$$ {\mathrm{Ohm}}^{'}\mathrm{s}\ \mathrm{law}:\operatorname{}\overrightarrow{J}=\sigma \kern0.2em \overrightarrow{E} $$

(1)

$$ {\mathrm{Gauss}}^{'}\ \mathrm{law}:\operatorname{}\overrightarrow{\nabla}\cdot \overrightarrow{E}=\frac{\rho }{\varepsilon_0} $$

(2)

$$ \mathrm{Equation}\ \mathrm{of}\ \mathrm{continuity}:\operatorname{}\overrightarrow{\nabla}\cdot \overrightarrow{J}+\frac{\partial \rho }{\partial \kern0.1em t}=0 $$

(3)

Since the current

*Ι* is constant in time,

$$ {\displaystyle \begin{array}{l}\frac{\partial \kern0.1em \overrightarrow{J}}{\partial \kern0.1em t}=0\kern0.36em \overset{(1)}{\Rightarrow}\kern0.36em \frac{\partial \overrightarrow{E}}{\partial \kern0.1em t}=0\kern0.36em \left( static\;\overrightarrow{E}\right)\kern0.36em \overset{(2)}{\Rightarrow}\kern0.36em \frac{\partial \rho }{\partial \kern0.1em t}=0\kern0.36em \overset{(3)}{\Rightarrow}\kern0.36em \overrightarrow{\nabla}\cdot \overrightarrow{J}=0\\ {}\kern4.319998em \overset{(1)}{\Rightarrow}\kern0.36em \overrightarrow{\nabla}\cdot \overrightarrow{E}=0\kern0.36em \overset{(2)}{\Rightarrow}\kern0.36em \rho =0\;\end{array}} $$

*Physical interpretation:* When the current in the wire is constant, the charge of the mobile electrons exactly counterbalances the opposite charge of the (stationary) ions of the metal so that the wire is electrically neutral everywhere in its interior.

*Comment:* When calculating the total charge density *ρ*, the electron must *always* be treated as *negatively* charged! This is because the quantity *ρ* is related only to the *presence*, not the *motion*, of charges. On the contrary, in the relation \( \overrightarrow{J}={\rho}_{\kappa}\kern0.2em \overrightarrow{\upsilon} \) for the *moving* electrons we are free to change the sign of *ρ*_{κ} provided that we simultaneously invert the direction of \( \overrightarrow{\upsilon} \), so that the current density \( \overrightarrow{J} \) is left unchanged.

- 2.
Prove *Kirchhoff’s first rule*: In a region *R* of space where the electric field is static, the total electric current through any closed surface is zero.

$$ \rho ={\varepsilon}_0\;\left(\overrightarrow{\nabla}\cdot \overrightarrow{E}\right)\kern0.48em \Rightarrow \kern0.48em \frac{\partial \rho }{\partial \kern0.1em t}={\varepsilon}_0\;\left(\overrightarrow{\nabla}\cdot \frac{\partial \overrightarrow{E}}{\partial \kern0.1em t}\right)=0,\forall \overrightarrow{r}\in R $$

Then, by the equation of continuity,

$$ \overrightarrow{\nabla}\cdot \overrightarrow{J}+\frac{\partial \rho }{\partial \kern0.1em t}=0\kern0.48em \Rightarrow \kern0.48em \overrightarrow{\nabla}\cdot \overrightarrow{J}=0,\forall \overrightarrow{r}\in R $$

The volume integral of the equation on the right, in a volume

*V* bounded by a closed surface

*S*, is

$$ {\int}_V\left(\overrightarrow{\nabla}\cdot \overrightarrow{J}\right)\; dv=0 $$

By using Gauss’ integral theorem, the volume integral is transformed into an integral over the closed surface

*S*. We thus have:

$$ {\oint}_S\overrightarrow{J}\cdot \overrightarrow{da}=0 $$

The surface integral represents the total current passing through the closed surface

*S* (since

\( \overrightarrow{da} \) is directed outward relative to

*S*, we are actually speaking of the total current

*exiting S*). Physically, the vanishing of this integral suggests that, as far as their absolute values are concerned, the total current going into

*S* equals the total current coming out of

*S* (a negative “outgoing” current is equivalent to a positive “ingoing” one). In the limit

*V* → 0, the closed surface

*S* degenerates to a point. In the case of an electric network, such a point corresponds to a

*junction* and Kirchhoff’s first rule expresses the conservation of charge at each junction of the system. (Another rule, called “Kirchhoff’s second rule”, expresses the conservation of energy along any closed path in the network.)