Speed of light c = 1 - √(ε₀ μ₀): Worked examples

This page is the "every scenario" drill for the speed-of-light note. We will not introduce new theory — instead we hit every kind of question this one formula can throw at you: forward (find $c$ ), backward (find a constant), media, degenerate limits, the $E = c B$ amplitude link, and an exam twist.

The scenario matrix

Before working anything, here is the full map of cases. Each worked example is tagged with the cell it fills, so you can see the coverage is complete.

#	Case class	What is unknown	Twist / edge
A	Forward — plug in $ε_{0}, μ_{0}$	find $c$	none (baseline)
B	Backward — measured $c$ known	find $μ_{0}$ (or $ε_{0}$ )	inverting the formula
C	Medium — $ε_{r}, μ_{r}$ given	find $v$ and $n$	non-trivial $μ_{r} \neq = 1$
D	Degenerate: $μ_{r} \to 1$	check reduces to $n = ε_{r}$	limiting behaviour
E	Degenerate: constant $\to 0$ or $\to \infty$	what happens to $c$ ?	zero/infinite "give"
F	Amplitude link — $E = c B$	find $B$ from $E$	units-driven
G	Word problem — real world	time / distance	Sun→Earth
H	Exam twist — " $c = ε_{0} μ_{0}$ ?"	spot the error by units	reciprocal trap

Cells covered: A→Ex1, B→Ex2, C→Ex3, D→Ex4, E→Ex5, F→Ex6, G→Ex7, H→Ex8. Every cell has an example.

Cell A — the forward baseline

Worked example Ex 1. Compute

$c$ from the two constants Given $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ and $μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1}$ , find $c$ .

Forecast: before reading on, guess: will the answer be near $3 \times 1 0^{8}$ ? Which order of magnitude?

Multiply the two constants. $ε_{0} μ_{0} = (8.85 \times 1 0^{- 12}) (1.2566 \times 1 0^{- 6}) = 1.112 \times 1 0^{- 17} .$ Why this step? The formula needs the product under the root; forming it first keeps the arithmetic in one clean number.
Take the square root. $1.112 \times 1 0^{- 17} = 3.335 \times 1 0^{- 9} .$ Why this step? $ε_{0} μ_{0}$ is the "slowness" (seconds per metre); we need it before inverting.
Invert. $c = \frac{1}{3.335 \times 1 0 ^{- 9}} = 3.00 \times 1 0^{8} m s^{- 1} .$ Why this step? $c = 1/ ε_{0} μ_{0}$ — the reciprocal turns slowness into speed.

Verify: units. $ε_{0} μ_{0}$ carries $s^{2} m^{- 2}$ , so its root is $s m^{- 1}$ (a slowness), and $1/ slowness = m s^{- 1}$ — a speed. ✓ Numerically it matches the optically measured $c$ . That match is the proof light is electromagnetic.

Cell B — running the formula backward

Worked example Ex 2. Recover

$μ_{0}$ from $c$ and $ε_{0}$ Suppose an experimenter measured $c = 3.00 \times 1 0^{8} m s^{- 1}$ and measured $ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}$ but has no value for $μ_{0}$ . Find $μ_{0}$ .

Forecast: should the answer come out near $1.26 \times 1 0^{- 6}$ ? Guess the exponent.

Start from $c = \frac{1}{ε _{0} μ _{0}}$ and square both sides. $c^{2} = \frac{1}{ε _{0} μ _{0}} .$ Why this step? Squaring removes the awkward square root so we can solve linearly for $μ_{0}$ .
Rearrange for $μ_{0}$ . $μ_{0} = \frac{1}{ε _{0} c ^{2}} .$ Why this step? $μ_{0}$ is the only unknown; isolate it algebraically.
Plug in. $μ_{0} = \frac{1}{( 8.85 \times 1 0 ^{- 12} ) ( 3.00 \times 1 0 ^{8} ) ^{2}} = \frac{1}{( 8.85 \times 1 0 ^{- 12} ) ( 9.0 \times 1 0 ^{16} )} = 1.255 \times 1 0^{- 6} .$ Why this step? Direct substitution of the two measured numbers.

Verify: the accepted $μ_{0} = 4 π \times 1 0^{- 7} = 1.2566 \times 1 0^{- 6} T m A^{- 1}$ . Our recovered value agrees to three figures. ✓ This is the historical logic in reverse: any two of $(c, ε_{0}, μ_{0})$ fix the third.

