Appendix

Examples of vector calculus in electromagnetism

1. For a continuous distribution of charge with charge density \(\rho\), the charge in a volume element \(d V\) is \(d Q=\rho d V\) and the charge in a volume \(\mathrm{V}\) is \(Q=\iiint_{V} \rho d V\).

   Gauss’s law (see VC4) states that for a closed surface \(S\), the flux of the electric field \(E\) through the surface is related to the charge Q enclosed by \(\iint_{S} \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{S}}=\frac{Q}{\epsilon_{0}}\) where \(\epsilon_{0}\) is a constant.

   Hence we can write \(\iint_{S} \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{S}}=\frac{1}{\epsilon_{0}} \iiint_{V} \rho d V\).

   Meanwhile the divergence theorem states \(\iint_{S} \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{S}}=\iiint \vec{\nabla} \cdot \overrightarrow{\boldsymbol{E}} d V\).

   Comparing these equations we obtain \(\vec{\nabla} \cdot \overrightarrow{\boldsymbol{E}}=\frac{\rho}{\epsilon_{0}}\), which is one of Maxwell’s equations.

   Since \(\overrightarrow{\boldsymbol{E}}\) is conservative, we can also define an electric potential \(\phi\) by \(\overrightarrow{\boldsymbol{E}}=-\overrightarrow{\boldsymbol{\nabla}} \phi\), and write \(\nabla^{2} \phi=\frac{\rho}{\epsilon_{0}}\).

2. Another of Maxwell’s equations is \(\vec{\nabla} \times \overrightarrow{\boldsymbol{E}}=-\frac{\partial B}{\partial t}\).

   In the absence of time-varying magnetic fields, this becomes \(\overrightarrow{\boldsymbol{\nabla}} \times \overrightarrow{\boldsymbol{E}}=\overrightarrow{\mathbf{0}}\).

   By Stokes’ theorem this is equivalent to saying \(\oint \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{r}}=0\) around any closed loop (an idea common used in analysing electrical circuits).

3. Ampere’s law states \(\oint_{C} \overrightarrow{\boldsymbol{B}} \cdot d \overrightarrow{\boldsymbol{r}}=\mu_{0} I\) where \(I\) is the net current passing through a surface bounded by a closed curve \(C\) and \(\mu_{0}\) is a constant.

   For example, a constant current \(I\) in a long wire produces a magnetic field tangential to any circle centered on and perpendicular to the wire, so by taking \(C\) to be such a circle of radius \(r\) we get \(\oint_{C} \overrightarrow{\boldsymbol{B}} \cdot d \overrightarrow{\boldsymbol{r}}=B \cdot 2 \pi r=\mu_{0} I\), and hence the magnetic field at a distance \(r\) from the wire is \(B=\frac{\mu_{0} I}{2 \pi r}\).

   See also https://www.youtube.com/watch?v=rB83DpBJQsE 6:00 - 7:40.