00:00 - 01:41 = Clear, 01:42-10:41 = POTUS 😮

If your concept is clear, answer my question above

Correct, you can calculate E from the gradient of the potential. In general, the electric field also includes a term for the time derivative of the magnetic vector potential, dA/dt.

@@BariScienceLab 0 At the quantum framework, everything in the world is made up by the particles in the standard model of particle physics and their interactions. All other theories can be shown mathematically to be emergent from this level. The macroscopic electric and magnetic fields are the result of the addition of a high number of charged quantum particles, what is a "force" at the quantum level is built up to "force" of classical level electric and magnetic fields. Anytime a charge is accelerated in the apparently static field of another charge, this acceleration violates the assumption of "static-ness" because it means that (classically) EM waves get radiated by the mutually accelerated charges. In the quantum field theoretical description these actual waves correspond to real photons. I don't know what your point is. If a charge experiences no acceleration, the electromagnetic field must be zero where it's located. Nothing magical there that would invalidate the general field concept. The superposition of two static fields to obtain a zero field strength at some location is an artificial construction that serves as an excuse for not knowing the underlying physics (i.e. QFT). -

@@BariScienceLab \hat{H} \psi(\mathbf{r}) = \left[ -\frac{\hbar^2}{2m} abla^2 + V(\mathbf{r}) ight] \psi(\mathbf{r}) = E \psi(\mathbf{r})

@@aninditabanerjee6862 That's the time-independent Schrödinger equation.

@@BariScienceLab ok

India ka badnam karega thu...konsi English he ye