# 6.9 Charge–Current Continuity Relation

Under static conditions, the charge density ρ v and the current density J at a given point in a material are totally independent of one another. This is no longer true in the time-varying case. To show the connection between ρ v and J, we start by considering an arbitrary volume υ bounded by a closed surface S (Fig. 6-14). The net positive charge contained in υ is Q. Since, according to the law of conservation of electric charge (Section 1-3.2), charge can neither be created nor destroyed, the only way Q can increase is as a result of a net inward flow of positive charge into the volume υ. By the same token, for Q to decrease, there has to be a net outward flow of charge from υ. The inward and outward flow of charge constitute currents flowing across the surface S into and out of υ, respectively. We define I as the net current flowing across S out of υ. Accordingly, I is equal to the negative rate of change of Q:

I=−d Q d t=−d d t∫υ ρ v d υ,(6.50)

where ρ v is the volume charge density in υ. According to Eq. (4.12), the current I is also defined as the outward flux of the current density J through the surface S. Hence,

∮S J⋅d s=−d d t∫υ ρ v d υ.(6.51)

# Figure 6-14 The total current flowing out of a volume υ is equal to the flux of the current density J through the surface S, which in turn is equal to the rate of decrease of the charge enclosed in υ.

By applying the divergence theorem given by Eq. (3.98), we can convert the surface integral of J into a volume integral of its divergence ∇⋅J, which then gives

∮S J⋅d s=∫υ∇⋅J d υ=−d d t∫υ ρ v d υ.(6.52)

For a stationary volume υ, the time derivative operates on ρ v only. Hence, we can move it inside the integral and express it as a partial derivative of ρ v:

∫υ∇⋅J d υ=−∫υ∂ρ v∂t d υ.(6.53)

In order for the volume integrals on both sides of Eq. (6.53) to be equal for any volume υ, their integrands have to be equal at every point within υ. Hence,

which is known as the charge–current continuity relation, or simply the charge continuity equation.

If the volume charge density within an elemental volume Δ υ (such as a small cylinder) is not a function of time (i.e., ∂ρ v/∂t=0), it means that the net current flowing out of Δ υ is zero or, equivalently, that the current flowing into Δ υ is equal to the current flowing out of it. In this case, Eq. (6.54) implies

∇⋅J=0,(6.55)

and its integral-form equivalent [from Eq. (6.51)] is

Let us examine the meaning of Eq. (6.56) by considering a junction (or node) connecting two or more branches in an electric circuit. No matter how small, the junction has a volume υ enclosed by a surface S. The junction shown in Fig. 6-15 has been drawn as a cube, and its dimensions have been artificially enlarged to facilitate the present discussion. The junction has six faces (surfaces), which collectively constitute the surface S associated with the closed-surface integration given by Eq. (6.56). For each face, the integration represents the current flowing out through that face. Thus, Eq. (6.56) can be cast as

∑i I i=0,(Kirchhoff's current law)(6.57)

where I i is the current flowing outward through the ith face. For the junction of Fig. 6-15, Eq. (6.57) translates into (I 1+I 2+I 3)=0. In its general form, Eq. (6.57) is an expression of Kirchhoff’s current law, which states that in an electric circuit the sum of all the currents flowing out of a junction is zero.

# Figure 6-15 Kirchhoff’s current law states that the algebraic sum of all the currents flowing out of a junction is zero.