Speed in a Medium & Point Source

PHYOrigin & Nature

In a material medium an EM wave slows to $v=c/\sqrt{\mu_r\varepsilon_r}$ , which for a non-magnetic dielectric ( $\mu_r=1$ ) gives $v=c/\sqrt{\varepsilon_r}$ and refractive index $n=\sqrt{\varepsilon_r}$ . A separate strand treats a point source radiating power $P$ uniformly: intensity falls as $1/r^2$ , from which the local peak field follows.

Both ideas are short, formula-driven numericals.

Speed in a medium

v = \dfrac{c}{\sqrt{\mu_r \varepsilon_r}}

\mu_r,\varepsilon_r

= relative permeability and permittivity; for a non-magnetic dielectric

\mu_r=1

v=c/\sqrt{\varepsilon_r}

Refractive index of medium

n = \dfrac{c}{v} = \sqrt{\mu_r \varepsilon_r}

Non-magnetic case (

\mu_r=1

n=\sqrt{\varepsilon_r}

(Maxwell's relation).

Intensity from a point source

I = \dfrac{P}{4\pi r^{2}}

P

= total radiated power (W),

r

= distance from source (m); assumes isotropic radiation, no absorption.

Peak field at distance r

E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{P}{2\pi \varepsilon_0 c\, r^{2}}}

Combine

I=P/4\pi r^2

with

I=\tfrac12\varepsilon_0 E_0^2 c

; then

B_0=E_0/c

Maxwell's relation $n=\sqrt{\mu_r\varepsilon_r}$ links optics and electromagnetism; for most dielectrics $\mu_r\approx1$ so $n\approx\sqrt{\varepsilon_r}$ .
Wave speed in a medium is always less than $c$ ( $n>1$ ), and the frequency is unchanged while $\lambda$ shrinks to $\lambda/n$ .
For a point source, intensity obeys the inverse-square law $I\propto 1/r^2$ because the same power spreads over a sphere of area $4\pi r^2$ .
The field amplitude therefore falls as $1/r$ (since $I\propto E_0^2$ ), not as $1/r^2$ .
To get $E_0$ at a distance, first find $I=P/4\pi r^2$ , then invert $I=\tfrac12\varepsilon_0 E_0^2 c$ .
In a medium replace $c$ by $v$ in the energy/intensity relations, and replace $\varepsilon_0$ by $\varepsilon=\varepsilon_r\varepsilon_0$ .
Use $\varepsilon_0=8.85\times10^{-12}\ \text{F m}^{-1}$ and $c=3\times10^8\ \text{m s}^{-1}$ for SI consistency.

Where the marks go

Forgetting the square root: $n=\sqrt{\varepsilon_r}$ , not $\varepsilon_r$ , for a non-magnetic medium.
Using $4\pi r$ or $\pi r^2$ instead of the spherical area $4\pi r^2$ in the point-source intensity.
Letting the field amplitude fall as $1/r^2$ ; it is the intensity that goes as $1/r^2$ , so $E_0\propto 1/r$ .
Mixing peak and rms when going from $I$ to $E_0$ — the $1/2$ factor in $I=\tfrac12\varepsilon_0 E_0^2 c$ uses the peak field.

How the board asks it

Numericalintensity from a point source
A point source radiates $P = 100\ \text{W}$ of electromagnetic power uniformly in all directions. Calculate the intensity of the wave and the peak value of the electric field $E_0$ at a distance of $2\ \text{m}$ from the source. (Take $\varepsilon_0 = 8.85\times10^{-12}\ \text{F m}^{-1}$ and $c = 3\times10^8\ \text{m s}^{-1}$ .)
Numericalspeed in a medium and refractive index
A non-magnetic dielectric has relative permittivity $\varepsilon_r = 4$ . Calculate the speed of an electromagnetic wave in this medium and its refractive index $n$ . (Take $c = 3\times10^8\ \text{m s}^{-1}$ .)
Give reasonsinverse-square law for intensity
For a point source, the intensity falls off as $1/r^2$ while the electric field amplitude falls off only as $1/r$ . Give reasons for this difference.
Define / stateMaxwell's relation $n = \sqrt{\mu_r\varepsilon_r}$
State Maxwell's relation connecting the refractive index of a medium to its electromagnetic properties, and write the form it takes for a non-magnetic dielectric ( $\mu_r = 1$ ).
Give reasonswavelength change on entering a medium
When an electromagnetic wave passes from vacuum into a medium of refractive index $n$ , account for the fact that its frequency stays unchanged while its wavelength reduces to $\lambda/n$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.