ARCHETYPE: Null Points & the Third-Charge Equilibrium

If you met the gravitational version of this problem — the point between the Earth and the Moon where the two pulls tie — you already own half of this article. Between two masses there is always exactly one such point, always in between, always nearer the lighter body. Electricity keeps the square-root machinery and then does something gravity never can: because charge comes in two signs, the null point can walk out of the segment entirely. Two unlike charges have no tie point between them at all — their fields reinforce there — and instead cancel at a point outside, beyond the weaker charge. Examiners adore this archetype because one setup carries three escalating questions. First: where is the field zero? (A balance equation and a square root.) Second: can a third charge be parked so that all three charges float in equilibrium? (Yes — one special position, one special magnitude, one forced sign.) Third: is that equilibrium worth anything? Here physics delivers one of its cleanest verdicts, Earnshaw's theorem: no arrangement of charges can hold a charge in stable equilibrium by electrostatic forces alone. The tower you carefully balanced is a house of cards — mathematically guaranteed to topple. Master the square-root shares, the inside/outside decision, and the third-charge recipe, and this becomes the most predictable "hard" question in electrostatics — plus a story about why levitation needs more than Coulomb.

Key takeaways

Between two like charges $q_1, q_2$ a distance $d$ apart, the field is zero at the in-between point with distance shares $\dfrac{r_1}{r_2} = \sqrt{\dfrac{q_1}{q_2}}$ — nearer the weaker charge, exactly as with two masses.
For two unlike charges there is no null between (fields reinforce); it sits outside the segment, beyond the smaller-magnitude charge, from $\dfrac{kq_{\text{small}}}{x^2} = \dfrac{kq_{\text{big}}}{(d+x)^2}$ .
Third-charge equilibrium: to freeze all three charges, $q_3$ must sit at the null point of the original pair and must be of opposite sign, with magnitude $q_3 = \dfrac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2}$ .
The equilibrium is always unstable — Earnshaw's theorem (1842) forbids stable rest in any electrostatic field; every null is a saddle, not a bowl.
Zero field does not mean zero potential — and for unlike charges the reverse trap appears too: points of zero potential where the field is decidedly non-zero.
Key idea: balance the two field magnitudes, take the square root, and let the signs of the charges decide whether the tie point lives inside or outside the segment.

The core insight

Superposition places the zero-field point of two charges where the field magnitudes tie — between like charges (shares $\sqrt{q_1}:\sqrt{q_2}$ , nearer the weaker) or outside unlike charges (beyond the weaker) — and a third charge of opposite sign and magnitude $q_1q_2/(\sqrt{q_1}+\sqrt{q_2})^2$ placed at that point freezes the whole system, though only in the unstable, Earnshaw-doomed sense.

Where this lives today

Earnshaw's verdict is why every modern device that does levitate charges is, in a sense, an elaborate cheat. The ion traps inside trapped-ion quantum computers hold single atoms at what looks exactly like our null point — but with a radio-frequency field that flips the saddle faster than the ion can fall off, a dynamic trick (the Paul trap) worth a Nobel Prize. Mass spectrometers steer charges through null regions rather than parking them there; accelerator "still points" are stabilised by motion and feedback, never statics. The humble interior null survives as a design target of the opposite kind: electrostatic lenses place electrodes so field zeros sit exactly where stray beams must feel no kick. Engineers still begin where this article began — balance two inverse-squares, take a square root — then spend their cleverness evading the theorem that closes it.

The setup

Two point charges are fixed on a line: $q_1$ at A, $q_2$ at B, separation $d$ . The question in all its costumes: where does their combined field vanish, and what can be parked there? A test point between the charges feels $q_1$ 's field pointing one way along the line and $q_2$ 's pointing — where? That depends on the signs, and this is the fork in the road that gravity never offered:

Like charges (both $+$ or both $-$ ). Between them the two fields point in opposite directions (each pushes a positive test charge away from itself, toward the other). Opposition means a tie is possible: one null point, in between, nearer the weaker charge — the weaker player needs home advantage. Outside the segment both fields point the same way; no cancellation.

