B Some Questions about Electric Current/Capacitors to help my understanding

[sadgirl]

TL;DR Summary

Need help with my understanding of the concepts below.

Q 1) In electric currents, in a battery, the positive charge starts at the negative terminal and gains energy through emf which forces the charge to go to the positive terminal of the battery, with plenty of energy. This voltage is the energy difference between the terminals. For the charges to flow down the circuit, is the potential energy converted to kinetic energy, or does it just flow down due to the difference in potential? Also can the potential difference be seen as potential energy per unit charge?

Q2) For capacitors, as the charges increase the electric field strength increases, which in turn means that the force will be stronger thus, the Work done on the charge causes it to gain a lot of potential energy. Does this potential energy mean a difference in voltage? And for the charges (let's say negative charges) to go to the negative capacitor plate, the force must be applied along the whole way the charge travels right?

Q3) Why is the potential energy with capacitors = 1/2QV?

 

[anuttarasammyak]

Q1) and Q2) seem all right to me.
Q3) In counting energy we have to think of process of accumulating charge.
Now the accumulated charge is q which produce voltage of q/C. For further accumulation of small charge dq, we input energy of q/C dq. For next pop-up the voltage becomes higher, q+dq /c which means we have to input more energy for next pop-up. To get enrgy to input for the process we should integate it from q=0 to final Q as
\int_0^Q \frac{q}{C}dq = \frac{1}{2C}Q^2=\frac{1}{2}QV= \frac{C}{2}V^2
1/2 comes from this integration.

 

[Tazerfish]

Most of my answer goes beyond what's immediately useful to the problem; nonetheless, I hope it improves your understanding of electricity and maybe you'll find it fascinating ;)

Q 1.1) For the charges to flow down the circuit, is the potential energy converted to kinetic energy, or does it just flow down due to the difference in potential?

You're mostly right!


When you're talking about kinetic energy, you're probably thinking of a rolling ball or some other example from classical mechanics.
Actually, that is sort of what happens. As the electrons move "down" the electrical potential, they accelerate and pick up kinetic energy, much like a ball rolling down a hill.

Strangely, they never get very fast!
Why not?
Well, I was kind of lying. They're erratically moving around the metal at mind numbing speeds, bouncing around aimlessly.
(This will all make much more sense if you know that heat is a form of microscopic motion, a vibrating and bouncing around of atoms and molecules.)
One moment an electron moves left, the next moment it moves right, back and forth and up and down.
In the end though, it all cancels out!
They never get very fast on average, moving fast and never getting anywhere.
What redirects them?
Collisions with the atoms.
It's as if that ball rolling down the hill was stuck in a pinball style maze of bouncy walls, moving very fast while hardly ever getting any further downhill.
On average, electrons actually drift so slowly in a conductor, that a snail could outpace them (no joke)! Because they are also extremely light (and slow), there is basically no "kinetic energy" in an electric current.


So where does the energy go if it's not making the electrons move from one place to the next?
Resistance!
Countless times a second, the electrons collide with the atoms that make up the metal; countless times a second, they transfer energy into the movement and vibrations of the material. In other words, they heat it up through resistance!

It's quite bizarre. The average snail-like drift velocity of electrons is about $10^{-3} \frac {m}{s}$, the random whizzing around they do because of heat is about $10^{5} \frac {m}{s}$ and the velocity with which signals travel through it is $3*10^{-8} \frac {m}{s}$ also know as the speed of light.
Crazy, right?

Also can the potential difference be seen as potential energy per unit charge?

Exactly right! 100%
That's what voltage is.

And for the charges (let's say negative charges) to go to the negative capacitor plate, the force must be applied along the whole way the charge travels right?

Interestingly enough, no.
Let's consider the analogy with the ball on the hill again.
The height of the ground is the potential.
However, the force required to push the ball isn't dependent on how high the ball is.
What is important is the steepness!
How fast does the height change with distance, that's what matters.

Going back to electricity, the height is the electrical potential and what matters for the force is the change in electrical potential over distance, the electric field.
(You may have heard of that.)
It's like the steepness.
When you have a big voltage (electrical potential difference) between two close points, then the electric field is strong.
If the voltage is low *or* if the distance between the two points is large, then the electric field is weak.


When you're looking at a capacitor in real life, most if the electric field is actually concentrated close to the capacitor itself. So it's going to require quite a bit of force to get it off/on the capacitor, but moving along the wire far away from it is basically "effortless".
There's hardly any electric field so far out from the capacitor and so there also isn't much force you have to push against.

I hope this was more enlightening than confusing :P

 

[Drakkith]

In electric currents, in a battery, the positive charge starts at the negative terminal and gains energy through emf which forces the charge to go to the positive terminal of the battery, with plenty of energy. This voltage is the energy difference between the terminals.

Are we talking about a battery discharging or charging? In discharge, we model things as positive charges going from the positive to the negative terminal (conventional current), though of course we really know that it is electrons flowing from negative to positive. For charging the battery it would be the reverse.

A few remarks on terminology (taken from wikipedia):

Voltage: the difference in electric potential between two points.
Electric potential: the amount of work/energy needed per unit of electric charge to move the charge from a reference point to a specific point in an electric field.
Volt: the unit of measurement of voltage. One volt is defined as the electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.

