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Maxwell's equations, named after James Clerk Maxwell who first expressed them together are a set of four equations from which all electromagnetic theory can be derived.
The integral form of Maxwell's equations in free space (ie., in the absence of dielectric or magnetic materials) are
$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ (Gauss's Law)
$\oint \vec{B}\cdot d\vec{A}=0$ (Magnetic equivalent of Gauss's Law)
$\oint\vec{E}\cdot d\vec{l}=\frac{d\Phi_{B}}{dt}$ (Faraday's Law)
$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ (Modified form of Ampere's Law).
When we want to find the gradient of a vector field we use an operator $\nabla$ which is read as del
$\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$
When applied to a scalar field we can use the word grad, so we would say “$E$ is equal to minus grad $V$”
$\vec{E}=\nabla V $
$\vec{E}=\hat{i}\frac{\partial V}{\partial x}\hat{j}\frac{\partial V}{\partial y}\hat{k}\frac{\partial V}{\partial z}$
The $\nabla$ operator can also be applied to a vector, but as it is a vector itself there are two operations, that can be performed, the scalar product “$\cdot$” and the vector product “$\times$”.
The scalar product of the $\nabla$ operator with a vector field is called the Divergence of the field, for example the divergence of the electric field is
$\nabla\cdot \vec{E}=\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}$
The divergence of a field gives a measurement of the magnitude of the source or sink of the field at a point, and we will discuss the relationship between this and charge in a moment.
The vector product of the $\nabla$ operator with a vector field,(e.g. $\nabla\times\vec{E}$) is called the Curl of the field, and gives a measurement of the rotation of the field.
This page has a nice description of the idea behind divergence and curl.
Using the divergence theorem
$\oint \vec{F}\cdot d\vec{A}=\int\nabla\cdot \vec{F}\, dV$
and Stokes' theorem
$\oint \vec{F}\cdot d\vec{l}=\int\nabla\times\vec{F}\cdot dA$
(Some links with a nice discussion of the idea behind the divergence theorem and Stokes' theorem).
Each of Maxwell's equations can also be written in differential form (the form that is most useful will depend on the problem to be solved!)
$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ → $\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$
$\oint \vec{B}\cdot d\vec{A}=0$ → $\nabla\cdot\vec{B}=0$
$\oint\vec{E}\cdot d\vec{l}=\frac{d\Phi_{B}}{dt}$ → $\nabla\times\vec{E}=\frac{\partial\vec{B}}{\partial t}$
$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ → $\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\frac{\partial \vec{E}}{\partial t}$
$\nabla\times\vec{E}=\left \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
E_{x} & E_{y} & E_{z} \end{array} \right = \hat{i}\left \begin{array}{cc}
\frac{\partial}{\partial y} & \frac{\partial}{\partial z}
E_{y} & E_{z} \end{array} \right

\hat{j}\left \begin{array}{cc}
\frac{\partial}{\partial x} & \frac{\partial}{\partial z}
E_{x} & E_{z} \end{array} \right
+
\hat{k}\left \begin{array}{cc}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y}
E_{x} & E_{y} \end{array} \right$
$=(\frac{\partial}{\partial y}E_{z}\frac{\partial}{\partial z}E_{y})\hat{i}+(\frac{\partial}{\partial z}E_{x}\frac{\partial}{\partial x}E_{z})\hat{j}+(\frac{\partial}{\partial x}E_{y}\frac{\partial}{\partial y}E_{x})\hat{k} $
If we consider a electric wave oscillating in the $y$ direction and moving in the $x$ direction then this reduces to $\frac{\partial E_{y}}{\partial x}\hat{k}$
The 3rd Maxwell equation
$\nabla\times\vec{E}=\frac{\partial\vec{B}}{\partial t}$
tells us that there must then be a time varying magnetic field $B_{z}$ as
$\frac{\partial E_{y}}{\partial x}\hat{k}=\frac{\partial B_{z}}{\partial t}\hat{k}$
A similar exercise with the 4th equation
$\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\frac{\partial \vec{E}}{\partial t}$
gives us
$\frac{\partial B_{z}}{\partial x}\hat{j}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}\hat{j}$
We now have two equations
$\frac{\partial E_{y}}{\partial x}=\frac{\partial B_{z}}{\partial t}$ and $\frac{\partial B_{z}}{\partial x}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}$
which can be used to obtain the wave equation for the electric and magnetic fields. To find the wave equation for $E$ we take $\frac{\partial}{\partial x}$ of the first equation and $\frac{\partial}{\partial t}$ of the second, giving.