Unlike charges. Between them both fields point the same way — from the positive charge toward the negative one — so they reinforce and no interior null exists. Step outside the segment and the two fields finally oppose. But a tie needs more than opposition: it needs the nearer charge to be the weaker one, so its stronger-by-proximity field can be matched by the bigger charge far away. So the null sits outside, beyond the smaller-magnitude charge, on the far side from the big one.

Figure

The sign fork — like charges tie inside the segment, unlike charges tie outside beyond the weaker

Hold the gravitational contrast explicitly — a favourite conceptual mark: for two masses the null is always between, because mass has one sign. One fact drives everything here: fields can only cancel where they oppose.

The master pattern

Move 1 — locate the null for like charges. At distance $r_1$ from $q_1$ (and $r_2 = d - r_1$ from $q_2$ ), set the magnitudes equal:

\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2} \quad\Longrightarrow\quad \frac{r_1}{r_2} = \sqrt{\frac{q_1}{q_2}} \quad\Longrightarrow\quad r_1 = \frac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}\;d \qquad (1)

where: $k$ cancels immediately, and the square root — not the charge ratio itself — sets the shares, because distance enters the field squared. Sanity reads: equal charges tie at the midpoint; $q_1 = 4q_2$ pushes the null to $\tfrac23 d$ , twice as far from the big charge. Nearer the weaker, always.

Move 2 — locate the null for unlike charges. Let $|q_2| < |q_1|$ and put the test point a distance $x$ beyond $q_2$ , on the side away from $q_1$ :

\frac{k|q_2|}{x^2} = \frac{k|q_1|}{(d+x)^2} \quad\Longrightarrow\quad \frac{d+x}{x} = \sqrt{\frac{|q_1|}{|q_2|}} \quad\Longrightarrow\quad x = \frac{d}{\sqrt{|q_1|/|q_2|}\; - 1} \qquad (2)

where: $x$ — measured outward from the weaker charge. The denominator tells the story: nearly equal magnitudes push the null far away ( $x \to \infty$ as the ratio $\to 1$ — two almost-balanced unlike charges look neutral only from infinity), while a very lopsided pair parks it just outside the weak charge.

Move 3 — freeze all three charges. Now the deluxe question: place a third charge $q_3$ so that every charge is in equilibrium. Two like charges $q_1, q_2$ repel each other — something must pull them back. The only candidate is $q_3$ , so $q_3$ must attract both: opposite sign, and it must sit between them — and not just anywhere between, but exactly where the pair's field vanishes, or $q_3$ itself would feel a force. Position from equation (1); magnitude from demanding equilibrium of $q_1$ (or $q_2$ — either gives the same answer):

q_3 = -\frac{q_1 q_2}{\left(\sqrt{q_1} + \sqrt{q_2}\right)^2} \qquad (3)

where: the minus sign is forced, not chosen. For the classic pair $q_1 = 4q$ , $q_2 = q$ : $q_3 = -\dfrac{4q^2}{(2\sqrt q + \sqrt q)^2/q} = -\dfrac{4q}{9}$ , parked at $\tfrac{2d}{3}$ from the $4q$ charge. Symmetric pair $q, q$ : $q_3 = -q/4$ at the midpoint — a one-mark classic in its own right.

Move 4 — pass the stability verdict. Every one of these equilibria fails the nudge test. Displace the middle negative charge along the line: it runs to the nearer positive charge. Displace a test charge at a like-pair null perpendicular to the line: the two repulsions sweep it away. Along one direction the point behaves like a valley, along another like a crest — a saddle — and a saddle holds nothing. That is Earnshaw's theorem, and quoting it (with the word "unstable") is usually the final mark of the question.

