Exercises — Superposition principle for forces

Kya galat hai: $14\text{mN}$ ek scalar sum hai. Lekin east aur north perpendicular directions hain — tum directly unke sizes nahi jod sakte, jaise "3 kadam aage" aur "4 kadam baayein" ko jodkar "7 kadam" nahi bol sakte.

Kya karna chahiye — figure mein right triangle dekho. Dono arrows legs banate hain; asli total diagonal hai (hypotenuse): $|\vec{F}|=\sqrt{6^2+8^2}\text{mN}=\sqrt{36+64}=\sqrt{100}=10\text{mN}$

Nahi. Superposition kehta hai har pair-force ko calculate karo jaise baaki exist hi nahi karte (dekho Coulomb's Law). Force ${\vec{F}}_{12}$ sirf $q_1$, $q_2$, aur unke separation par depend karta hai — isme koi $q_3$ term nahi hai.

Jo cheez badlegi wo hai $q_1$ par net force: ab use ${\vec{F}}_{13}$ bhi feel hoga, ek alag joda gaya arrow. Purana ${\vec{F}}_{12}$ bilkul intact rehta hai; uske upar ek naya arrow stack ho jaata hai.

$q_1$ se (distance $1\text{m}$, same signs → repel → $q_0$ ko $q_1$ se door push karta hai, yaani $+x$): $F_{01}=\frac{k(2\times 10^{-6})(1\times 10^{-6})}{1^2}=9\times 10^9\cdot 2\times 10^{-12}=1.8\times 10^{-2}\text{N}(+x)$ $q_2$ se (distance $2\text{m}$, opposite signs → attract → $q_0$ ko $q_2$ ki taraf $x=3$ par kheenchta hai, yaani $+x$): $F_{02}=\frac{k(3\times 10^{-6})(1\times 10^{-6})}{2^2}=\frac{9\times 10^9\cdot 3\times 10^{-12}}{4}=6.75\times 10^{-3}\text{N}(+x)$ Dono $+x$ direction mein hain, toh ye ek aisa case hai jahan hum directly sizes jod sakte hain (wo collinear hain): $F_{\text{net}}=1.8\times 10^{-2}+6.75\times 10^{-3}=2.475\times 10^{-2}\text{N}=24.75\text{mN}\text{along }+x$

Beech mein kyun? Dono source charges positive hain, toh ek positive test charge ko $q_1$ right ki taraf push karega aur $q_2$ left ki taraf. Sirf unke beech hi ye dono pushes oppose aur cancel ho sakti hain. (Bahar, dono ek hi direction mein point karengi — koi cancellation nahi.)

Maano test charge $q_1$ se $x$ distance par hai, toh $q_2$ se $(3-x)$ door hai. Cancellation ka matlab hai equal magnitudes (common factors $k$ aur test charge cancel ho jaate hain): $\frac{q_1}{x^2}=\frac{q_2}{(3-x{)}^2}\Rightarrow \frac{4}{x^2}=\frac{1}{(3-x{)}^2}$ Dono sides ka square root lo (dono distances positive hain, toh $+$ root rakho): $\frac{2}{x}=\frac{1}{3-x}\Rightarrow 2(3-x)=x\Rightarrow 6=3x\Rightarrow x=2\text{m}$ Answer: $q_1$ se $2\text{m}$ door (yaani $q_2$ se $1\text{m}$). Ye point kamzor charge ke zyada paas hai — sensible hai, kyunki use chhote charge ke paas rehna hoga taaki uski push dur waale bade wale jaisi strong ho.

Figure dekho — do source charges ek square ke do corners par hain, target teesre corner par.

