Worked examples — Superposition principle for forces

Before anything, one anchor picture. A "force" here is an arrow: its length is how hard the push is, its direction is which way it pushes. Adding arrows means laying them tip-to-tail — never adding their lengths unless they lie on one line.

Throughout, $k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^9{\text{N⋅m}}^2/{\text{C}}^2$ is the Coulomb constant (from Coulomb's Law), and $\mu \text{C}$ means $10^{-6}$ coulomb.

The scenario matrix

Every problem in this topic is one (or a blend) of these cells. The worked examples below each tag the cell they fill.

Cross-cutting checks every example must pass:

• Signs: like charges repel (arrow points away from the source), unlike attract (arrow points toward the source).
• Units: every $F$ comes out in newtons (N).
• Direction: never add magnitudes of non-collinear arrows.

Case A — Collinear, forces add

Forecast: guess first — will the two forces help each other or fight? (One repels, one attracts... but from which side?)

Step 1 — Force from $q_1$. Why this step? Superposition says treat $q_1$ alone. Distance $r_1=|2-0|=2\text{m}$. $F_{01}=\frac{k|q_1{q}_0|}{r_1^2}=\frac{(9\times 10^9)(3\times 10^{-6})(1\times 10^{-6})}{2^2}=6.75\times 10^{-3}\text{N}$ $q_1$ and $q_0$ are both positive ⇒ repel. $q_1$ sits to the left, so it shoves $q_0$ to the +x.

Step 2 — Force from $q_2$. Why this step? Now treat $q_2$ alone. $r_2=|4-2|=2\text{m}$. $F_{02}=\frac{(9\times 10^9)(2\times 10^{-6})(1\times 10^{-6})}{2^2}=4.5\times 10^{-3}\text{N}$ $q_2$ is negative, $q_0$ positive ⇒ attract. $q_2$ sits to the right, pulling $q_0$ to the +x.

Step 3 — Add (they're collinear and same way). $F_{\text{net}}=6.75+4.5=11.25\times 10^{-3}\text{N}=11.25\text{mN}(+x)$

Case B — Collinear, forces oppose (partial cancel)

Forecast: both neighbours are positive and push $q_0$ away — but from opposite sides. Which side wins?

Step 1 — From $q_1$ (left, repels ⇒ pushes +x). $r_1=1\text{m}$. $F_{01}=\frac{(9\times 10^9)(4\times 10^{-6})(2\times 10^{-6})}{1^2}=7.2\times 10^{-2}\text{N}(+x)$

Step 2 — From $q_2$ (right, repels ⇒ pushes $-x$). $r_2=2\text{m}$. $F_{02}=\frac{(9\times 10^9)(1\times 10^{-6})(2\times 10^{-6})}{2^2}=4.5\times 10^{-3}\text{N}(-x)$

Step 3 — Subtract (opposite directions on one line). Why subtract? Opposite arrows on the same line partly cancel; keep the sign of the larger. $F_{\text{net}}=7.2\times 10^{-2}-0.45\times 10^{-2}=6.75\times 10^{-2}\text{N}(+x)$

Case C — Collinear, forces exactly cancel (equilibrium point)

Forecast: the null point sits closer to the smaller charge or the larger one? Guess before solving.

Step 1 — Set the two magnitudes equal. Why this step? "Zero net force" between two like charges means the two pushes (opposite directions here) have equal size. Let the point be at $x$, $0<x<1$. Distances: $x$ from $q_1$, $(1-x)$ from $q_2$. $\frac{k{q}_1{q}_0}{x^2}=\frac{k{q}_2{q}_0}{(1-x{)}^2}$

Step 2 — Cancel $k{q}_0$ and take square roots. Why cancel? The test charge and $k$ appear on both sides identically. $\frac{9}{x^2}=\frac{4}{(1-x{)}^2}\Rightarrow \frac{3}{x}=\frac{2}{1-x}$

Step 3 — Solve the linear equation. $3(1-x)=2x\Rightarrow 3=5x\Rightarrow x=0.6\text{m}$

The null point is at $x=0.6\text{m}$ — closer to the smaller charge $q_2$, because a weaker charge must be approached to feel as strong a push as the far bigger one.