Derivation AThe third charge that freezes the pair

We derive the position and magnitude of the charge $q_3$ that puts a pair of like charges $q_1$ and $q_2$ (separation $d$ ) into complete three-body equilibrium — equation (3). The strategy: equilibrium of $q_3$ pins its position (the pair's null point, from move 1); equilibrium of $q_1$ then pins $q_3$ 's magnitude and sign; equilibrium of $q_2$ comes free as a consistency check, which we run explicitly. Keep signs in words (attract/repel) until the last line — pushing signed symbols through Coulomb's law is where solutions go to die.

Step 1. Position of $q_3$ : for $q_3$ to feel no force, it must sit where the pair's field vanishes. For like charges that is the interior null of equation (1): $r_1 = \dfrac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}\,d$ from $q_1$ , and $r_2 = d - r_1 = \dfrac{\sqrt{q_2}}{\sqrt{q_1}+\sqrt{q_2}}\,d$ .

Step 2. Sign of $q_3$ : charges $q_1$ and $q_2$ repel each other with $F_{12} = \dfrac{kq_1q_2}{d^2}$ . For $q_1$ to be in equilibrium, $q_3$ must pull it back toward $q_2$ — an attraction. So $q_3$ is opposite in sign to the pair.

Step 3. Magnitude from $q_1$ 's equilibrium: attraction by $q_3$ at distance $r_1$ must balance the repulsion by $q_2$ at distance $d$ :

\frac{k\,q_1|q_3|}{r_1^2} = \frac{k\,q_1q_2}{d^2} \quad\Longrightarrow\quad |q_3| = q_2\,\frac{r_1^2}{d^2} = \frac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2}

Why the last equality: substitute $r_1/d = \sqrt{q_1}/(\sqrt{q_1}+\sqrt{q_2})$ from Step 1, so $r_1^2/d^2 = q_1/(\sqrt{q_1}+\sqrt{q_2})^2$ .

Step 4. Consistency check on $q_2$ : attraction by $q_3$ at distance $r_2$ must balance the repulsion by $q_1$ at distance $d$ :

\frac{k\,q_2|q_3|}{r_2^2} = \frac{k\,q_2\,q_1}{d^2}\;?\quad |q_3| = q_1\frac{r_2^2}{d^2} = \frac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2} \;✓

The same magnitude satisfies both — one charge really does freeze the whole system. Why it had to work: with $q_3$ at the null point, the pair's mutual repulsion is the only thing left to balance, and it is a third-law pair — fixing one side fixes the other. *Verified: symmetric case $q_1 = q_2 = q$ gives $r_1 = d/2$ and $|q_3| = q^2/(2\sqrt q)^2 = q/4$ — the textbook "−q/4 at the midpoint." Classic case $4q, q$ : $|q_3| = 4q^2/(3\sqrt q)^2 = 4q/9$ at $2d/3$ from the larger — matching worked example 3's numeric check of 32.4 N balancing 32.4 N.*

Figure

The frozen triple — every charge in equilibrium, every equilibrium doomed

Limiting cases

case	result	meaning
$q_1 = q_2$ (like)	null at midpoint	symmetry answers before algebra
$q_1 = 4q_2$ (like)	null at $2d/3$ from $q_1$	shares $\sqrt4:\sqrt1 = 2:1$
$q_1 \gg q_2$ (like)	null hugs the weak charge	the heavyweight owns the line
unlike, $\\|q_1\\| \to \\|q_2\\|$	exterior null recedes to infinity	a balanced $\pm$ pair cancels nowhere nearby
unlike, $\\|q_1\\| = 9\\|q_2\\|$	null at $x = d/2$ beyond the weak charge	$(d+x)/x = 3$
third charge, pair $q,q$	$q_3 = -q/4$ at midpoint	the symmetric classic
any equilibrium above	unstable (saddle)	Earnshaw: no electrostatic parking

Common trap“The field is zero between unlike charges, somewhere nearer the weaker one”

Between unlike charges the two fields point the same way — off the positive, onto the negative — so they add and can never cancel there. No interior null exists, full stop; the null sits outside the segment, beyond the smaller-magnitude charge. The confusion comes from importing the like-charge (or gravitational) picture on autopilot. Run the two-second direction check before any algebra: sketch the two field arrows at a candidate point; only if they oppose can a tie exist. And its energy cousin deserves equal care: between unlike charges there is a point of zero potential (scalars of opposite sign happily cancel) — but the field there is strong, not zero. Zero field and zero potential are different animals living at different addresses; for two like charges $V$ never vanishes at the null, and for unlike charges $V = 0$ points are exactly where $E \ne 0$ .

