Coulomb's law — force, comparison with gravity: Worked examples

The scenario matrix

Everything Coulomb's law can throw at you falls into one of these cells. We will hit every single one.

#	Cell (case class)	What makes it tricky	Hit by
A	Two charges, opposite signs	direction = attraction	Ex 1
B	Two charges, like signs	direction = repulsion	Ex 2
C	Scaling (change $r$ or $q$ )	inverse-square vs linear	Ex 3
D	One charge is zero (degenerate)	force must vanish	Ex 4
D′	Coincident charges $r \to 0$ (forbidden/limit)	formula blows up; domain	Ex 4b
E	Three charges in a line (vector add, 1-D)	signs set left/right	Ex 5
F	Non-collinear, all like signs (2-D)	components + symmetry	Ex 6
F′	Non-collinear, mixed signs (2-D)	one push, one pull	Ex 6b
G	Equilibrium point (where net force = 0)	limiting / algebra	Ex 7
H	Real-world word problem	translate words → numbers	Ex 8
I	Exam twist (compare to gravity / ratios)	conceptual, $r$ cancels	Ex 9

Before we start, one reminder of the tool we lean on constantly:

Cell A — opposite signs (attraction)

Worked example Example 1 — a plus and a minus

Charges $q_{1} = + 4 μ C$ and $q_{2} = - 5 μ C$ sit $r = 0.20 m$ apart. Find the force magnitude and direction.

Forecast: signs are opposite, so before any arithmetic — will they pull together or push apart? (Guess: pull.)

Step 1 — Convert to SI, and handle the sign deliberately. $q_{1} = + 4 μ C = 4 \times 1 0^{- 6} C$ and $q_{2} = - 5 μ C = - 5 \times 1 0^{- 6} C$ . The magnitude formula needs $∣ q_{1} q_{2} ∣$ , so we take sizes only: $∣ q_{1} ∣ = 4 \times 1 0^{- 6}$ , $∣ q_{2} ∣ = 5 \times 1 0^{- 6}$ , $r = 0.20 m$ . Why this step, and why is dropping the minus sign allowed? Converting µC → C is just a scale change ( $\times 1 0^{- 6}$ ); it never touches the sign. We drop $q_{2}$ 's minus not because it doesn't matter, but because we've split the job into two halves: the magnitude formula answers "how big?", and by construction it only ever uses $∣ q_{1} q_{2} ∣$ , so a negative number has no place in it. The minus sign is not thrown away — it is parked and brought back in Step 4 to decide direction. This is the sign-vs-magnitude discipline the parent note insists on.

Step 2 — Magnitude formula, using $∣ q_{1} q_{2} ∣$ . $F = (8.99 \times 1 0^{9}) \frac{( 4 \times 1 0 ^{- 6} ) ( 5 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}}$ Why magnitudes here? We compute size first; the sign never belongs in the magnitude formula.

Step 3 — Crunch it. Numerator inside: $(4 \times 1 0^{- 6}) (5 \times 1 0^{- 6}) = 20 \times 1 0^{- 12}$ . Divide by $(0.20)^{2} = 0.04$ : $20 \times 1 0^{- 12} /0.04 = 5 \times 1 0^{- 10}$ . Times $k$ : $8.99 \times 1 0^{9} \times 5 \times 1 0^{- 10} = 4.495 N$ . Why break it into these three sub-steps? Powers of ten are where sign-and-exponent slips happen; grouping the charge product, then the distance square, then the multiply-by- $k$ keeps each exponent visible and checkable rather than juggling everything at once.

Step 4 — Bring the parked sign back for direction. $q_{1} q_{2} = (+) (-) < 0 \Rightarrow$ attractive. The forecast holds.

Answer: $F \approx 4.50 N$ , pulling the two charges toward each other.

Verify: Units: $\frac{N\cdotm ^{2}}{C ^{2}} \cdot \frac{C ^{2}}{m ^{2}} = N$ ✓. Magnitude is a few newtons for microcoulombs at $20 cm$ — the right ballpark.

