Question

A particle with charge q and mass m moves in static external fields E0(7) and Bo(r). If Φ。(%) is the electrostatic potential at the position of the particle, and there is negligible radiation, the usual statement of conservation of energy for this system is Show that this equation agrees with the power balance represented by Poynting's theorem only if one adds to the latter the power supplied by an external energy source to maintain Eo() and Bo(F) against the work done on their sources by the fields produced by the particle. Why is the latter work no n-zero?

Answer 1

The theorem is - partial differential u/partial differential t = nabla s + J E where S = E_0 (r) times H_0 (r) u = 1/2 (elementof_0 E E + B/mu_0 B) The eqn 1 is for unit volume Total mechanical energy due to em - field is U_M = uv = 1/2 (elementof_0 E E + B/mu_0 B) v where v = volume Now force, F = q_1 q_2/4 pi elementof_0 p^2 almostequalto q^2/4 pi elementof_0 l^2 almostequalto q^2/elementof l^2 and from Biot savart law, B = mu_0/4 pi J/p^2 B/mu_0 almostequalto I^0 l/r^2 so, 1/elementof_0 = Fl^2/q^2 and B = F/qv = F = 4E u_M = 1/2 (q^2/Fl^2 E^2 + I degree l/l^2 F/qv)v = 1/2(F/l^2 v + 1 Fl) = 1/2 [Fl + Fl] = Fl = work done = change in kinetic energy = 1/2 mv^2 and power supplied per unit energy is J E = I/A P/l = q/t Q/v [E = Q/l volume V = Al] \r\n so, energy