Worked examples

Board level 1 — the like-charge null (CBSE pattern, 2–3 marks)

Charges $+4\ \mathrm{\mu C}$ and $+1\ \mathrm{\mu C}$ are fixed $30\ \mathrm{cm}$ apart. Find the point on the line joining them where the electric field is zero, and verify your answer.

Step 1. Like charges: the null is between them, nearer the weaker ( $+1\ \mathrm{\mu C}$ ). Shares from equation (1): $\sqrt{4} : \sqrt{1} = 2 : 1$ .

Step 2. Split $30$ cm in ratio $2:1$ : the null P sits $\boxed{20\ \mathrm{cm\ from\ the\ } +4\ \mathrm{\mu C}\ \text{charge}}$ (10 cm from the $+1\ \mathrm{\mu C}$ ).

Step 3. Verify the tie. Field of $4\ \mathrm{\mu C}$ at $0.20$ m: $E_1 = \dfrac{(9.0\times10^9)(4\times10^{-6})}{(0.20)^2} = 9.0\times10^5\ \mathrm{N/C}$ . Field of $1\ \mathrm{\mu C}$ at $0.10$ m: $E_2 = \dfrac{(9.0\times10^9)(1\times10^{-6})}{(0.10)^2} = 9.0\times10^5\ \mathrm{N/C}$ ✓ — equal magnitudes, opposite directions.

Step 4. State the direction logic for the method mark: between like charges the two fields oppose, so cancellation is possible only there; outside, both point the same way.

Board level 2 — the unlike-charge null (CBSE/ISC pattern, 3 marks)

Charges $+9\ \mathrm{\mu C}$ and $-4\ \mathrm{\mu C}$ are fixed $20\ \mathrm{cm}$ apart. Locate the point where the net electric field vanishes.

Step 1. Unlike charges: no null between (fields reinforce there). The null lies outside, beyond the weaker charge — the $-4\ \mathrm{\mu C}$ side, away from $+9\ \mathrm{\mu C}$ .

Step 2. Let the null be $x$ beyond the $-4\ \mathrm{\mu C}$ charge. Balance, equation (2): $\dfrac{9}{(0.20+x)^2} = \dfrac{4}{x^2}$ .

Step 3. Square-root both sides: $\dfrac{0.20+x}{x} = \dfrac{3}{2} \Rightarrow 0.40 + 2x = 3x \Rightarrow \boxed{x = 0.40\ \mathrm{m}}$ beyond the $-4\ \mathrm{\mu C}$ charge (i.e. $0.60$ m from $+9\ \mathrm{\mu C}$ ).

Step 4. Verify: $E_9 = \dfrac{(9.0\times10^9)(9\times10^{-6})}{(0.60)^2} = 2.25\times10^5\ \mathrm{N/C}$ ; $E_4 = \dfrac{(9.0\times10^9)(4\times10^{-6})}{(0.40)^2} = 2.25\times10^5\ \mathrm{N/C}$ ✓ — and they oppose out there: the $+9$ pushes outward, the $-4$ pulls inward.

Competitive level 1 — freeze all three (JEE-Main pattern, ~3 minutes)

Charges $+36\ \mathrm{\mu C}$ and $+9\ \mathrm{\mu C}$ are fixed $30\ \mathrm{cm}$ apart. Where should a third charge be placed — and what should it be — so that the entire system is in equilibrium? Verify the equilibrium of one of the fixed charges.