$q_1$ se origin par: separation vector $(2,0)-(0,0)=(2,0)$, distance $2\text{m}$, unit vector $(1,0)$. Repulsion → $+x$ ke along: $F_{01}=\frac{k(2)(1)\times 10^{-12}}{2^2}=\frac{9\times 10^9\cdot 2\times 10^{-12}}{4}=4.5\times 10^{-3}\text{N}\Rightarrow (4.5\times 10^{-3},0)$ $q_2$ se $(0,2)$ par: separation vector $(2,0)-(0,2)=(2,-2)$, distance $\sqrt{2^2+2^2}=2\sqrt{2}\approx 2.828\text{m}$. Unit vector $=(2,-2)/(2\sqrt{2})=(0.707,-0.707)$ — ye source $q_2$ se target $q_0$ ki taraf point karta hai, toh repulsion $q_0$ ko isi direction mein push karta hai (neeche-aur-daayein). $F_{02}=\frac{k(2)(1)\times 10^{-12}}{(2\sqrt{2}{)}^2}=\frac{18\times 10^{-3}}{8}=2.25\times 10^{-3}\text{N}$ ${\vec{F}}_{02}=2.25\times 10^{-3}(0.707,-0.707)=(1.591\times 10^{-3},-1.591\times 10^{-3})$ Components jodo (isliye resolve karte hain — $45^{\circ }$ par angle banaate arrows ki sizes nahi jod sakte): $F_x=4.5\times 10^{-3}+1.591\times 10^{-3}=6.091\times 10^{-3}$ $F_y=0-1.591\times 10^{-3}=-1.591\times 10^{-3}$ Dobara combine karo: $|\vec{F}|=\sqrt{6.091^2+1.591^2}\times 10^{-3}=6.29\times 10^{-3}\text{N}=6.29\text{mN}$ $\varphi =\arctan \frac{-1.591}{6.091}=-14.65^{\circ }$ Answer: $\approx 6.29\text{mN}$, $+x$ axis se $14.7^{\circ }$ neeche tilt — ye downward tilt $q_2$ ka component hai. (Machinery ke liye dekho Vector Addition and Resolution.)

Figure dekho — upar waale charge ko pick karo. Dono neeche waale charges use repel karte hain; symmetry se unke horizontal parts cancel ho jaate hain aur vertical parts add ho jaate hain.

Har pair force ki magnitude hai $F=\frac{k{q}^2}{a^2}$ (saare sides $a$ ke barabar).

Directions: dono neeche waale charges upar waale charge ko side ke along outward push karte hain. Har side vertical ke saath $30^{\circ }$ banati hai (kyunki interior angles $60^{\circ }$ hain; dono arrows vertical ke baare mein symmetric hain, $60^{\circ }$ apex angle ko $30^{\circ }$ $30^{\circ }$ mein todke).

Horizontal parts: equal aur opposite → cancel. ✅ (Isliye components choose karte hain — symmetry ek axis ko khatam kar deti hai.)

Vertical parts add hote hain: har ek $F\cos 30^{\circ }$ contribute karta hai, aur do hain: $F_{\text{net}}=2F\cos 30^{\circ }=2\cdot \frac{k{q}^2}{a^2}\cdot \frac{\sqrt{3}}{2}=\sqrt{3}\frac{k{q}^2}{a^2}$ Answer: $F_{\text{net}}=\sqrt{3}\frac{k{q}^2}{a^2}$, seedha triangle ke centre se door point karta hua.

Set-up. Centre se har corner tak ki distance half diagonal hai: $r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}},r^2=\frac{a^2}{2}$ Har corner $+Q$ par ek magnitude ki force produce karta hai: $F=\frac{kqQ}{r^2}=\frac{kqQ}{a^2/2}=\frac{2kqQ}{a^2}$ Trick — imagine karo saare char $+q$ hain. Tab char repulsive forces char corners ki taraf outward point karengi, perfectly balanced opposite pairs mein → net zero.

Hamara case hai "saare $+q$" plus ek correction: ek corner $-q$ hai $+q$ ki jagah. $+q\to -q$ karna ek corner par ek extra $-2q$ superpose karne ke barabar hai (kyunki $+q+(-2q)=-q$). Superposition se hum inhe alag treat kar sakte hain:

• "Saare $+q$" arrangement → zero.
• Ek corner par extra $-2q$ → $+Q$ ko us corner ki taraf attract karta hai magnitude ke saath: $F_{\text{net}}=\frac{k(2q)Q}{r^2}=\frac{k(2q)Q}{a^2/2}=\frac{4kqQ}{a^2}$ Answer: $F_{\text{net}}=\frac{4kqQ}{a^2}$, centre se negative corner ki taraf directed. Superposition ne hume ek mushkil 4-body sum ko "zero + ek easy term" mein tod diya.

Zero tilt ke liye condition: total $y$-component zero hona chahiye, $F_y=0$.

