Worked examples — Superposition principle for forces

1.8.3 · D3 · Physics › Electromagnetism › Superposition principle for forces

Shuru karne se pehle, ek anchor picture. Yahan "force" ek arrow hai: uski length bataati hai push kitni strong hai, aur uski direction bataati hai push kis taraf hai. Arrows add karne ka matlab hai unhe tip-to-tail lagana — kabhi unki lengths mat joḍo jab tak woh ek hi line par na hon.

Poori discussion mein, $k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^9{\text{N⋅m}}^2/{\text{C}}^2$ Coulomb constant hai (from Coulomb's Law), aur $\mu \text{C}$ matlab $10^{-6}$ coulomb.

Scenario matrix

Is topic ka har problem in cells mein se ek hai (ya inka blend). Neeche ke worked examples mein har ek apna cell tag karta hai.

Cross-cutting checks jo har example ko pass karni chahiye:

• Signs: like charges repel (arrow source se door point karta hai), unlike attract (arrow source ki taraf point karta hai).
• Units: har $F$ newtons (N) mein aata hai.
• Direction: non-collinear arrows ki magnitudes kabhi mat jodo.

Case A — Collinear, forces add

Forecast: pehle guess karo — kya dono forces ek doosre ki help karenge ya fight karenge? (Ek repel karta hai, ek attract... lekin kis side se?)

Step 1 — $q_1$ se force. Yeh step kyun? Superposition kehta hai $q_1$ ko akele treat karo. Distance $r_1=|2-0|=2\text{m}$. $F_{01}=\frac{k|q_1{q}_0|}{r_1^2}=\frac{(9\times 10^9)(3\times 10^{-6})(1\times 10^{-6})}{2^2}=6.75\times 10^{-3}\text{N}$ $q_1$ aur $q_0$ dono positive hain ⇒ repel. $q_1$ left pe baitha hai, isliye $q_0$ ko +x ki taraf dhakelta hai.

Step 2 — $q_2$ se force. Yeh step kyun? Ab $q_2$ ko akele treat karo. $r_2=|4-2|=2\text{m}$. $F_{02}=\frac{(9\times 10^9)(2\times 10^{-6})(1\times 10^{-6})}{2^2}=4.5\times 10^{-3}\text{N}$ $q_2$ negative hai, $q_0$ positive ⇒ attract. $q_2$ right pe baitha hai, $q_0$ ko +x ki taraf khiinchta hai.

Step 3 — Add karo (collinear hain aur same direction mein hain). $F_{\text{net}}=6.75+4.5=11.25\times 10^{-3}\text{N}=11.25\text{mN}(+x)$

Case B — Collinear, forces oppose (partial cancel)

Forecast: dono neighbours positive hain aur $q_0$ ko door dhakelte hain — lekin opposite sides se. Kaun sa side jeetega?

Step 1 — $q_1$ se (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 — $q_2$ se (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 karo (opposite directions on one line). Subtract kyun? Ek hi line par opposite arrows partly cancel ho jaate hain; bade wale ka sign rakho. $F_{\text{net}}=7.2\times 10^{-2}-0.45\times 10^{-2}=6.75\times 10^{-2}\text{N}(+x)$

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

Forecast: null point chhote charge ke paas hoga ya bade ke paas? Solve karne se pehle guess karo.

Step 1 — Dono magnitudes equal set karo. Yeh step kyun? "Zero net force" do like charges ke beech matlab hai dono pushes (yahan opposite directions) ki equal size honi chahiye. Point ko $x$ par maano, $0<x<1$. Distances: $q_1$ se $x$, $q_2$ se $(1-x)$. $\frac{k{q}_1{q}_0}{x^2}=\frac{k{q}_2{q}_0}{(1-x{)}^2}$

Step 2 — $k{q}_0$ cancel karo aur square root lo. Cancel kyun? Test charge aur $k$ dono sides par identically aate hain. $\frac{9}{x^2}=\frac{4}{(1-x{)}^2}\Rightarrow \frac{3}{x}=\frac{2}{1-x}$

Step 3 — Linear equation solve karo. $3(1-x)=2x\Rightarrow 3=5x\Rightarrow x=0.6\text{m}$

Null point $x=0.6\text{m}$ par hai — chhote charge $q_2$ ke zyada paas, kyunki ek kamzor charge ko itne strong push feel karne ke liye paas aana padta hai jitna door wala bada charge deta hai.

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

Forecast: kya net force straight $+x$ ki taraf point karega, ya usse neeche jhukka hoga?

