You say "There is no electric field before closing the switch". I don't think this is true. Here is the same diagram with two negative charges electrons sitting between the plates of your capacitor(s). What way do they move? and why do they move, and what provides the energy for them to move?
________________ common plate
+ + + + + + +
- -
_______/ ________
-----------
Also the positive charges are way too close together. Their mutual repulsion will cause them to spread out over the plate pretty much uniformly.
Also the unconnected wire will have potentials at either end, as one end is closer to a large static charge.
That is incorrect and I noticed the same wrong explanation in Derek's video. The electrons and holes will be equal and they will extend just up to the switch even on the common plate.
Only after the switch is closed electrons and holes will move symmetrically on top and bottom plate.
Keep in mind the drawing is nowhere near to scale. There may be just 0.1mm between plates and plates may be a few meters long for the charged capacitor and then for the discharged capacitor that represents the transmission line the distance between plates may be order of magnitude higher but even if the same 0.1mm there charge will end at the switch both on top and bottom plates.
Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...
You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).
Lol - if that is the case, when you close the switch, what makes the electrons move across the switch? It seems you are saying that they are quite happy being all bunched up in a huddle on the left...
You never did say what happens to the two charges floating in the middle of the capacitor (assuming they are free to move if they have any force on them).
Yes that is exactly what I'm saying and I'm fairly sure that I'm correct. They will move across the switch when switch is closed and symmetrically on the top plate.
After the right side is charged they will be also uniformly distributed across the entire new capacitor that has now higher capacity.
This is assuming that distance between plates is the same on the left side and right side else if say on the right side the distance between plates is larger (lower capacity) there will be proportionally less electrons on that side.
No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.
________________ common plate
+++++++
- - << These guys which way do the move and why.
_______/ ________
-----------
If they are free to move, which way to they go and why?
No, the charges floating in the middle between the plates - the extra '-' signs... I'll go back to your original diagram, so I'm not changing two things at once.
________________ common plate
+++++++
- - << These guys which way do the move and why.
_______/ ________
-----------
If they are free to move, which way to they go and why?
Not sure why you had to draw those ones as they are already on the diagram. And yes they are incorrectly drawn under the bottom plate (limitations of this CAD tool ).
The ++++ and ------ symbols in my diagram represents holes and electrons in the plates/wires as that is what is of interest in this discussion.
They will move when switch is closed and they will move symmetrically on the top and bottom plate's.
To maybe help you understand that there is symmetry between the charges on the top and bottom plate imagine there is also a switch on the top plate exactly above the one on the bottom and now thing what will happen if you start with both switches open and only close the top one.
Do you think there will be any current trough that closed top switch ? If not that means charges remain arranged the same as they where before you closed that switch.
They will not move symmetrically, because there is no symmetry to begin with.
Electrostatics is relatively easy so I wrote a quick solver for distribution of 50 + charges and 50 - charges on 10m and 5m wires one 1m apart. (so a 10:1 aspect ratio).
You can make what you want of this, but the net horizontal force on each charge (other than those at the ends of the 'wires', where they are constrained) is zero.
Yes, one capacitor can charge another, and there is energy that can be taken from the system, one way or another (e.g. heating of components). Does it have any utility here? No, not that I can see.
In fact, I've lost track of exactly what you were trying to prove... I only joined in again as you were commenting that there was no electric field before the switch was closed, when there clearly one is present.
Can't be bothered. Enjoy whatever.
+------------+
| |
---- 0V ---- 0V
---- ----
| |
Gnd +----/ ------+
You charge up the capacitor on the left hand side, Gnd to the switch side, +10V to the wire common between the two capacitors. +------------+
| |
---- 10V ---- 0V
---- ----
| |
GND +----/ ------+
10V
+------------+
| |
---- 5V ---- 5V
---- ----
| |
GND +------------+
0V
You now have 5V across both caps - and half the energy goes out of the system as heat. Hontas Farmer is back still saying the Derek is both right and wrong acording to QFT/QED
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Double-posting, Dave?
https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432
I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
Double-posting, Dave?
https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432
Yes, because there are now two threads, and many people are not reading the other thread any more.
Q1) How did that 10V measured over the switch get there ? My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.
I've measured this all on the bench - there is 10V across the switch.
You close the switch a small spark is heard and the 10V of potential difference is now gone.Code: [Select]+------------+
You now have 5V across both caps - and half the energy goes out of the system as heat.
| |
---- 5V ---- 5V
---- ----
| |
GND +------------+
0V
What part am I missing?
Hontas Farmer is back still saying the Derek is both right and wrong according to QFT/QED
So let me double post this from the other thread, because I have yet to find a satisfying answer:
Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?
Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?
I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat. Let's say I 'generated' 1 joule of energy.
Has this energy traveled along the 1000 meters path?
If we take the Poynting style conservation of energy argument, we know that energy is created in the volume of space around where you 'generated' it. Energy was 'dissipated' in the volume of space around where you dropped it. So we can say energy flowed through the space between. But we can't say exactly the path that the energy took.
Edit:
However, if we decide that the potential energy is located where the mass of the rock is located (like we do by saying that the fields have energy), then energy flowed along with the rock. How much potential energy the rock contains is a problem without knowing the baseline zero potential energy of the system.
I hate analogies. Water analogy, rock analogy, whatever.
If "energy doesn't flow in wires" why he even needs the wires ?