For a 12-volt source, each coulomb of charge does 12 joules of work as it passes through the circuit, and for a 24-volt source each coulomb does 24 joules of work. Note the definition of a volt. A 12-volt source passing 1-amp of current through a resistor would dissipate 12 watts of power, which is 12 joules-per-second. A 24-volt source passing a 1-amp current through a resistor would dissipate 24 watts, or 24 joules-per-second.

For the charges to flow down the circuit, is the potential energy converted to kinetic energy, or does it just flow down due to the difference in potential?

The type of energy that the potential energy is converted to depends on the circuit. A circuit consisting purely of resistive elements would convert nearly all of the potential energy to thermal energy, which would be lost to the circuit's environment. A voltage source connected to an electric motor would convert the potential energy into kinetic energy, rotational energy, another form of potential energy, or almost any other form, it just depends on what the motor is connected to. A voltage source connected to an LED would convert the potential energy into radiant energy, or light.

I would say that the charges flow because there is an electromagnetic field that forces them to flow. It is often described in a somewhat simpler way by saying that the charges flow from high energy to low energy, or because there is a difference in electric potential.

Also can the potential difference be seen as potential energy per unit charge?

I think so, as that basically follows directly from one of the definitions of voltage: the work done per unit of charge to move it against the field, which also means the potential energy changes accordingly. A coulomb of charge placed at point A with a 10-volt difference between points A and B can be said to have 10 joules of potential energy relative to point B.

Q2) For capacitors, as the charges increase the electric field strength increases, which in turn means that the force will be stronger thus, the Work done on the charge causes it to gain a lot of potential energy. Does this potential energy mean a difference in voltage?

It means that the voltage across the plates of the capacitor increases, yes. I'm not sure if that's exactly what you were asking. Notice that, until charged, a capacitor's plates are always at a lesser voltage than the source. So a capacitor charged to 8-volts placed in series with a 10-volt source will charge identically to a discharged capacitor (0-volts) placed in series with a 2-volt source. The difference between the two is that the first capacitor is, ultimately, charged to a higher voltage and when discharged each unit of charge will dissipate more energy. At least until the first capacitor discharges enough to get down to the same voltage as the 2nd capacitor.

And for the charges (let's say negative charges) to go to the negative capacitor plate, the force must be applied along the whole way the charge travels right?

In real circuits, yes, as there is always some resistance, even in conductors. Though talking about 'force' in the context of an electric circuit is tricky, as you will virtually never talk about the force on a charge in the context of electric circuits. Obviously it's there, otherwise the charges wouldn't move against a resistance, but nothing in my basic EE class ever talked about force at this level. I suspect this is because of two reasons:

1. Force is sort of 'baked in' to the definition of voltage and electric potential and such.
2. Trying to find the force on any one of potentially trillions of trillions of charges in a conductor or other component is meaningless, as it likely varies in both space and time at the subatomic scale.

Circuit theory simply doesn't get down far enough in scale to worry about the force on individual charges flowing in a circuit. Much more useful to think about voltage drops, current, impedance, etc.

 

[sophiecentaur]

TL;DR Summary: Need help with my understanding of the concepts below.

Q3) Why is the potential energy with capacitors = 1/2QV?

Simple answer: Putting one Joule of energy into a capacitor by any method will always involve 'wasting' 1J of energy somewhere in the system. Something has to get hot, something must get moved or distorted or some EM energy must get radiated however you design your system.

 

[Gavran]

and the velocity with which signals travel through it is $3*10^{-8} \frac {m}{s}$ also know as the speed of light.

$3\cdot10^{-8}\frac{m}{s}$ is not correct.
$3\cdot10^{8}\frac{m}{s}$ is correct.

 

[Gavran]

Also can the potential difference be seen as potential energy per unit charge?

The electric potential difference can be seen as the change of electric potential energy per unit charge experienced by a test charge that is so small that the disturbance of the field under consideration is negligible.

 

[sophiecentaur]

TL;DR Summary: Need help with my understanding of the concepts below.

For the charges to flow down the circuit, is the potential energy converted to kinetic energy, or does it just flow down due to the difference in potential? Also can the potential difference be seen as potential energy per unit charge?

Thinking in terms of a wire just carrying a current is to forget the Work involved, which will be very little for a piece of copper wire.
When current flows 'usefully' from one terminal to another then significant Energy will be transferred. This can just be heating a wire or driving an electric motor which does some work. The kinetic energy imparted to the electrons is (as already discussed) extremely tiny because the average velocity is around 1mm per second and the total mass of all the electrons is a tiny fraction of the mass of the wire. So the Power transferred must be more than just the KE change.

Here's a really simple analogy. A bicycle chain has a mass of a few (say 200) grams and moves at about 0.5m/s; that's 0.2X0.25/2 J = 0.025J of KE. The Power delivered by a cyclist will be 100W (or more)or 100J/s. The chain would need to be going round very fast to have that much KE transferred every second. The Work done is via the tension of the chain. Power = tension X chain speed. Analogously, it has to be the magnetic force (torque) of the motor that transfers all the mechanical power. That's caused by the electric fields which cause magnetic fields to drive the motor round. For heating the wire, the interaction between the electrons and the atoms does the job.