Step 1 — $q_1$ se force. $r_1=4$ 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 — $q_2$ se force — pehle direction find karo. Direction pehle kyun? 2-D mein arrow kisi axis ke saath nahi hota; hamen uska unit vector chahiye. Source $q_2$ se target $q_0$ tak separation hai $(4,0)-(0,3)=(4,-3)$, length $\sqrt{4^2+3^2}=5$. Unit vector $=(0.8,-0.6)$. Repulsion isi direction mein push karta hai ($q_2$ se door). $F_{02}=\frac{(9\times 10^9)(10^{-12})}{25}=3.6\times 10^{-4}\text{N}$ ${\vec{F}}_{02}=3.6\times 10^{-4}(0.8,-0.6)=(2.88\times 10^{-4},-2.16\times 10^{-4})$

Step 3 — Components add karo (isliye resolve karte hain). $F_x=(5.625+2.88)\times 10^{-4}=8.505\times 10^{-4},F_y=-2.16\times 10^{-4}$

Step 4 — Magnitude aur angle mein recombine karo.  ​ aur arctan kyun? Arrow ki magnitude uske legs se Pythagoras se milti hai; tilt woh angle hai jiska tangent (opposite $F_y$)/(adjacent $F_x$) hota hai — dekho 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 aur repulsion

Forecast: ek neighbour push karta hai, ek pull — kya arrows reinforce karte hain ya fight karte hain?

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

Step 2 — $q_2$ se (unlike ⇒ attract, isliye arrow $q_2$ ki TARAF point karta hai). Taraf kyun? Attraction $q_0$ ko source ki taraf kheenchta hai. 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 — Components add karo. $F_x=0+4.32\times 10^{-4}=4.32\times 10^{-4}$ $F_y=11.25\times 10^{-4}-5.76\times 10^{-4}=5.49\times 10^{-4}$

Step 4 — Magnitude aur 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: symmetry se, net force kis direction mein honi chahiye? Compute karne se pehle guess karo.

Step 1 — Har pair force ki magnitude (dono equal). Equal kyun? 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 — Geometry set up karo. Target ko $(2,0)$ par rakho, baaki do ko $(0,0)$ aur $(1,\sqrt{3})$ par. Dono forces repulsive hain, target ko har source se door push karti hain.

• $(0,0)$ se: direction $(2,0)-(0,0)=(1,0)$ ⇒ push $+x$: $(9\times 10^{-3},0)$.
• $(1,\sqrt{3})$ se: 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 — Components add karo. $F_x=(9+4.5)\times 10^{-3}=13.5\times 10^{-3},F_y=-7.794\times 10^{-3}$

Step 4 — Magnitude. $|\vec{F}|=\sqrt{13.5^2+7.794^2}\times 10^{-3}=1.559\times 10^{-2}\text{N}$ Symmetry check: net triangle ke centre se bahar bisector ke saath point karta hai — do equal forces $60^{\circ }$ apart milke $|\vec{F}|=\sqrt{3}F=1.732\times 9\times 10^{-3}=1.559\times 10^{-2}\text{N}$ dete hain. ✓

Case G — Limiting / degenerate inputs

Forecast: agar source charge vanish ho jaaye, uska force term bhi vanish hona chahiye — kya formula aisa karta hai?

Step 1 — $q_2=0$ set karo. Yeh kyun check karo? Ek robust formula gracefully degrade honi chahiye. $F_{02}=\frac{k{q}_1\cdot 0}{r^2}=0$. Sirf $q_1$ bachta hai: ${\vec{F}}_{\text{net}}=(5.625\times 10^{-4},0)\text{N}$ Net pure $+x$ pe snap back karta hai — exactly ek-charge ka answer. Zero term ke saath Superposition = us term ko drop karo.

Step 2 — $r\to \infty$ karo. ${\lim }_{r\to \infty }\frac{k{q}_1{q}_0}{r^2}=0$ Kyun matter karta hai: bahut door ke charges negligibly contribute karte hain. Isliye Continuous Charge Distributions mein charge ke door-door bits safely small hote hain, aur sum ek integral $\int d\vec{F}$ ban sakta hai.

Step 3 — Degenerate distance $r\to 0$. Formula blow up karta hai ($\propto 1/r^2\to \infty$). Physically do point charges kabhi truly touch nahi karte; model break ho jaata hai aur hum us se pehle ruk jaate hain.

Case H — Word problem (real-world framing)

Forecast: middle ball do equal opposite pushes ke beech hai — calculate karne se pehle, net kya hoga?

Step 1 — Words ko numbers mein translate karo. Yeh step kyun? "Midway" ⇒ har fixed ball middle se $3\text{cm}=0.03\text{m}$ door hai. Dono $+5\mu \text{C}$ hain, dono middle ball ko repel karte hain.

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

Step 3 — Vectors ki tarah add karo. Left ball middle ko right push karta hai ($+250$ N), right ball use left push karta hai ($-250$ N). Ek line par equal aur opposite: $F_{\text{net}}=250-250=0\text{N}$

Middle ball ko zero net force feel hoti hai — symmetry se woh (unstable) equilibrium mein hai. Har 250 N push real aur badi hai; woh sirf cancel ho jaati hain.

Active Recall

Connections

• Coulomb's Law — single-pair force jisse har example shuru hota hai.
• Vector Addition and Resolution — Cases D–F mein use hone wala resolve-then-recombine engine.
• Electric Field — same superposition, forces per unit charge.
• Continuous Charge Distributions — Case G ki $r\to \infty$ smallness sum ko integral banne deti hai.
• Electric Potential — scalar cousin (numbers jodo, arrows nahi).
• Principle of Linear Superposition (Waves) — same "contributions add karo" idea physics mein kahin aur.

Case Map