Cell B — like signs (repulsion)

Cell C — scaling (inverse-square)

Worked example Example 3 — change distance and charge together

Start from Ex 1 ( $F = 4.50 N$ ). Now triple the distance and double $q_{1}$ . New force?

Forecast: doubling a charge multiplies $F$ by $2$ ; tripling $r$ divides $F$ by... $3$ ? or $9$ ? (Guess before reading — the inverse-square is the trap.)

Step 1 — Write $F$ as a product of ratios. Since $F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$ , scaling $q_{1}$ by factor $a$ and $r$ by factor $b$ gives $F_{new} = F_{old} \cdot \frac{a}{b ^{2}} .$ Why this step? Ratios let us reuse the old answer instead of recomputing from scratch — and they expose exactly where the square lives.

Step 2 — Insert $a = 2$ (double $q_{1}$ ), $b = 3$ (triple $r$ ). $F_{new} = 4.50 \times \frac{2}{3 ^{2}} = 4.50 \times \frac{2}{9} .$ Why $3^{2}$ and not $3$ ? Distance enters as $r^{2}$ , so its factor gets squared. This is the whole point of the cell.

Step 3 — Evaluate. $4.50 \times 2/9 = 4.50 \times 0.2222 = 1.00 N$ .

Answer: $F_{new} \approx 1.00 N$ .

Verify: $\frac{2}{9} \approx 0.222$ ; $4.50 \times 0.222 = 1.00$ ✓. If you'd wrongly used $\frac{2}{3}$ you'd get $3.0 N$ — three times too big. The squared distance is why.

Cell D — a zero charge (degenerate case)

Worked example Example 4 — what if one charge is zero?

$q_{1} = + 7 μ C$ , $q_{2} = 0$ , distance $r = 0.05 m$ . Force?

Forecast: an object with no charge — does Coulomb's law even apply? What must $F$ be? (Guess: exactly zero.)

Step 1 — Substitute $q_{2} = 0$ . $F = k \frac{∣ q _{1} \cdot 0∣}{r ^{2}} = k \frac{0}{r ^{2}} = 0.$ Why this step? The product $q_{1} q_{2}$ contains the zero factor, killing the whole numerator regardless of $q_{1}$ or $r$ .

Step 2 — Interpret. No charge means nothing for $q_{1}$ to grab. The force is identically zero, and its direction is undefined (a zero vector has no direction). Why mention direction? So you don't waste time asking "attract or repel?" — neither; there is no force at all.

Answer: $F = 0 N$ , for any distance $r > 0$ .

Verify: $∣ q_{1} \cdot 0∣ = 0$ , so $F = 0$ ✓. The limiting behaviour: as $q_{2} \to 0$ , $F \to 0$ continuously — no surprises.

Cell D′ — coincident charges, $r \to 0$ (forbidden input / limit)

Worked example Example 4b — what happens as the charges touch?

Keep $q_{1} = + 7 μ C$ , $q_{2} = + 2 μ C$ but shrink the gap: compute $F$ at $r = 0.10$ , then $0.010$ , then $0.0010 m$ , and ask what $r = 0$ would give.

Forecast: as $r$ gets smaller, does $F$ settle down to some finite value, or does it grow without bound? (Guess before computing.)

Step 1 — The charge product is fixed. $k ∣ q_{1} q_{2} ∣ = (8.99 \times 1 0^{9}) (7 \times 1 0^{- 6}) (2 \times 1 0^{- 6}) = 0.1259 N\cdotm^{2}$ (this whole lump is a constant here). Why isolate it? Only $r$ is changing, so pulling the constant out shows cleanly that $F = 0.1259/ r^{2}$ .

Step 2 — Evaluate at shrinking $r$ . $r = 0.10 : F = \frac{0.1259}{0.01} = 12.6 N; r = 0.010 : F = \frac{0.1259}{1 0 ^{- 4}} = 1.26 \times 1 0^{3} N; r = 0.0010 : F = \frac{0.1259}{1 0 ^{- 6}} = 1.26 \times 1 0^{5} N .$ Why a sequence? Each tenfold shrink of $r$ multiplies $F$ by $100$ (because of $r^{2}$ ). The pattern makes the trend unmistakable.