Step 1. Position: the third charge must feel no force, so it sits at the pair's null. Shares $\sqrt{36} : \sqrt{9} = 2 : 1$ , so $q_3$ goes $20$ cm from $+36\ \mathrm{\mu C}$ , $10$ cm from $+9\ \mathrm{\mu C}$ .

Step 2. Sign and magnitude, equation (3): $|q_3| = \dfrac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2} = \dfrac{36\times9}{(6+3)^2}\ \mathrm{\mu C} = \dfrac{324}{81} = 4\ \mathrm{\mu C}$ , and it must attract both: $\boxed{q_3 = -4\ \mathrm{\mu C}\ \text{at } 20\ \mathrm{cm from the } 36\ \mathrm{\mu C}}$ .

Step 3. Verify on the $+36\ \mathrm{\mu C}$ : repulsion from $+9\ \mathrm{\mu C}$ at $0.30$ m: $F = \dfrac{(9.0\times10^9)(36\times10^{-6})(9\times10^{-6})}{0.09} = 32.4\ \mathrm{N}$ outward. Attraction from $-4\ \mathrm{\mu C}$ at $0.20$ m: $F = \dfrac{(9.0\times10^9)(36\times10^{-6})(4\times10^{-6})}{0.04} = 32.4\ \mathrm{N}$ inward ✓.

Step 4. Verdict for the last mark: the equilibrium is unstable (Earnshaw) — displace $q_3$ along the line and the nearer attraction wins; the arrangement disintegrates. Timing note: shares first ( $\sqrt{36}:\sqrt 9$ — clean perfect squares are the examiner's gift), then formula (3) by substitution. Budget 3 minutes including the verification; if the given charges aren't perfect squares, expect the answer in surd form and don't reach for the calculator.

Competitive level 2 — oscillation on the bisector (JEE-Advanced flavour, ~4 minutes)

Two charges $+4\ \mathrm{\mu C}$ each are fixed at $(\pm 30\ \mathrm{cm}, 0)$ . A small ball of mass $10\ \mathrm{g}$ carrying charge $-1\ \mathrm{\mu C}$ is placed at the origin and displaced slightly along the y-axis, then released. Show the motion is simple harmonic for small displacements and find its frequency. (Ignore gravity.)

Step 1. Geometry: at height $y$ on the bisector, each $+4\ \mathrm{\mu C}$ attracts the ball toward itself, at distance $r = \sqrt{a^2 + y^2}$ with $a = 0.30$ m. Horizontal components cancel by symmetry; the vertical components add and point back toward the origin — a restoring force.

Step 2. The net restoring force. Each attraction has magnitude $kQq/r^2$ and vertical fraction $y/r$ :

F_y = -2\,\frac{kQq}{r^2}\cdot\frac{y}{r} = -\frac{2kQq\,y}{(a^2+y^2)^{3/2}} \approx -\frac{2kQq}{a^3}\,y \quad (y \ll a)

Step 3. Read off the effective spring constant: $k_{\text{eff}} = \dfrac{2kQq}{a^3} = \dfrac{2(9.0\times10^9)(4\times10^{-6})(1\times10^{-6})}{(0.30)^3} = 2.67\ \mathrm{N/m}$ . Linear restoring force ⇒ SHM.

Step 4. Frequency: $\omega = \sqrt{k_{\text{eff}}/m} = \sqrt{2.67/0.010} = 16.3\ \mathrm{rad/s}$ , so $f = \dfrac{\omega}{2\pi} = \boxed{2.6\ \mathrm{Hz}}$ (period $0.38$ s).

Step 5. Reconcile with Earnshaw: stable along y, yes — but displace the same ball along the x-axis and the nearer charge's attraction wins: unstable. Stability in one direction is always paid for by instability in another; the origin is a saddle, and true three-dimensional trapping remains forbidden. Timing note: the only real work is Step 2's geometry — write the vertical fraction $y/r$ immediately and binomially drop $y^2$ against $a^2$ . The $(a^2+y^2)^{3/2} \to a^3$ move is the entire "show SHM" step; budget 4 minutes. A negative charge on the bisector of a like pair (or a positive charge between an unlike pair) is the standard rigged-for-SHM configuration.

