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Veritasium "How Electricity Actually Works"
SandyCox:
--- Quote from: electrodacus on May 26, 2022, 07:17:02 am ---[How will energy travel outside wire if you shield the wire from electric fields?
--- End quote ---
Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.
How long will it now take for the bulb to turn on?
electrodacus:
--- Quote from: SandyCox on May 26, 2022, 01:20:25 pm ---
--- Quote from: electrodacus on May 26, 2022, 07:17:02 am ---[How will energy travel outside wire if you shield the wire from electric fields?
--- End quote ---
Let's think of the following setup:
1. Place both the battery and bulb in separate sealed boxes made of a perfect conductor.
2. Replace the two cables with coaxial cable made of perfect conductors and a lossless dielectric.
3. Connect the shields of the two coaxial cables to the two sealed boxes. (No pigtails allowed)
4. Connect the inner conductors to the bulb, switch, and battery, like in Vertasium's original video.
How long will it now take for the bulb to turn on?
--- End quote ---
You did not mentioned connecting the shield to anything.
Just connect the shield to say one side of the lamp.
But you do not even need all that complication to shield everything.
All you need is to shield a quarter of the wire so 10m of shield in Derek's experiment and connect that to the opposite side of the lamp/resistor.
______________________[resistor]_________________________
| ______________________________| |
| |
| |
| |
|___________________/ ____- +____________________________|
Hope you can see the diagram. The shield is only electrically connected to the right side of the resistor/lamp
What will happen is that the shield will now be the second plate of the capacitor formed with the lower wire and current for charging the capacitor will travel through that shield and connected to the opposite side of the resistor.
So there will still be current through the shield on left side and the line on the right side as the line capacitance charge but no more current through the resistor thus no more power and no energy until after the 65ns or so the electron wave needs to get to resistor through wire.
So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire.
SandyCox:
You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.
electrodacus:
--- Quote from: SandyCox on May 26, 2022, 03:52:57 pm ---You are answering a different question. The shield is connected to the two boxes. Connect the shield to ground (mother earth) if you want to.
--- End quote ---
That will not be called a shield if it is not connected. Earth has nothing to do with this isolated system so connecting the shield to earth will do nothing unless you want to shield against external influence.
The shield I proposed is the simplest that will shield against electric field and so it shows that even with blocking the electric field from getting to Lamp/resistor the energy travels through wires and gets to resistor.
To "shield" against energy traveling through wire all you need to do is cut a gap of say 1mm in the wire :)
SandyCox:
What is your answer to my original question? How long will it take?
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