Step 3 — The limit and the forbidden point. As $r \to 0$ , $F = 0.1259/ r^{2} \to + \infty$ : the force diverges. At exactly $r = 0$ the expression $1/ r^{2}$ is undefined (division by zero), so Coulomb's law gives no answer at all — $r = 0$ is outside its domain, as stated above. Why does this not break physics? Two ideal point charges at the same point is a mathematical fiction; real charged objects have size and can't occupy one point, and at tiny separations other physics (quantum effects, the strong force in a nucleus) takes over. Coulomb's law is a large-scale, point-charge law and simply does not apply at $r = 0$ .

Answer: $F \to \infty$ as $r \to 0$ ; the value at $r = 0$ is undefined (forbidden input).

Verify: $0.1259/0.01 = 12.59$ ; $0.1259/1 0^{- 4} = 1259$ ; each step $\times 100$ ✓ — a genuine blow-up, confirming the singularity.

Cell E — three charges in a line (1-D superposition)

Worked example Example 5 — a charge feels two others

Three charges on a straight line (positions in metres): $q_{A} = + 2 μ C$ at $x = 0$ , $q_{B} = + 3 μ C$ at $x = 0.30$ , and our target $q_{C} = + 1 μ C$ at $x = 0.10$ . Find the net force on $q_{C}$ .

Figure (Cell E). The horizontal axis is the line the three charges sit on. The blue dot is $q_{A}$ , the pink dot is $q_{B}$ , the yellow dot in the middle is our target $q_{C}$ . The two coloured arrows leaving $q_{C}$ are the two pushes: the long blue arrow (from the near charge $q_{A}$ ) points right; the short pink arrow (from the far charge $q_{B}$ ) points left. The point of the picture is length = strength — you can literally see the blue arrow outreach the pink one, so before any arithmetic the diagram tells you the net force is rightward.

Forecast: $q_{C}$ sits between two positive charges. $q_{A}$ (left) pushes it right; $q_{B}$ (right) pushes it left. Which wins — and so which way does $q_{C}$ end up moving? (Guess before computing.)

Step 1 — Force from $q_{A}$ on $q_{C}$ . Separation $r_{A C} = 0.10 m$ . $F_{A C} = (8.99 \times 1 0^{9}) \frac{( 2 \times 1 0 ^{- 6} ) ( 1 \times 1 0 ^{- 6} )}{( 0.10 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{2 \times 1 0 ^{- 12}}{0.01} = 1.798 N .$ Both positive → repulsion → this pushes $q_{C}$ in the $+ x$ direction (to the right). Why this step? Superposition says treat each source alone — pretend $q_{B}$ isn't there — and compute the plain two-charge force first. We'll fold $q_{B}$ in next; doing them one at a time is what stops the signs from tangling.

Step 2 — Force from $q_{B}$ on $q_{C}$ . Separation $r_{B C} = 0.30 - 0.10 = 0.20 m$ . $F_{B C} = (8.99 \times 1 0^{9}) \frac{( 3 \times 1 0 ^{- 6} ) ( 1 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{3 \times 1 0 ^{- 12}}{0.04} = 0.674 N .$ Both positive → repulsion → this pushes $q_{C}$ in the $- x$ direction (to the left). Why this step? Same routine as Step 1 but for the other source. Notice $q_{B}$ is bigger ( $3$ vs $2 μ C$ ) yet twice as far; the $1/ r^{2}$ penalty is what we're testing against the raw charge advantage.

Step 3 — Add as signed 1-D vectors (take right = $+$ ). $F_{net} = + 1.798 - 0.674 = + 1.124 N .$ Why subtract? The two pushes point opposite ways along the same line, so their signed values add — a subtraction in disguise.

Answer: $F_{net} \approx 1.12 N$ directed to the right ( $+ x$ ). The nearer charge $q_{A}$ wins.

Verify: $1.798 - 0.674 = 1.124$ ✓. Sanity: $q_{A}$ is closer ( $0.10$ vs $0.20$ ), and closeness matters as $1/ r^{2}$ , so it dominates despite $q_{B}$ being larger — exactly what the arrows in the figure show.