Case D — 2-D right angle (resolve then recombine)

Forecast: will the net force point straight along $+x$, or dip below it?

Step 1 — Force from $q_1$. $r_1=4$ m, repulsion straight $+x$. $F_{01}=\frac{(9\times 10^9)(10^{-12})}{16}=5.625\times 10^{-4}\text{N}\Rightarrow (5.625\times 10^{-4},0)$

Step 2 — Force from $q_2$ — find the direction first. Why direction first? In 2-D the arrow isn't along an axis; we need its unit vector. The separation from source $q_2$ to target $q_0$ is $(4,0)-(0,3)=(4,-3)$, length $\sqrt{4^2+3^2}=5$. Unit vector $=(0.8,-0.6)$. Repulsion pushes along it (away from $q_2$). $F_{02}=\frac{(9\times 10^9)(10^{-12})}{25}=3.6\times 10^{-4}\text{N}$ ${\vec{F}}_{02}=3.6\times 10^{-4}(0.8,-0.6)=(2.88\times 10^{-4},-2.16\times 10^{-4})$

Step 3 — Add components (this is why we resolve). $F_x=(5.625+2.88)\times 10^{-4}=8.505\times 10^{-4},F_y=-2.16\times 10^{-4}$

Step 4 — Recombine into magnitude and angle. Why  ​ and arctan? Magnitude of an arrow from its legs is Pythagoras; the tilt is the angle whose tangent is (opposite $F_y$)/(adjacent $F_x$) — see Vector Addition and Resolution. $|\vec{F}|=\sqrt{8.505^2+2.16^2}\times 10^{-4}=8.78\times 10^{-4}\text{N}$ $\varphi =\arctan \frac{-2.16}{8.505}=-14.3^{\circ }\text{(below }+x)$

Case E — 2-D mix of attraction and repulsion

Forecast: one neighbour pushes, one pulls — do the arrows reinforce or fight?

Step 1 — From $q_1$ (like ⇒ repel). Separation $q_1\to q_0=(0,4)$, length $4$, unit $(0,1)$. Repel ⇒ push along $+y$. $F_{01}=\frac{(9\times 10^9)(2\times 10^{-6})(1\times 10^{-6})}{16}=1.125\times 10^{-3}\text{N}\Rightarrow (0,1.125\times 10^{-3})$

Step 2 — From $q_2$ (unlike ⇒ attract, so arrow points TOWARD $q_2$). Why toward? Attraction pulls $q_0$ to the source. Direction $=q_0\to q_2=(3,0)-(0,4)=(3,-4)$, length $5$, unit $(0.6,-0.8)$. $F_{02}=\frac{(9\times 10^9)(2\times 10^{-6})(1\times 10^{-6})}{25}=7.2\times 10^{-4}\text{N}$ ${\vec{F}}_{02}=7.2\times 10^{-4}(0.6,-0.8)=(4.32\times 10^{-4},-5.76\times 10^{-4})$

Step 3 — Add components. $F_x=0+4.32\times 10^{-4}=4.32\times 10^{-4}$ $F_y=11.25\times 10^{-4}-5.76\times 10^{-4}=5.49\times 10^{-4}$

Step 4 — Magnitude and angle. $|\vec{F}|=\sqrt{4.32^2+5.49^2}\times 10^{-4}=6.98\times 10^{-4}\text{N}$ $\varphi =\arctan \frac{5.49}{4.32}=51.8^{\circ }\text{above }+x$

Case F — Symmetric arrangement (equilateral triangle)

Forecast: by symmetry, which direction must the net force point? Guess before computing.