$\frac{\partial^{2} E_{y}}{\partial x^{2}}=\frac{\partial^{2} B_{z}}{\partial t \partial x}$ and $\frac{\partial^{2} B_{z}}{\partial t \partial x}=\mu_{0}\varepsilon_{0}\frac{\partial^{2} E_{y}}{\partial t^{2}}$
which can be combined to give
$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$
To find the wave equation for $B$ we take $\frac{\partial}{\partial t}$ of the first equation and $\frac{\partial}{\partial x}$ of the second, giving.
$\frac{\partial^{2} E_{y}}{\partial x \partial t}=\frac{\partial^{2} B_{z}}{\partial t^{2}}$ and $\frac{\partial^{2} B_{z}}{\partial x^{2}}=\mu_{0}\varepsilon_{0}\frac{\partial^{2} E_{y}}{\partial x \partial t}$
which can be combined to give
$\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$
We can remind ourselves that the differential equation that we call the wave equation is
$v^2\frac{\partial^{2} D}{\partial x^2}=\frac{\partial^{2} D}{\partial t^2}$
which has a solution
$D(x,t)=A\sin\frac{2\pi}{\lambda}(xvt)$
where $D$ is some kind of displacement.
Our equations
$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$ and $\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$
thus give us two waves in phase with one another, but at right angles to one another
$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$
$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$
The equations
$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$
$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$
tell us that electromagnetic waves have velocity
$v=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}=3.00\times10^{8}\,\mathrm{m/s}=c$
At the time Maxwell derived this result it was known that light had wavelike properties, but the speed of light was not known. The speed of light was first measured by Michelson, who is also famous for one of the most influential “failed” experiments, the Michelson Morley experiment which aimed to find the ether, the reference frame in which light moves at the speed of light, but ended up showing that there was no ether and that light has the same speed in all reference frames.
Either of the equations
$\frac{\partial E_{y}}{\partial x}=\frac{\partial B_{z}}{\partial t}$ and $\frac{\partial B_{z}}{\partial x}=\mu_{0}\varepsilon_{0}\frac{\partial E_{y}}{\partial t}$
can be used to show that
$\frac{E_{0}}{B_{0}}=c\to\frac{E}{B}=c$
The equation we used for waves in general $v=f\lambda$ can be applied to electromagnetic waves, $c=f\lambda$, to find the frequency from the wavelength and vice versa. The full electromagnetic spectrum is much broader than the relatively narrow range we can see.
The energy stored in an electric field is
$u_{E}=\frac{1}{2}\varepsilon_{0}E^{2}$
and in a magnetic field
$u_{B}=\frac{1}{2}\frac{B^{2}}{\mu_{0}}$
The total energy of an EM wave is
$u=u_{E}+u_{B}=\frac{1}{2}\varepsilon_{0}E^{2}+\frac{1}{2}\frac{B^{2}}{\mu_{0}}$
and using
$\frac{E}{B}=c=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}$
$u=\varepsilon_{0}E^{2}=\frac{B^{2}}{\mu_{0}}=\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}EB$
In an electromagnetic wave the fields are moving with velocity $c$ the amount of energy passing through a unit area at any given time is
$S=\varepsilon_{0}cE^{2}=\frac{cB^{2}}{\mu_{0}}=\frac{1}{\mu_{0}}EB$
More generally, the Poynting vector is a vector $\vec{S}$ which represents the flux of energy in an electromagnetic field
$\vec{S}=\frac{1}{\mu_{0}}\vec{E}\times\vec{B}$
NOTE: In the lecture I said some incorrect things about the need for a square root in the last equation. What is shown above is now correct!
As an EM wave carries energy it should be able to exert a force also. The force per unit area exerted by an EM wave is called radiation pressure and was predicted by Maxwell. If light is absorbed by a material the change in momentum which is transferred to the material is
$\Delta p =\frac{\Delta U}{c}$
whereas if it is fully reflected
$\Delta p =\frac{2\Delta U}{c}$
Energy is transferred to the object at a rate
$\frac{dU}{dt}=\bar{S}A$
The force is
$F=\frac{dp}{dt}$
so the pressure is
$P=\frac{1}{A}\frac{dp}{dt}$
When the light is absorbed
$P=\frac{1}{Ac}\frac{dU}{dt}=\frac{\bar{S}}{c}$
When it is reflected
$P=\frac{2}{Ac}\frac{dU}{dt}=\frac{2\bar{S}}{c}$