Cell F — non-collinear, all like signs (2-D vector addition)

Worked example Example 6 — the right-triangle setup

$q_{1} = + 6 μ C$ at the origin $(0, 0)$ , and $q_{2} = + 6 μ C$ at $(0.40, 0) m$ . A test charge $q_{3} = + 2 μ C$ sits at $(0.20, 0.15) m$ — above the midpoint. Find the net force on $q_{3}$ .

Figure (Cell F). The two lower dots ( $q_{1}$ blue, $q_{2}$ pink) and the raised dot ( $q_{3}$ yellow) form an isosceles triangle. Each dashed line is a $0.25 m$ leg. The blue and pink arrows are the two pushes on $q_{3}$ , each pointing straight away from its source; they lean outward symmetrically. The yellow arrow is their sum, standing dead vertical. The picture makes the cancellation visible: the two slanted arrows are mirror images, so their sideways parts point opposite ways and delete, while their upward parts stack — this is why the net arrow is purely vertical.

Forecast: by symmetry $q_{1}$ and $q_{2}$ are mirror images about the vertical through $q_{3}$ . When we add their two pushes, which components cancel and which reinforce? (Guess: horizontals cancel, verticals add — net force straight up.)

Step 1 — Distances from each source to $q_{3}$ . $q_{1} \to q_{3}$ : $Δ = (0.20, 0.15)$ , so $r_{1} = 0.2 0^{2} + 0.1 5^{2} = 0.0625 = 0.25 m$ . By symmetry $r_{2} = 0.25 m$ too. Why this step? We need $r$ before we can use $1/ r^{2}$ ; the geometry is a $3$ - $4$ - $5$ triangle scaled by $0.05$ .

Step 2 — Magnitude of each force (they're equal by symmetry). $F_{1} = (8.99 \times 1 0^{9}) \frac{( 6 \times 1 0 ^{- 6} ) ( 2 \times 1 0 ^{- 6} )}{( 0.25 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{12 \times 1 0 ^{- 12}}{0.0625} = 1.726 N .$ Why compute only one? The setup is a mirror image, so $F_{2}$ has the identical magnitude; computing it twice would waste effort. We'll differ them only by direction in the next step.

Step 3 — Break each force into components. All three charges are positive, so each force pushes $q_{3}$ away from its source. Force from $q_{1}$ points along $(0.20, 0.15) /0.25 = (0.8, 0.6)$ . $F_{1 x} = 1.726 \times 0.8 = 1.381, F_{1 y} = 1.726 \times 0.6 = 1.036 N .$ Force from $q_{2}$ points along $(0.20 - 0.40, 0.15) /0.25 = (- 0.8, 0.6)$ : $F_{2 x} = - 1.381, F_{2 y} = + 1.036 N .$ Why components? Vectors in different directions can't be added as numbers; we split each into $x$ and $y$ , add those separately, then recombine. This is the only reliable 2-D method.

Step 4 — Sum components. $F_{x} = 1.381 + (- 1.381) = 0, F_{y} = 1.036 + 1.036 = 2.072 N .$ The forecast holds: horizontals cancel, verticals reinforce.

Answer: $F_{net} \approx 2.07 N$ straight up ( $+ y$ ).

Verify: $F_{x} = 0$ by exact symmetry ✓; $2 \times 1.036 = 2.072$ ✓. The magnitude $0^{2} + 2.07 2^{2} = 2.072 N$ is less than $2 F_{1} = 3.45 N$ , because only the vertical slices survive — sensible.

Cell F′ — non-collinear, mixed signs (one push, one pull) in 2-D

Worked example Example 6b — a pull and a push added as vectors

Same geometry as Ex 6, but now flip $q_{2}$ 's sign: $q_{1} = + 6 μ C$ at $(0, 0)$ , $q_{2} = - 6 μ C$ at $(0.40, 0) m$ , test charge $q_{3} = + 2 μ C$ at $(0.20, 0.15) m$ . Find the net force on $q_{3}$ .

