# 6.10 Free-Charge Dissipation in a Conductor

We stated earlier that current flow in a conductor is realized by the movement of loosely attached electrons under the influence of an externally applied electric field. These electrons, however, are not excess charges; their charge is balanced by an equal amount of positive charge in the atoms’ nuclei. In other words, the conductor material is electrically neutral, and the net charge density in the conductor is zero (ρ v=0). What happens then if an excess free charge q is introduced at some interior point in a conductor? The excess charge gives rise to an electric field, which forces the charges of the host material nearest to the excess charge to rearrange their locations, which in turn cause other charges to move, and so on. The process continues until neutrality is reestablished in the conductor material and a charge equal to q resides on the conductor’s surface.

How fast does the excess charge dissipate? To answer this question, let us introduce a volume charge density ρ v o at the interior of a conductor and then find out the rate at which it decays down to zero. From Eq. (6.54), the continuity equation is given by

∇⋅J=−∂ρ v∂t.(6.58)

In a conductor, the point form of Ohm’s law given by Eq. (4.73) states that J=σ E. Hence,

σ∇⋅E=−∂ρ v∂t.(6.59)

Next, we use Eq. (6.1), ∇⋅E=ρ v/ε, to obtain the partial differential equation

∂ρ v∂t+σ ε ρ v=0.(6.60)

Given that ρ v=ρ v o at t=0, the solution of Eq. (6.60) is

ρ v(t)=ρ v o e−(σ/ε)t=ρ v o e−t/τ r (C/m 3),(6.61)

where τ r=ε/σ is called the relaxation time constant. We see from Eq. (6.61) that the initial excess charge ρ v o decays exponentially at a rate τ r. At t=τ r, the initial charge ρ v o will have decayed to 1/e≈37% of its initial value, and at t=3 τ r, it will have decayed to e−3≈5% of its initial value at t=0. For copper, with ε≈ε 0=8.854×10−12 F/m and σ=5.8×10 7 S/m, τ r=1.53×10−19 s. Thus, the charge dissipation process in a conductor is extremely fast. In contrast, the decay rate is very slow in a good insulator. For a material like mica with ε=6 ε 0 and σ=10−15 S/m, τ r=5.31×10 4 s, or approximately 14.8 hours.

Concept Question 6-11: Explain how the charge continuity equation leads to Kirchhoff’s current law.

Concept Question 6-12: How long is the relaxation time constant for charge dissipation in a perfect conductor? In a perfect dielectric?