Figure 2.35 summarizes the laws of electrostatics in a "triangle diagram" relating the source $\mathbf{\left(}\mathbf{\rho }\mathbf{\right)}$, the field ,$\mathbf{\left(}\mathbf{E}\mathbf{\right)}$ and the potential $\mathbf{\left(}\mathbf{V}\mathbf{\right)}$. Figure 5.48 does the same for magnetostatics, where the source is $\mathit{J}$, the field is$\mathit{B}$ , and the potential is $\mathit{A}$. Construct the analogous diagram for electrodynamics, with sources $\mathit{\rho }$and$\mathit{J}$ (constrained by the continuity equation), fields $\mathit{E}$and$\mathit{B}$ , and potentials$\mathit{V}$and$\mathit{A}$ (constrained by the Lorenz gauge condition). Do not include formulas for $\mathit{V}$and$\mathit{A}$ in terms of $\mathit{E}$androle="math" localid="1657970465123" $\mathit{B}$ .

The quantities of electrostatics.

The source charge distribution is$\rho$.

The field is$E$.

The scaler potential is $V$.

The quantities of magnetostatics

The current density is$J$.

The vector potential is$A$.

The field is$B$.

$r$The expression for the current density is given by,

localid="1658117699866" $J=\frac{1}{{\mu }_0}(\nabla \times B)$

Herelocalid="1658117712093" ${\mu }_0$is the permeability of the free space.

The current density can also be expressed as,

localid="1658118201653" $J=-{\epsilon }_0\frac{\partial E}{\partial t}$

Here localid="1658117721372" ${\epsilon }_0$is the permeability of free space.

The electric field strength is given by,

localid="1658117726324" $E=\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{r^2}\widehat{r}dr$

Here,$\widehat{r}$is the unit vector of the position vectorlocalid="1658117731358" $r$.

Form the Poisson’s equation is given by,

localid="1658118216349" $\nabla {}^{2}v=\frac{p}{{\epsilon }_0}$

The expression for the scaler potential is given by,

localid="1658117736537" $V=\frac{1}{4\pi {\epsilon }_0}\int \frac{\rho }{r}dr$

The relationship between electric field and scaler potential is given by,

localid="1658117743380" $E=-\nabla V$

The scalar potential in term of line integral of electrical field is given by.

localid="1658117758220" $V=-\int Edl$

From the maxwell’s equation of electromagnetism is given by,

localid="1658117764321" $\nabla \cdot B=-\frac{1}{c^2}\frac{\partial V}{\partial t}$

The electric field in terms of vector potentiallocalid="1658407444461" $A$is given by,

localid="1658117772742" $E=\nabla V-\frac{\partial A}{\partial t}$

Therefore, triangle diagram for electrodynamics analogous to triangle diagram of electrostatics withsource localid="1658407429337" $J$, localid="1658407433898" $\rho$and field localid="1658407449348" $E$andlocalid="1658407454000" $B$, potentiallocalid="1658407459615" $V$andlocalid="1658407464380" $A$is shown below.