density po(r) is introduced at time t=0. The charge density must obey the continuit V·J(r,t)= dp(r, t) since =gE. we have dp(r, t) aV·E(r,1)= By Gausss law, V.E can be eliminated dp(r, t) p(r, t) Solving this differential equation for the unknown p(r, t) we have p(r, t)=po(r)e-at/e (322) The charge density within a homogeneous, isotropic conducting body decreases exponen tially with time, regardless of the original charge distribution and shape of the body. o course, the total charge must be constant, and thus charge within the body travels to the surface where it distributes itself in such a way that the field internal to the body approaches zero at equilibrium. The rate at which the volume charge dissipates is deter mined by the relazation time e/o; for copper(a good conductor) this is an astonishingly small 10-19 s. Even distilled water, a relatively poor conductor, has E/o=10-6s.Thus we see how rapidly static equilibrium can be approached 3.1.3 Steady current Since time-invariant fields must arise from time-invariant sources, we have from the continuity equation V·J(r)=0. In large-scale form this is J·ds=0. (3.24) A current with the property(3.23)is said to be a steady current. By(3. 24), a steady current must be completely lineal (and infinite in extent)or must form closed loops. However, if a current forms loops then the individual moving charges must undergo acceleration(from the change in direction of velocity). Since a single accelerating particle radiates energy in the form of an electromagnetic wave, we might expect a large steady loop current to produce a great deal of radiation. In fact, if we superpose the fields produced by the many particles comprising a steady current, we find that a steady current produces no radiation 91. Remarkably, to obtain this result we must consider the exact relativistic fields, and thus our finding is precise within the limits of our macroscopic If we try to create a steady current in free space, the flowing charges will tend to disperse because of the Lorentz force from the field set up by the charges, and th resulting current will not form closed loops. a beam of electrons or ions will produce both an electric field(because of the nonzero net charge of the beam)and a magnetic field (because of the current). At nonrelativistic particle speeds, the electric field produces an outward force on the charges that is much greater than the inward (or pinch) force produced by the magnetic field. Application of an additional, external force will allow ②2001 by CRC Press LLCdensity ρ0(r) is introduced at time t = 0. The charge density must obey the continuity equation ∇ · J(r, t) = −∂ρ(r, t) ∂t ; since J = σE, we have σ∇ · E(r, t) = −∂ρ(r, t) ∂t . By Gauss’s law, ∇ · E can be eliminated: σ ρ(r, t) = −∂ρ(r, t) ∂t . Solving this differential equation for the unknown ρ(r, t) we have ρ(r, t) = ρ0(r)e−σt/ . (3.22) The charge density within a homogeneous, isotropic conducting body decreases exponen￾tially with time, regardless of the original charge distribution and shape of the body. Of course, the total charge must be constant, and thus charge within the body travels to the surface where it distributes itself in such a way that the field internal to the body approaches zero at equilibrium. The rate at which the volume charge dissipates is deter￾mined by the relaxation time /σ; for copper (a good conductor)this is an astonishingly small 10−19 s. Even distilled water, a relatively poor conductor, has /σ = 10−6 s. Thus we see how rapidly static equilibrium can be approached. 3.1.3 Steady current Since time-invariant fields must arise from time-invariant sources, we have from the continuity equation ∇ · J(r) = 0. (3.23) In large-scale form this is S J · dS = 0. (3.24) A current with the property (3.23)is said to be a steady current. By (3.24), a steady current must be completely lineal (and infinite in extent)or must form closed loops. However, if a current forms loops then the individual moving charges must undergo acceleration (from the change in direction of velocity). Since a single accelerating particle radiates energy in the form of an electromagnetic wave, we might expect a large steady loop current to produce a great deal of radiation. In fact, if we superpose the fields produced by the many particles comprising a steady current, we find that a steady current produces no radiation [91]. Remarkably, to obtain this result we must consider the exact relativistic fields, and thus our finding is precise within the limits of our macroscopic assumptions. If we try to create a steady current in free space, the flowing charges will tend to disperse because of the Lorentz force from the field set up by the charges, and the resulting current will not form closed loops. A beam of electrons or ions will produce both an electric field (because of the nonzero net charge of the beam)and a magnetic field (because of the current). At nonrelativistic particle speeds, the electric field produces an outward force on the charges that is much greater than the inward (or pinch)force produced by the magnetic field. Application of an additional, external force will allow