$q_1$ x-axis par target ke saath hai → koi $y$-component contribute nahi karta. Toh saari $y$-force $q_2$ se aati hai, aur ye sirf tab zero ho sakti hai agar $q_2=0$ ho... jab tak hum koi fourth charge allow na karein — lekin sirf in do ke saath, $q_2$ ki force mein hamesha ek downward $y$-part $(-)$ hoga jab tak $q_2\ne 0$.

${\vec{F}}_{02}=\frac{k{q}_0{q}_2}{(L^2+d^2)}\cdot \frac{(L,-d)}{\sqrt{L^2+d^2}},F_{02,y}=-\frac{k{q}_0{q}_{2}d}{(L^2+d^2{)}^{3/2}}$ Ye zero hoga sirf agar $q_2=0$ ya $d=0$ ho. Toh: Answer: koi esa nonzero ratio nahi hai jo tilt hataa sake — $q_1$ on-axis hone se wo kabhi vertically contribute nahi karta, isliye koi bhi off-axis $q_2$ ek residual tilt chhodta hai. Sirf do escape hain: $q_2=0$ (koi second charge nahi) ya $d=0$ (dono sources axis par, collinear ban jaata hai). Ye ek degenerate-case check hai: seedhi soch "ratio tune karo" fail hoti hai, aur kyun fail hoti hai ye samajhna hi point hai.

Integral kyun? Rod ek charge nahi balki infinitely many infinitesimal charges ka collection hai. Superposition kehta hai total force har tiny piece ki force ka sum hai — aur ek continuum par sum hi integral hota hai (ye exactly wahi move hai jo Continuous Charge Distributions mein hai).

Step 1 — rod ko kaato. Uniform charge ka matlab hai linear density $\lambda =Q/L$ (charge per metre). $q_0$ se $x$ distance par $dx$ width ka ek slice contain karta hai: $dq=\lambda dx=\frac{Q}{L}dx$ Coordinates set karo: $q_0$ origin par, rod $x=a$ (near end) se $x=a+L$ (far end) tak jaati hai.

Step 2 — ek slice ki force. Har slice $q_0$ ke saath ek hi line par hai, isliye saari forces collinear hain (koi components ki zarurat nahi — geometry ki ek meherbani): $dF=\frac{k{q}_{0}dq}{x^2}=\frac{k{q}_{0}Q}{L}\cdot \frac{dx}{x^2}$ Step 3 — superpose (integrate) near end se far end tak: $F=\frac{k{q}_{0}Q}{L}{\int }_a^{a+L}\frac{dx}{x^2}=\frac{k{q}_{0}Q}{L}{\left[-\frac{1}{x}\right]}_a^{a+L}=\frac{k{q}_{0}Q}{L}\left(\frac{1}{a}-\frac{1}{a+L}\right)$ Fractions combine karo: $\frac{1}{a}-\frac{1}{a+L}=\frac{(a+L)-a}{a(a+L)}=\frac{L}{a(a+L)}$ $\boxed{F=\frac{k{q}_{0}Q}{L}\cdot \frac{L}{a(a+L)}=\frac{k{q}_{0}Q}{a(a+L)}}$ Sanity check (limiting behaviour): agar rod bahut chhoti hai, $L\to 0$, toh $a+L\to a$ aur $F\to \frac{k{q}_{0}Q}{a^2}$ — exactly ek point charge $Q$ at distance $a$. ✅ Formula sabse simple case mein collapse karta hai, jaisa kisi bhi sahi derivation ko karna chahiye.

$F=\frac{k{q}_{0}Q}{a(a+L)}=\frac{(9\times 10^9)(2\times 10^{-6})(5\times 10^{-6})}{0.5(0.5+1)}$ Numerator: $9\times 10^9\cdot 2\times 10^{-6}\cdot 5\times 10^{-6}=9\times 10^9\cdot 10^{-11}=9\times 10^{-2}$. Denominator: $0.5\times 1.5=0.75$. $F=\frac{9\times 10^{-2}}{0.75}=0.12\text{N}$ Answer: $F=0.12\text{N}$, rod se door directed (like charges repel).

• L1 ::: vector vs scalar ko pehchanna, aur pairs ki non-interference.
• L2 ::: collinear cases ke liye Coulomb's law plug in karna; null-point equal-magnitude trick.
• L3 ::: components mein resolve karna aur signs se quadrant padhna.
• L4 ::: ek mushkil configuration ko superposition se todna ("zero + correction").
• L5 ::: continuous distribution ke liye discrete sum ko integral mein baadalna.