Comments on: Poynting theorem and derivation

By: Dragon

Dragon — Tue, 14 Feb 2023 13:49:27 +0000

let us consider the term,
=> d/dt (H.H) = H.dH/dt + H.dH/dt
=> d/dt H^2 = 2(H.dH/dt)
=> H.dH/dt = 1/2 d/dt H^2
Similarly,
=> E.dH/dt = 1/2 d/dt E^2

Dragon — Tue, 14 Feb 2023 13:49:05 +0000

let us consider the term,
=> d/dt (H.H) = H.dH/dt + H.dH/dt
=> d/dt H^2 = 2(H.dH/dt)
=> H.dH/dt = 1/2 d/dt H^2
Similarly,
=> E.dH/dt = 1/2 d/dt E^2

Roshan verma — Sun, 16 Jun 2019 05:39:51 +0000

In reply to Avirup Mullick. because d/dt [H^2] = 2H.dH/dt

Mudit Srivastava — Wed, 10 Oct 2018 07:36:38 +0000

d/dt(H^2) =d/dt(H.H)= H.d/dt(H) +H.d/dt(H) => 2H.d/dt(H) = d/dt(H^2)

Lone Shehzan — Wed, 03 Oct 2018 08:34:24 +0000

In reply to Avirup Mullick. As we k/t dH^2/dt=2HdH/dt = HdH/dt =1/2 dH^2/dt

amsh — Sat, 03 Aug 2013 10:17:40 +0000

In reply to Avirup Mullick. please explain the question.

Avirup Mullick — Sat, 03 Aug 2013 06:09:02 +0000

How these two identities used? H. dH/dt = 1/2 dH2/dt E. dE/dt = 1/2 dE2/dt