains potential energy. Taking the potential energy to be zero at the point of launch and since $U=mgy$, we have $\Delta U=mgh-0=mgh$. Answer: The rock gains potential energy $mgh$.

What is actually going on but you don't get to see
The potential energy of the Earth-rock system is $U=-\dfrac{GM_e \ m}{r}.$ Here, $M_e$ is the mass of the Earth and $r$ is the distance from the Earth's center to the rock. Initially, the rock is on the surface of the Earth, i.e. at distance $r=R_e$, where $R_e$ is the radius of the Earth. Finally, it is at $r=R_e+h$. Therefore the change in potential energy of the Earth-rock system is
$$\Delta U=-GM_e \ m \left(\frac{1}{R_e+h}-\frac{1}{R_e} \right)=\frac{GM_e \ m \ h}{R_e(R_e+h)}$$
So far the expression is exact. Considering that the radius of the Earth is much much larger than any height any man throw a rock straight up we can approximate $R_e+h \approx R_e$, in which case $$\Delta U \approx \dfrac{GM_e \ m \ h}{R_e^2}$$ Now the force with which the Earth attracts the rock is$$F=\frac{GM_e \ m}{R_e^2}=mg~\rightarrow~g=\frac{GM_e}{R_e^2}$$Therefore, to a very good approximation we can write the change in potential energy of the Earth-rock system as$$\Delta U=mgh.$$Now here is the important point: Both the Earth and the rock move farther away from their initial positions when the rock reaches height $h$ as required by momentum conservation. However, because the Earth is $\rm{10^{25}}$ times more massive than your typical rock, it moves by distance $\rm{10^{-25}}h$, many orders of magnitude less than the size of an atom. So to a good approximation, which you don't get to see when gravitational potential energy is first introduced, we say that the transfer of energy from potential to kinetic and back concerns only the rock.

It should be clear then that if you have two hands of equal mass, any gain or loss of potential energy involves both hands equally, whether we are talking about gravitational electric or magnetic forces between them.