Forecast: $q_{1}$ (like sign) pushes $q_{3}$ away; $q_{2}$ (opposite sign) pulls $q_{3}$ toward it. Their magnitudes are still equal. When we add a push and a pull that are mirror images, which components now survive — the vertical or the horizontal? (Guess before computing — it flips from Ex 6.)

Step 1 — Magnitudes unchanged. Distances are still $0.25 m$ and $∣ q_{1} q_{2} ∣ = ∣ q_{1} q_{3} ∣$ pairs are the same sizes, so each force magnitude is again $F_{1} = F_{2} = 1.726 N$ . Why reuse them? Magnitude depends only on $∣ q ∣$ and $r$ , both unchanged; only directions differ. This isolates the new physics (mixed signs) cleanly.

Step 2 — Direction of the force from $q_{1}$ (push, away from $q_{1}$ ). Same as before: unit vector $(0.8, 0.6)$ . $F_{1 x} = + 1.381, F_{1 y} = + 1.036 N .$ Why this direction? $q_{1}$ and $q_{3}$ are both positive → repulsion → the force points from $q_{1}$ toward $q_{3}$ , i.e. up-and-to-the-right.

Step 3 — Direction of the force from $q_{2}$ (pull, toward $q_{2}$ ). $q_{2}$ is negative, $q_{3}$ positive → attraction → the force on $q_{3}$ points from $q_{3}$ toward $q_{2}$ , i.e. along $(0.40 - 0.20, 0 - 0.15) /0.25 = (0.8, - 0.6)$ . $F_{2 x} = 1.726 \times 0.8 = + 1.381, F_{2 y} = 1.726 \times (- 0.6) = - 1.036 N .$ Why does the sign flip only the $y$ -part? Attraction reverses the arrow relative to Ex 6: it now aims downward toward $q_{2}$ instead of upward away. The horizontal parts happen to line up; the verticals now oppose.

Step 4 — Sum components. $F_{x} = 1.381 + 1.381 = 2.762, F_{y} = 1.036 + (- 1.036) = 0 N .$ Exactly the opposite survivor from Ex 6: now the horizontals reinforce and the verticals cancel.

Answer: $F_{net} \approx 2.76 N$ pointing straight right ( $+ x$ , from $q_{1}$ 's side toward $q_{2}$ 's side).

Verify: $F_{y} = 0$ by the mirror symmetry of a push-plus-pull ✓; $2 \times 1.381 = 2.762$ ✓. Physical sense: $q_{3}$ is shoved away from the $+$ charge and pulled toward the $-$ charge — both effects point the same horizontal way, so they add. This is the classic "field points from + to −" behaviour of the Electric Field.

Cell G — the equilibrium point (net force = 0)

Worked example Example 7 — where does a test charge feel nothing?

Two fixed charges: $q_{1} = + 9 μ C$ at $x = 0$ and $q_{2} = + 4 μ C$ at $x = 1.0 m$ . Where on the line between them can a small charge sit with zero net force?

Figure (Cell G). The blue dot ( $q_{1}$ , the bigger charge) is on the left, the pink dot ( $q_{2}$ ) on the right, and the yellow square marks the balance point. The two short arrows leaving that square point in opposite directions and are drawn equal in length — that equal length is the whole condition "net force = 0." Crucially the yellow square sits past the midpoint, closer to the smaller pink charge: the diagram shows why the balance must drift toward the weaker charge, so that its shorter distance compensates for its smaller charge.

Forecast: both fixed charges are positive, so a test charge between them is pushed away from each. Equilibrium sits where the two pushes balance. Since $q_{1}$ is bigger, will the balance point be nearer the big charge or the small one? (Guess: nearer the smaller charge $q_{2}$ , so the bigger one's extra distance weakens it.)

Step 1 — Set the two magnitudes equal. Let the test charge be at distance $x$ from $q_{1}$ , so distance $(1 - x)$ from $q_{2}$ . The $k$ and the test charge $q$ appear on both sides and cancel: $\frac{q _{1}}{x ^{2}} = \frac{q _{2}}{( 1 - x ) ^{2}} .$ Why set them equal? "Zero net force" means the leftward push equals the rightward push in magnitude — that's an equation for $x$ .