Exam traps

Importing the "between" habit to unlike charges. The null for unlike charges is outside the segment, beyond the weaker charge. Direction-check the two field arrows before any algebra: cancellation needs opposition. (Options placing the null between $+q$ and $-q$ are the paper's most reliable wrong answer.)

The quadratic detour and its ghost root. Cross-multiplying $\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}$ into a quadratic produces two roots; one is a point where the magnitudes tie but the directions don't oppose — a phantom. Square-root the balance directly and keep the physically placed root (inside for like, outside-beyond-weaker for unlike).

Shares without the square root. Charges $4q$ and $q$ split the line $2:1$ , not $4:1$ — distance enters Coulomb's law squared, so shares go as $\sqrt q$ . Ten-second check: the null must sit nearer the weaker charge; if yours cuddles the stronger one, the ratio is upside down.

Third charge at the midpoint by reflex. $q_3$ belongs at the null point, which is the midpoint only for equal charges. And it is not optional decoration: its sign must be opposite (it has to pull the pair together against their mutual repulsion) and its magnitude is fixed by equation (3) — three separate marks, routinely.

"In equilibrium" sold as "stable". Every electrostatic equilibrium in this archetype is unstable — Earnshaw's theorem. If the question asks about stability, the answer is a saddle: stable along some directions, unstable along others, never trappable. The word "unstable" with one line of nudge-reasoning is a standard allocated mark.

Zero field ⇒ zero potential (and the converse). At a like-pair null, $V$ is a sum of two positive terms — decidedly non-zero. Between unlike charges, $V = 0$ happens where $E$ is large. Keep the vector story (fields, can cancel) and the scalar story (potentials, add with sign) in separate ledgers.

Sign-juggling inside Coulomb's law. Work with magnitudes and directions-in-words ("attracts," "repels"), then assemble. Substituting signed charges into $F = kq_1q_2/r^2$ and trusting algebra to sort directions is the fastest route to a reversed answer.

Origins · 1 minSamuel Earnshaw and how this was discovered

Samuel Earnshaw

1805–1888 · Cambridge & Sheffield, England

Earnshaw was a Cambridge-trained mathematician and clergyman who, in an 1842 paper on the luminiferous ether, proved a result far sturdier than the theory it served: no charge can rest in stable equilibrium under inverse-square forces alone. Inverse-square fields in empty space have no "bowls," only saddles — every equilibrium leaks in at least one direction. The theorem ended two centuries of electrostatic-levitation dreams and still polices engineering: maglev trains, ion traps and levitating displays all smuggle in something his proof excludes — motion, feedback, oscillating fields or diamagnetism. His name survives on exam papers in one word, "unstable," attached to every balanced arrangement of charges, including this article's three-charge tower.

History

1785

— Coulomb's torsion balance establishes the inverse-square law behind every balance argument here.

1813

— Poisson's equation formalises how charge shapes the field.

1842

— Samuel Earnshaw proves no stable electrostatic equilibrium exists — balanced charge towers are mathematically doomed.

1953

— Wolfgang Paul's oscillating-field ion trap dodges Earnshaw with time-dependence (Nobel 1989).

1997

— a live frog is levitated diamagnetically at Nijmegen — a loophole outside Earnshaw's premise.

Going further

Texts. NCERT Physics Class 12, Ch. 1 (Electric Charges and Fields — equilibrium exercises) · HC Verma, Concepts of Physics Vol. 2, Ch. 29 · the gravitational twin of this archetype in the Gravitation chapter (two-mass null points).