Cell C — light inside a real material

Worked example Ex 3. Speed and index in a magnetic dielectric

A hypothetical medium has relative permittivity $ε_{r} = 4.0$ and relative permeability $μ_{r} = 2.25$ . Find its refractive index $n$ and the wave speed $v$ .

Forecast: with both constants above 1, will light be faster or slower than in vacuum? By how much roughly?

Replace vacuum constants by material ones in the derivation: $v = \frac{1}{ε μ}$ with $ε = ε_{r} ε_{0}, μ = μ_{r} μ_{0}$ . Why this step? The wave equation is identical in a medium; only the constants change, so the form of $v$ is unchanged. See Refractive Index and n = c/v.
Factor out the vacuum part. $v = \frac{1}{ε _{r} μ _{r}} \cdot \frac{1}{ε _{0} μ _{0}} = \frac{c}{ε _{r} μ _{r}} .$ Why this step? This exposes $n = ε_{r} μ_{r}$ cleanly and reuses the known $c$ .
Compute $n$ . $n = 4.0 \times 2.25 = 9.0 = 3.0.$ Why this step? $n$ is the single number that tells you the slow-down factor.
Compute $v$ . $v = \frac{3.00 \times 1 0 ^{8}}{3.0} = 1.00 \times 1 0^{8} m s^{- 1} .$ Why this step? $v = c / n$ turns the index directly into a speed.

Verify: both $ε_{r}, μ_{r} > 1$ ⇒ medium "gives" more sluggishly than vacuum ⇒ $v < c$ . Indeed $1 \times 1 0^{8} < 3 \times 1 0^{8}$ . ✓ And $n > 1$ as required for a real passive medium.

Cell D — the degenerate limit $μ_{r} \to 1$

Worked example Ex 4. Non-magnetic glass (

$μ_{r} = 1$ ) Glass has $ε_{r} \approx 2.25$ and, being non-magnetic, $μ_{r} \approx 1$ . Show the index formula collapses to $n = ε_{r}$ and find $v$ .

Forecast: guess $n$ for glass — is it near 1.5?

Start from $n = ε_{r} μ_{r}$ and set $μ_{r} = 1$ . $n = ε_{r} \cdot 1 = ε_{r} .$ Why this step? Most optical materials are non-magnetic, so this is the case you meet 99% of the time; it is worth seeing the general formula reduce to it, not memorising a separate rule.
Evaluate. $n = 2.25 = 1.5.$ Why this step? Plug the one remaining constant.
Speed. $v = \frac{c}{n} = \frac{3.00 \times 1 0 ^{8}}{1.5} = 2.00 \times 1 0^{8} m s^{- 1} .$ Why this step? $v = c / n$ again.

Verify: this is the textbook glass value ( $n = 1.5$ , $v = 2 \times 1 0^{8}$ ). ✓ Note the limit is smooth: as $μ_{r} \to 1$ the full formula and the reduced one give the identical number, so nothing "breaks" at the boundary.

Cell E — zero and infinite "give"

Worked example Ex 5. What if a constant went to

$0$ or $\infty$ ? Purely as a limiting thought-experiment, examine $c = 1/ ε_{0} μ_{0}$ as one constant is dialled toward $0$ or toward $\infty$ . What happens to $c$ ? (No plugging-in — reasoning about limits.)

Forecast: if space offered zero resistance to fields ( $ε_{0} μ_{0} \to 0$ ), do you expect light infinitely fast or infinitely slow?

Case $ε_{0} μ_{0} \to 0^{+}$ (vanishing "stiffness" of space). $c = \frac{1}{ε _{0} μ _{0}} ⟶ \frac{1}{0 ^{+}} = + \infty.$ Why this step? The denominator shrinks to zero from above, so its reciprocal blows up. Physically: if space resisted fields not at all, the field "dominoes" would fall infinitely fast.
Case $ε_{0} μ_{0} \to \infty$ (infinitely "stiff" space). $c = \frac{1}{\infty} ⟶ 0^{+} .$ Why this step? A huge product under the root makes the reciprocal collapse to zero: infinitely stiff space would freeze the wave.
Reality check on signs: $ε_{0} > 0$ and $μ_{0} > 0$ always (they measure a physical "give," never negative), so $ε_{0} μ_{0} > 0$ strictly and the root is real. There is no case where $c$ is imaginary or negative. Why this step? This closes the sign question — the analogue of "checking all quadrants." Every physical input keeps $c$ real and positive.

