# Electric potential and kinetic energy in any flowing charge

We just started with electricity in school(grade 10) and I have some confusions regarding electric potential.

According to my book,

The potential at a point is defined as the amount of work done per unit charge in bringing a positive test charge from infinity to that point. In the given figure, a test charge $$+Q$$ is brought from infinity to a point $$P$$ in the vicinity of a positively charged body. If $$W$$ joule of work is done in bringing the test charge $$Q$$ coulomb from infinity to the point $$P$$ in the vicinity of a positively charged body then the electrical potential at that point is given as $$V=\frac WQ$$.

Now my question is:

• Based on my understanding from another question which I had asked, suppose we are lifting a ball kept on the ground upwards. The ball exerts a force $$-mg$$ on us and we exert an opposing force $$F_o$$ to it, out of which the net work done on the ball by the net force acting on it ($$F_o - mg$$) which causes chage in the kinetic energy of the ball. The change in potential energy is caused by the external force equal to the force of gravity on the ball which will be equal to $$-(-mg)=mg$$.

• Similarly for electric potential, I thought that to get the unit charged object moving towards the bigger charged body, we need to apply a force(say, $$F_e$$) greater than the force of repulsion(say, $$-F_r$$) between them, out of which the net force $$F_e+(-F_r)$$ provides the kinetic energy to the test charge, and $$F_r$$ adds up to the potential of the charge $$Q$$.

Is the last point correct? Did I misunderstand anything? I have another question too, but I would reserve that for another post considering I am unsure of the above points which I have mentioned. My book mentions absolutely nothing about kinetic energy, so I am unsure of its presence and condition in charged bodies.

Yes, TO ACTUALLY MAKE IT MOVE a net force must be greater, however let's say I provide a force infinitesimal greater, it will have some velocity. And then moving along that distance I will have done X amount of work against gravity.

The way you should think of it, is that IF an object moves through a distance, the Electric field is going to either be doing positive or negative work per unit charge.

Meaning if lets say in moving through a distance A to B the field does -W work per unit charge. The total amount of energy I need to provide to overcome that negative work, will be "W" aka the negative of the work done by the field. As W+(-W) = 0 so if I give W work over a distance, then the field will equally "Steal" that work from me, causing no net gain of KE by the object

Its all about measuring how much work is either done or lost on an object by the field IF it were to move along a certain path. The negative of this is the total amount of energy that "i" put in, to overcome this

Thought experiment:

Given an object moves vertically with an initial velocity V, as the object moves gravity will do negative work on it, such that at a height "H" the total amount of negative work done by gravity is equal to the objects kinetic energy, aka the object STOPS at height H

now imagine that now, the object is at the ground with the same initial velocity. But this time, I am applying an EQUAL BUT OPPOSITE force to gravity. The net force is zero, so the object will continue to move upwards at a constant velocity. When the object reaches a certain distance upwards "A", I will then STOP that applied force, clearly now the object is at a height "A" and will then continue to move upwards by the same amount as the first scenario and then stop.

Meaning that I have now applied a force -F over a distance A but this time the object has now has a maximum height A+h

Aka, A meters higher. Clearly the work that I have put in, has caused the object to move A meters heigher, than the previous scenario.

Aka, that is the total amount of work I have to do in order to overcome the force of gravity through a distance A.

More mathematically.

Consider the equation

$$1/2 m v^2 + \int_{A}^{B} F \cdot dr = 0$$

Aka, an object with an initial Ke, moves through the distance a to b, in the presence of a force field, such that when it reaches B, it stops. Well how much energy do I need to give the object such that in the presence of F, it stops when it moves through a to b?

Clearly

$$1/2 m v^2 = - \int_{A}^{B} F \cdot dr$$

this is exactly the same as if I apply the work to the object over the distance, instead of at the very start in the form of Ke

• Alright, but even that small acceleration adds on to the energy of the electron, right? As it was initially at rest and now has a velocity. I read that $W_{external}=\Delta KE + \Delta PE$. If that's the case then is it safe to say that say, in an ideal cell connected to an ideal circuit (with no internal resistance or resistance whatsoever), the energy provided by the cell will be more than the electromotive force(which will be equal to terminal voltage here due to no resistance) as specified by the cell manufacturer? Commented Dec 21, 2021 at 2:39
• Just on that last comment you said. No, there are no opposing forces so the emf is exactly equal to emf provided by the cell. And The first thing, you said. I mean technically if I DID give it an initial kinetic energy then obviously the total energy of the particle is higher. But that doesn't mean I need to do more work against gravity. I think you should forget the idea of potential being the amount of energy I need to Actually move something from A to B. And instead think of it as how much work would I have to do on an object against gravity IF something WERE to move in that path. Commented Dec 21, 2021 at 3:13
• If I give something an initial KE, and it moves a distance, and stops. Then that is the total amount of energy gravity "steals" from that particle. Lets say I run a race, work is being done by my muscles to move me and give me a kinetic energy. Now imagine as I'm running alot of wind is pushing me backwards, and it is slowing me down. How much energy do I need to "give " to the wind? Well if moving from A to B the wind is adding "-E" energy then the total amount of energy I would say I need to "give" the wind would be "E" aka the negative of how much work the wind would do on me, Commented Dec 21, 2021 at 3:19
• I get it. So you mean when stopping, the body loses as much KE as it gained but has an overall gain in PE right? Commented Dec 21, 2021 at 3:44
• Well the principle of work is experimentally derived, but the derivation of Kinetic energy from $$\int \vec{F} \cdot \vec{dr}$$ Is as follows, $$\int \vec{F} \cdot \vec{dr}$$ $$\int m\frac{d\vec{v}}{dt} \cdot \vec{dr}$$ $$\int m\frac{d\vec{v}}{dt} \cdot \vec{v} dt$$ use inverse chain rule $$1/2 m v^2$$ Commented Dec 21, 2021 at 14:39

Bringing in the test charge requires only a momentary acceleration. After that the applied force can be equal to the repelling electric force. The work done during acceleration can be recovered when the test charge comes to rest at the position under consideration. At that point it has no kinetic energy.

• So the net kinetic energy is $0$? Commented Dec 20, 2021 at 16:35
• Also, this post says that the kinetic energy is so less that it can be considered $0$. physics.stackexchange.com/questions/562552/… Commented Dec 20, 2021 at 16:38
• How can you explain potential difference across a circuit being the work done by a cell with the net kinetic energy being $0$, when the electrons don't come to rest and are constantly moving, and ignoring resistance, even accelerating? Commented Dec 20, 2021 at 16:40
• Your confusing yourself alot. In the presence of a force field. I must do work against that force. Potential is the work that I would have to put in. EMF/potential in the context of circuit theory is the work done BY the field Commented Dec 21, 2021 at 11:36
• In the classical model of current flow, the free electrons (which have very little mass) are rapidly bouncing around in thermal equilibrium with the atoms of a conductor. The energy they gain by being accelerated between bounces by the E field is dissipated in the random collisions. The small drift velocity associated with the current results from all the accelerations being in the same direction. Commented Dec 21, 2021 at 14:35