Recent content by lightlightsup

https://www.patreon.com/posts/least-action-q-126148989 It seems that they're aware that this laser experiment should be repeated with more exacting instrumentation. I am surprised that there is no one on YouTube with a repetition of this experiment.

The guy in this thread seems to have limited the laser leakage, except, I can't tell how long his cardboard goes on for, relative to the starting point of the laser light.

Does anyone have a link or reference to the Feynman lecture video that this Veritasium video uses? I'd rather hear this from the horse's mouth.

I am surprised that this test is not being replicated with low-leakage lasers.

Please let me know if this experiment is 'legit'. Can leakage or something else other than "light considers all paths" explain this phenomenon?

These don't carry the explanation I'm looking for. Why doesn't the positive test charge (cat, here) experience an attraction to the wire if the electrons are getting length contracted?

Are you able to explain this phenomenon, then?

(1) https://www.youtube.com/results?search_query=relativity+and+electromagnetism (2) If I understand this correctly, then, well, shouldn't all current-carrying wires exhibit a small amount of positive charge?

What does "F(r,θ) = 0" mean here?

JACKPOT. That's what I was missing. Thank You! I should have thought of it as a displacement of Q.

Net charge is still zero for the circuit before and after. It's just the potential difference caused by the charge separation on C1 that leads to the different Qs you see in the end state?

Voltage drop on C1 led to Q gained on C2 and C3?

So, it may seem like we went from 48mL excess to 64mL excess and violated conservation of mass. But, in reality, the water is in the pipes as well? Just at a different pressure? There's some misconception I have that's causing the math to not add up: LH: 48mL -> 32mL (-16mL) RH: 0mL -> 32mL...

+Q will accumulate on one side, -Q on the other. Net charge will be zero. Let's say I have two capacitors in series. If I am asked what the total charge accumulated on any 1 plate on both capacitors is, what would the answer be? Q+Q = 2Q? Q of the equivalent combined capacitor = Q on any of the...

There must be something I'm not understanding about capacitors in series. I know that we can treat them as one equivalent capacitor with: (1) with 1/Ceq, (2) same Q as anyone of the capacitors, (3) and add up the Vs for the sum total V across them. If the equivalent capacitor (Ceq) would...