Provenance. All results are standard exam canon; every number recomputed. A verifier should recheck: example 1's tie ( $9.0\times10^5$ N/C both sides at 20 cm/10 cm); example 2's $x = 0.40$ m and $2.25\times10^5$ N/C tie; example 3's $q_3 = -4\ \mathrm{\mu C}$ and the 32.4 N balance on both fixed charges; example 4's $k_{\text{eff}} = 2.67$ N/m, $\omega = 16.3$ rad/s, $f = 2.6$ Hz.

Prerequisites. Coulomb's Law · Superposition: Forces & Fields of Many Charges · Field Null Points (the gravitational twin, mechanics track). Leads to. Symmetry Kills Options (technique) · equilibrium-of-charges variants throughout the E&M archetype gym.

Why do the distance shares involve square roots of the charges?

Because charge and distance fight with different powers in $E = kq/r^2$ : charge enters once, distance twice. To balance a charge ratio of 9, the distance ratio only needs to be $\sqrt9 = 3$ — every metre of extra distance counts double (squared) against the field. This is the same square-root shares rule as the two-mass gravitational null, and it generalises: any inverse-square balancing act — intensities of two lamps, sounds of two sources — splits distances as square roots of source strengths.

How do I quickly decide inside versus outside?

One sentence of field-direction reasoning beats memorisation. Between: like charges push a test charge opposite ways (null inside possible); unlike push the same way (no tie). Outside: like charges push the same way (no tie); unlike oppose, and the tie lands beyond the weaker charge — only there does proximity boost the weak field enough to match the strong one. Sketch two arrows at a candidate point, then do algebra only where a tie is possible.

Why must the third charge sit exactly at the null point and not somewhere else with a bigger magnitude?

Because $q_3$ 's own equilibrium is non-negotiable. Anywhere off the null the pair's field is non-zero and $q_3$ feels a force — no choice of its magnitude fixes that (magnitude scales the forces it exerts, not the force it feels). Position is pinned by $q_3$ 's equilibrium; magnitude and sign by the fixed charges'. Three conditions, three unknowns — uniquely solvable.

Doesn't the SHM example contradict Earnshaw's theorem?

No — it illustrates it. The bisector oscillation is stable along y only; the same configuration is unstable along x. Earnshaw forbids equilibria stable in all three directions at once; saddles with some stable directions are legal, and exam problems milk exactly those for SHM. A genuine contradiction would be a charge trapped statically from every side — which never happens.

Is there a null point off the line joining the two charges?

For two point charges, no. Off the axis the two field vectors are not antiparallel — each points radially from its own charge — so they cannot cancel exactly. Cancellation needs collinear fields, which happens only on the line through the charges; "on the line joining them" in the question is physics, not filler.

What happens to the null point if both charges are doubled?

Nothing — it stays put. The balance $\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}$ cares only about the ratio $q_1/q_2$ , so scaling both charges (or changing $k$ , or immersing everything in a uniform dielectric) leaves the location untouched; only the ratio moves the null. The field everywhere else doubles, and the third-charge magnitude in the freeze problem doubles too ( $q_3 \propto q_1q_2/(\sqrt{q_1}+\sqrt{q_2})^2$ scales linearly) — but the geometry is scale-blind.

In the gravitational version the null was always between the bodies. What exactly changed?

One fact: sources now come in two signs. Mass is always positive, so two fields between two masses always oppose — interior null, end of story. Electric charge's four sign combinations collapse to two cases: like pairs behave gravitationally (interior null, square-root shares); unlike pairs are the new physics (reinforcement inside, cancellation outside-beyond-the-weaker). Everything else — balance equation, square root, instability verdict — transfers unchanged. Solve the Earth–Moon problem and you have done two-thirds of this archetype.

Do these three-charge equilibria exist anywhere in nature?

Not statically — Earnshaw sees to that. Nature's stable structures (atoms, molecules, crystals) rest on physics beyond pure electrostatics: quantum mechanics keeps electrons off nuclei, and lattice stability needs exclusion. The classical three-charge tower is a beautiful unstable solution — a pencil balanced on its tip. The configurations still matter: finding force-free points is daily work in electron-optics design; engineers just never expect anything to stay there unaided.