Step 1 — Magnitude of each pair force (both equal). Why equal? Same charges, same side length $2\text{m}$. $F=\frac{(9\times 10^9)(2\times 10^{-6}{)}^2}{2^2}=9\times 10^{-3}\text{N}$

Step 2 — Set up geometry. Put the target at $(2,0)$, the other two at $(0,0)$ and $(1,\sqrt{3})$. Both forces are repulsive, pushing the target away from each source.

• From $(0,0)$: direction $(2,0)-(0,0)=(1,0)$ ⇒ push $+x$: $(9\times 10^{-3},0)$.
• From $(1,\sqrt{3})$: direction $(2,0)-(1,\sqrt{3})=(1,-\sqrt{3})$, length $2$, unit $(0.5,-\frac{\sqrt{3}}{2})$ ⇒ $9\times 10^{-3}(0.5,-0.866)=(4.5\times 10^{-3},-7.794\times 10^{-3})$.

Step 3 — Add components. $F_x=(9+4.5)\times 10^{-3}=13.5\times 10^{-3},F_y=-7.794\times 10^{-3}$

Step 4 — Magnitude. $|\vec{F}|=\sqrt{13.5^2+7.794^2}\times 10^{-3}=1.559\times 10^{-2}\text{N}$ Symmetry check: the net points along the bisector away from the triangle's centre — the two equal forces at $60^{\circ }$ apart combine into $|\vec{F}|=\sqrt{3}F=1.732\times 9\times 10^{-3}=1.559\times 10^{-2}\text{N}$. ✓

Case G — Limiting / degenerate inputs

Forecast: if a source charge vanishes, its force term should vanish too — does the formula obey?

Step 1 — Set $q_2=0$. Why check this? A robust formula must degrade gracefully. $F_{02}=\frac{k{q}_1\cdot 0}{r^2}=0$. Only $q_1$ survives: ${\vec{F}}_{\text{net}}=(5.625\times 10^{-4},0)\text{N}$ The net snaps back to pure $+x$ — exactly the one-charge answer. Superposition with a zero term = drop that term.

Step 2 — Let $r\to \infty$. ${\lim }_{r\to \infty }\frac{k{q}_1{q}_0}{r^2}=0$ Why it matters: very distant charges contribute negligibly. This is why in Continuous Charge Distributions far-away bits of charge are safely small, and the sum can become an integral $\int d\vec{F}$.

Step 3 — Degenerate distance $r\to 0$. The formula blows up ($\propto 1/r^2\to \infty$). Physically two point charges never truly touch; the model breaks and we stop before that.

Case H — Word problem (real-world framing)

Forecast: the middle ball sits between two equal pushes from opposite sides — before calculating, what's the net?

Step 1 — Translate words to numbers. Why this step? "Midway" ⇒ each fixed ball is $3\text{cm}=0.03\text{m}$ from the middle. Both are $+5\mu \text{C}$, both repel the middle ball.

Step 2 — Compute each force magnitude. $F=\frac{(9\times 10^9)(5\times 10^{-6}{)}^2}{(0.03{)}^2}=\frac{(9\times 10^9)(25\times 10^{-12})}{9\times 10^{-4}}=250\text{N}$

Step 3 — Add as vectors. Left ball pushes the middle right ($+250$ N), right ball pushes it left ($-250$ N). Equal and opposite on one line: $F_{\text{net}}=250-250=0\text{N}$

The middle ball feels zero net force — it is in (unstable) equilibrium by symmetry. Each 250 N push is real and large; they merely cancel.

Active Recall

Connections

• Coulomb's Law — the single-pair force each example starts from.
• Vector Addition and Resolution — the resolve-then-recombine engine used in Cases D–F.
• Electric Field — same superposition, forces per unit charge.
• Continuous Charge Distributions — Case G's $r\to \infty$ smallness lets the sum become an integral.
• Electric Potential — the scalar cousin (add numbers, no arrows).
• Principle of Linear Superposition (Waves) — identical "add contributions" idea elsewhere in physics.

Case Map