Verify: monotonic and consistent — larger stiffness ⇒ slower light; smaller stiffness ⇒ faster. The finite real value we actually measure ( $3 \times 1 0^{8}$ ) sits between these extremes because vacuum has a finite, positive stiffness. ✓

Cell F — the $E = c B$ amplitude link

Worked example Ex 6. Magnetic field of sunlight

A plane light wave has electric-field amplitude $E_{0} = 900 V m^{- 1}$ . Find the magnetic-field amplitude $B_{0}$ .

Forecast: because $c$ is huge, do you expect $B_{0}$ to be a large or tiny number of tesla?

Use the plane-wave relation from Electromagnetic Waves: $E = c B$ , so $B_{0} = E_{0} / c$ . Why this step? In a single EM wave $E$ and $B$ are locked by Maxwell's equations; the ratio is exactly $c$ , not a free choice.
Substitute. $B_{0} = \frac{900}{3.00 \times 1 0 ^{8}} = 3.0 \times 1 0^{- 6} T .$ Why this step? Direct division; the factor $c \approx 3 \times 1 0^{8}$ shrinks the number enormously.

Verify: units. $E_{0} / c$ has $\frac{V m ^{- 1}}{m s ^{- 1}} = V s m^{- 2} = T$ (since $1 T = 1 V s m^{- 2}$ ). ✓ And $3 μ T$ is comparable to Earth's field — the magnetic part of light is genuinely tiny in SI numbers, exactly because it is divided by $c$ .

Cell G — real-world word problem

Worked example Ex 7. How long does sunlight take to reach us?

The Sun is $1.50 \times 1 0^{11} m$ from Earth. Sunlight travels through vacuum. How many seconds — and minutes — does light take to arrive?

Forecast: guess in minutes before computing. More or less than 10?

Light in vacuum moves at $c$ ; time $=$ distance $/$ speed. $t = \frac{d}{c} = \frac{1.50 \times 1 0 ^{11}}{3.00 \times 1 0 ^{8}} .$ Why this step? Uniform speed ⇒ simplest kinematics; no $μ_{0}, ε_{0}$ needed once $c$ is known.
Compute seconds. $t = 5.00 \times 1 0^{2} s = 500 s .$ Why this step? Divide the powers of ten: $1 0^{11} /1 0^{8} = 1 0^{3}$ , and $1.5/3.0 = 0.5$ .
Convert to minutes. $t = \frac{500}{60} = 8.33 min .$ Why this step? Minutes are the intuitive unit for this famous "eight-minute" fact.

Verify: the well-known answer is "about 8 minutes 20 seconds." $8.33 min = 8 min 20 s$ . ✓ Units: $m / (m s^{- 1}) = s$ . ✓

Cell H — the exam twist (spot the trap)

Worked example Ex 8. Is

$c = ε_{0} μ_{0}$ ever right? A test offers the formula $c = ε_{0} μ_{0}$ . Without computing $c$ , decide by units alone whether it can possibly be a speed, then state the fix.

Forecast: does the raw (non-reciprocal) form give metres-per-second, or its opposite?

Recall from Ex 1 that $ε_{0} μ_{0}$ has units $s^{2} m^{- 2}$ . Why this step? Units decide the whole question without any arithmetic — the cheapest possible check.
Take the root as written. $ε_{0} μ_{0} has units s^{2} m^{- 2} = s m^{- 1} .$ Why this step? $s m^{- 1}$ is a slowness (seconds per metre), the reciprocal of a speed.
Numerically this is $3.335 \times 1 0^{- 9} s m^{- 1}$ — absurd as a "speed." Why this step? Shows the magnitude is also wrong, not just the dimension.
Fix: invert. $c = 1/ ε_{0} μ_{0} = 3.00 \times 1 0^{8} m s^{- 1}$ . Why this step? Only the reciprocal has units $m s^{- 1}$ .

Verify: the offered form gives $3.335 \times 1 0^{- 9}$ , the correct form gives $3.00 \times 1 0^{8}$ ; they are reciprocals, differing by the factor $ε_{0} μ_{0}$ . ✓ Units caught the error before any number was trusted.

Recall Self-test on the matrix

Which cell asks you to invert the formula? ::: Cell B (Ex 2) — solve for $μ_{0}$ . Which cell shows the general index reducing to $n = ε_{r}$ ? ::: Cell D (Ex 4), when $μ_{r} \to 1$ . Why can $c$ never be imaginary? ::: Because $ε_{0} μ_{0} > 0$ always (both are positive "gives"), so the root is real (Ex 5). What single check exposes the $c = ε_{0} μ_{0}$ error instantly? ::: Units — the raw form is s/m, a slowness (Ex 8).