By P. Hammond

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50). t This section could well be omitted on a first reading of the book. ' ELECTROMAGNETIC FIELDS A N D THEIR SOURCES 51 Consider first eqn. 49) : div F = -div grad φ + div curl W = - Ψ φ = s. 54) This is Poisson's equation and is satisfied by eqn. 52). Consider next eqn. 50): curl F = curl curl W = grad div W - V2W = с. 55) We shall be able to show that grad div W is zero. Assuming this result for the moment we have V2W = - c , which is Poisson's equation and is satisfied by eqn. 53). Hence eqns.

_jg^Dxł£5, FIG. 4. Divergence of a vector field. Consider Fig. 4 which shows the flow from a small volume δχ ôy δζ. Consider first the flow in the x-direction. On one side the flux entering the volume is Dx ôy δζ; on the other side the flux leaving the volume is [Dx + (dDJdx) ôx] òy δζ. Hence the net outflow in the x-direction is given t The symbol ρ is used also for resistivity. The context generally makes clear which quantity is under discussion. ELECTROMAGNETIC FIELDS A N D THEIR SOURCES 41 (DX + Ë^JL δχ] ду δζ - Dx ду dz = ^L δχ òyòz.

Force on a conducting body.