Step 2 — Plug charges (in µC — the units cancel in a ratio). $\frac{9}{x ^{2}} = \frac{4}{( 1 - x ) ^{2}} .$

Step 3 — Take square roots (both sides positive, so this is safe): $\frac{3}{x} = \frac{2}{1 - x} .$ Why square-root now? It turns the messy squared equation into a clean linear one.

Step 4 — Cross-multiply and solve. $3 (1 - x) = 2 x \Rightarrow 3 - 3 x = 2 x \Rightarrow 3 = 5 x \Rightarrow x = 0.6 m$ .

Answer: the balance point is at $x = 0.60 m$ from $q_{1}$ (so $0.40 m$ from $q_{2}$ ).

Verify: $\frac{9}{0. 6 ^{2}} = \frac{9}{0.36} = 25$ ; $\frac{4}{0. 4 ^{2}} = \frac{4}{0.16} = 25$ ✓. And $0.60 > 0.50$ means it is nearer the smaller charge $q_{2}$ — forecast confirmed. (The other algebraic root, $x = 3$ , lies outside the segment and would need the sign check to reject — between two like charges only the inside root is a real balance.)

Cell H — a real-world word problem

Worked example Example 8 — the static-cling balloon

You rub a balloon on your hair; it picks up a charge of $q = - 8 \times 1 0^{- 8} C$ . You bring it $d = 3.0 cm$ from a small charged wall spot of $Q = + 5 \times 1 0^{- 8} C$ . What force holds the balloon to the wall, and which way?

Forecast: rubbing gives the balloon negative charge, the spot is positive — opposite signs. Attract or repel? And will the force be small (fraction of a newton) or large? (Guess before computing.)

Step 1 — Translate words to SI, and split off the signs. $∣ q ∣ = 8 \times 1 0^{- 8} C$ , $∣ Q ∣ = 5 \times 1 0^{- 8} C$ , $d = 0.030 m$ . Record the actual signs separately for later: $q$ is $-$ , $Q$ is $+$ . Why this step, and why split the sign off? "3.0 cm" and nanocoulomb-scale numbers must become metres and coulombs before $k$ can be used. The magnitude formula $F = k ∣ q Q ∣/ d^{2}$ takes only sizes — if you fed it the negative $q$ you'd get a "negative force," which is meaningless (a force size can't be negative). So we deliberately set the signs aside now, do the size calculation, and bring the signs back in the final step to decide direction. This is exactly the "sign vs magnitude" discipline from Electric Field and the parent note.

Step 2 — Magnitude formula. $F = (8.99 \times 1 0^{9}) \frac{( 8 \times 1 0 ^{- 8} ) ( 5 \times 1 0 ^{- 8} )}{( 0.030 ) ^{2}} .$

Step 3 — Crunch, one factor at a time. Numerator: $(8 \times 1 0^{- 8}) (5 \times 1 0^{- 8}) = 40 \times 1 0^{- 16} = 4.0 \times 1 0^{- 15}$ . Divide by $(0.030)^{2} = 9.0 \times 1 0^{- 4}$ : $4.0 \times 1 0^{- 15} /9.0 \times 1 0^{- 4} = 4.44 \times 1 0^{- 15 + 4} = 4.44 \times 1 0^{- 12}$ . Times $k$ : $8.99 \times 1 0^{9} \times 4.44 \times 1 0^{- 12} = 3.99 \times 1 0^{- 2} N$ . Why group it this way? The exponents ( $- 8$ twice, then $- 4$ , then $+ 9$ ) are the danger zone; handling the charge product, then the distance square, then $k$ in separate moves keeps each power of ten in view so a slip is caught immediately.

Step 4 — Bring the signs back for direction. $q$ is negative and $Q$ is positive → opposite signs → attraction; the balloon is pulled toward the wall.

Answer: $F \approx 0.040 N$ (about $4$ grams-force), attractive — the balloon sticks to the wall.

Verify: $\approx 0.040 N$ ; the weight this could support $= F / g = 0.040/9.8 \approx 0.0041 kg = 4.1 g$ ✓ — a balloon is a few grams, so it plausibly sticks. Realistic.

