Proving the continuity equation

• We start with the charge density and the current flux through a surface

  $$ Q=\int\text dq=\int \rho (\vec r ,t)\,\text dV \qquad I=\int \text dI=\int \vec j(\vec r,t)\cdot \text d\vec S $$

• Conservation implies $\dot Q=-I$

  $$ \int_V\ \dot \rho \, \text dV=\dot Q =-\int_S \vec J \cdot \text d \vec S=-\int_V \vec \nabla \cdot \vec j \,\text dV $$

  which can be rearranged to:

  $$ \int_V \left \{ \dot{\rho} + \vec \nabla \cdot \vec j \right \} \,\text dV=0 $$

• This must hold for any volume so

  $$ \dot \rho +\vec \nabla \cdot \vec j =0 $$

  ✨ This is the continuity equation

Current $\vec j$ through a surface, with surface element $\text d\vec S=\hat n\text dA$

• $\int\text dS$ is the area of the surface (m$^2$)
• The minus arises as the current is defined in the $+$ve charge the electron drift velocity is in the opposite direction

We can define a velocity field $\vec{\mathrm{v}}$ such that $\vec j = \rho \vec{\mathrm v}$ which corresponds to the local velocity of the positive charge carriers

$$ \begin{aligned} I&=\int \text dI \\ &=\int \vec J \cdot \text d \vec S \\ &=\int \rho \vec{\mathrm v} \cdot \hat n \,\text dS \\ &= -n\,e\,v_\text{drift}\int \text dS

\end{aligned} $$

• $v_\text{drift}=-\vec{\mathrm v}\cdot \hat n$ is the electron drift velocity through the surface
• $n$ is the number density (m$^{-3}$)
• $e$ is the electron charge ($e=1.6\times 10^{-19}$C)

Definitions

1. ⚡ Electric field or electric flux density [Vm$^{-1}$], $\vec E(\vec r,t)$

2. 🧲 Magnetic field or magnetic flux density [T], $\vec B(\vec r,t)$

3. 🔌 Electric flux through a surface [Vm]

   $$ \Phi_\text E=\int_S \vec E \cdot \text d\vec S $$

4. 🛩️ Magnetic flux through a surface units [Wb]=[Tm$^2$]=[Vs]

   $$ \Phi_\text B=\int_S \vec B\cdot \text d \vec S $$

5. 👊 The force per unit charge [NC$^{-1}$] is computed via the Lorentz force law

   $$ \vec f=\vec E+\vec {\mathrm v} \times \vec B $$

6. 🧹 The electromagnetic energy density [Jm$^{-3}$] in vacuum is

   $$ u=\frac 12 \epsilon_0 |\vec E |^2 + \frac{1}{2\mu_0} |\vec B|^2 $$

7. 🌐 The total electromagnetic energy in a volume $V$ is [J]

   $$ U_\text{tot}=\int_V u\,\text dV $$

8. 🏋️ The electromotive force [V] is

   $$ \mathcal{E}=\int \vec E\cdot\text d\vec l $$

Proving laws:

🧄 Gauss’ law

• Experimentally we found

  $$ \int_S \vec E\cdot \text d\vec S=\Phi_\text{E}=\frac{Q}{\epsilon_0}=\frac{1}{\epsilon_0}\int_V \rho\,\text d V $$

• Using the divergence theorem we can find

  $$ \int_S\vec E\cdot \text d\vec S=\int_V \vec \nabla \cdot \vec E\,\text dV =\frac{1}{\epsilon_0}\int_V\rho\,\text dV $$

• so that

  $$ \int_V \left ( \vec \nabla\cdot \vec E-\frac{1}{\epsilon_0}\rho \right )\,\text dV=0 $$

• this integral must hold over any integration volume $V$ and therefore must be zero

  $$ \vec \nabla \cdot \vec E=\frac{1}{\epsilon_0}\rho $$

  which is the differential form of Gauss’ law