Cell I — the exam twist (compare to gravity)

Worked example Example 9 — electric vs gravity for two protons

For two protons ( $q = 1.6 \times 1 0^{- 19} C$ , $m = 1.67 \times 1 0^{- 27} kg$ ), find the ratio of their electric repulsion to their gravitational attraction. ( $G = 6.67 \times 1 0^{- 11}$ .)

Forecast: the parent note found $\sim 1 0^{42}$ for two electrons. Protons are heavier, so gravity is a bit stronger — will the ratio be bigger or smaller than for electrons? (Guess: smaller, since heavier means more gravity.)

Step 1 — Write both forces and take the ratio. $\frac{F _{e}}{F _{g}} = \frac{k q ^{2} / r ^{2}}{G m ^{2} / r ^{2}} = \frac{k q ^{2}}{G m ^{2}} .$ Why this step? Both forces carry the identical $1/ r^{2}$ ; taking the ratio cancels $r$ entirely — this is the whole trick of the twist. The answer is a pure number, independent of separation. (See Newton's Law of Gravitation.)

Step 2 — Plug numbers. $\frac{F _{e}}{F _{g}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 6.67 \times 1 0 ^{- 11} ) ( 1.67 \times 1 0 ^{- 27} ) ^{2}} .$ Why keep top and bottom separate? Each side is a product of very different powers of ten; evaluating the numerator fully, then the denominator fully, then dividing once, keeps the exponent bookkeeping honest.

Step 3 — Evaluate top and bottom. Top: $8.99 \times 1 0^{9} \times 2.56 \times 1 0^{- 38} = 2.30 \times 1 0^{- 28}$ . Bottom: $6.67 \times 1 0^{- 11} \times 2.79 \times 1 0^{- 54} = 1.86 \times 1 0^{- 64}$ . Ratio: $2.30 \times 1 0^{- 28} /1.86 \times 1 0^{- 64} = 1.24 \times 1 0^{36}$ .

Answer: $\frac{F _{e}}{F _{g}} \approx 1.2 \times 1 0^{36}$ .

Verify: The exponent $36 < 42$ (the electron value) ✓ — protons are heavier, so gravity closes the gap a little. Forecast confirmed. Distance never appeared: the ratio is the same at $1 fm$ or across a galaxy.

Recall Which cell is this? (quick self-test)

A charge sits at the corner of a square with three others. Which method? ::: 2-D vector addition (Cell F/F′): components, add $x$ and $y$ separately, recombine — watch each charge's sign for push vs pull. You're told one charge is $0$ . What is the force? ::: Exactly zero, any distance (Cell D). Two charges are placed at the same point, $r = 0$ . What does the formula give? ::: Undefined — $r = 0$ is forbidden; $F \to \infty$ as $r \to 0$ (Cell D′). "Where is the net force zero?" ::: Equilibrium point (Cell G): set magnitudes equal, square-root, solve. Distance triples, charge doubles — factor on $F$ ? ::: $\times 2/9$ (Cell C): the distance factor is squared.

Active-Recall Flashcards

What sets the direction (attract/repel) of a Coulomb force?

The sign of the product

q_{1} q_{2}

: positive → repel, negative → attract. Magnitude uses

∣ q_{1} q_{2} ∣

Why is

r = 0

forbidden in Coulomb's law?

The

r^{2}

in the denominator makes

F

undefined (division by zero); as

r \to 0

F \to \infty

. Point charges can't coincide.

Three charges in a line — how do you get the net force on one?

Superposition: compute each pairwise force, assign it a sign for its direction along the line, add the signed values.

Non-collinear charges — the method?

Split each force into

x

and

y

components, add components separately, recombine; direction of each force depends on that pair's sign (push if like, pull if unlike).

Condition for the equilibrium point between two like charges?

k q_{1} / x^{2} = k q_{2} / (1 - x)^{2}

(test charge and

k

cancel); square-root and solve.

Why does

r

cancel in an electric-to-gravity ratio?

Both

F_{e}

and

F_{g}

carry

1/ r^{2}

; dividing removes it, leaving a pure constant.