Electric field — definition, field lines, superposition: Visual walkthrough

1.8.4 · D2 · Physics › Electromagnetism › Electric field — definition, field lines, superposition

Step 1 — Stage draw karo: do charges, measure karne ke liye ek point

KYA. Ek positive charge $+ q$ ko flat ground line par centre ke daayein $+ a$ position par rakho, aur ek identical $+ q$ ko baayein $- a$ par. Hum ek point $P$ par electric field jaanna chahte hain jo centre ke seedha upar, height $y$ par float kar raha hai.

YEH choices KYUN. Hum ek point vertical centre-line par pick karte hain kyunki yeh setup left-to-right mirror image hai. Symmetry baad mein humara aadha kaam free mein kar degi — lekin tabhi jab hum exactly mirror line par baithe hoon. Agar hum ek taraf hat ke baithe, kuch bhi cancel nahi hota aur algebra blast ho jaata.

PICTURE. Figure dekho. Do red dots charges hain. Green dot $P$ hai. Letter $a$ centre se har charge tak ka horizontal distance hai; $y$ $P$ ki seedhi-upar height hai.

Figure — Electric field — definition, field lines, superposition

Step 2 — Har charge $P$ se kitna door hai? (Pythagoras)

KYA. Ek charge (maano daaya wala) se $P$ tak ki straight-line distance $r$ nikalo. Woh charge ground-level par $a$ sideways distance par baitha hai, aur $P$ $y$ upar hai. Yeh dono legs right angle par milte hain, toh diagonal hai $r = right triangle ka diagonal a^{2} + y^{2} .$

Squares ke sum ka square root KYUN. Horizontal leg ( $a$ ) aur vertical leg ( $y$ ) ek right triangle ki do chhoti sides banate hain; $r$ uski hypotenuse hai. Pythagoras ( $leg^{2} + leg^{2} = hyp^{2}$ ) wahi tool hai jo do perpendicular distances ko ek straight distance mein convert karta hai — exactly yahi sawaal hum pooch rahe hain.

PICTURE. Blue triangle: bottom leg $a$ , vertical leg $y$ , slanted hypotenuse $r$ right charge se $P$ tak jaati hui. Left-right symmetry se left charge ka ek identical triangle hai, mirror-flipped — toh dono charges $P$ se same distance $r$ par hain.

Step 3 — EK charge ke field ki strength (Coulomb's Law)

KYA. Coulomb's Law se, distance $r$ par ek single point charge ka field magnitude hota hai $E_{1} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}} = k \frac{q}{r ^{2}} = \frac{k q}{a ^{2} + y ^{2}} .$ Isse piece by piece padho:

$k = \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}$ ::: universal strength constant.
$q$ ::: bada charge → zyada strong field (fraction ke upar).
$r^{2} = a^{2} + y^{2}$ ::: zyada door → kamzor field, aur yeh distance ke square ke saath kamzor hota hai (neeche).

Squared KYUN, sirf $r$ nahi? Ek point charge apna field sabhi directions mein equally baahir bhejta hai, ek sphere par phailta hai. Sphere ka surface area $4 π r^{2}$ hota hai. Usi "amount of field" ka bade sphere par phatna matlab strength $\propto 1/ r^{2}$ . Isliye exponent 2 hai, aur yeh seedha geometry se aata hai.

PICTURE. Har red charge se ek arrow $E_{1}$ directly baahir ki taraf point karta hai (dono charges positive hain, toh yeh ek positive test charge ko baahir dhakelte hain). Har arrow apni hypotenuse line par lie karta hai. Dono arrows ki length same hai kyunki dono $r$ equal hain.

Step 4 — Ek tedha arrow actually do arrows hota hai ( $x$ aur $y$ mein split karo)

KYA. Har $E_{1}$ arrow diagonally point karta hai. Hum isse ek horizontal piece $E_{1 x}$ aur ek vertical piece $E_{1 y}$ mein tod dete hain. Usi blue triangle ka use karke, angle $θ$ $P$ par vertical aur hypotenuse ke beech measure hota hai, aur $E_{1 y} = E_{1} cos θ, E_{1 x} = E_{1} sin θ .$

Cosine KYUN yahan? Cosine ek specific sawaal ka jawaab deta hai: "is slanted length mein se kitna us direction mein lie karta hai jis direction ki mujhe parwaah hai (vertical)?" Yeh define hota hai $cos θ = \frac{adjacent}{hypotenuse}$ ke roop mein. Angle $θ$ ke adjacent side (vertical leg) ki length $y$ hai, aur hypotenuse $r$ hai, toh $cos θ = \frac{y}{r} = \frac{y}{a ^{2} + y ^{2}} .$ Hum cosine use karte hain sine nahi kyunki woh piece jo hum ultimately chahte hain (vertical) angle ke saath baithta hai, uske saamne nahi.

PICTURE. Diagonal arrow ko L-shape ke roop mein redraw kiya gaya hai: ek green vertical component upar jaata hua, ek yellow horizontal component sideways jaata hua. Angle $θ$ $P$ par vertical dashed line aur arrow ke beech mark kiya gaya hai.

Step 5 — Do fields add karo (Superposition) aur horizontals ko marte dekho

KYA. Superposition Principle se total field dono ka plain vector sum hai. Pieces alag-alag add karo: $E_{x}^{tot} = E_{1 x}^{(left)} + E_{1 x}^{(right)}, E_{y}^{tot} = E_{1 y}^{(left)} + E_{1 y}^{(right)} .$

Horizontals cancel hote hain. Right charge $P$ ko upar-aur-left ki taraf push karta hai? Nahi — upar-aur-right ki taraf (apne aap se door). Left charge upar-aur-left ki taraf push karta hai. Yeh dono horizontal pieces size mein equal hain aur direction mein opposite hain, toh $E_{x}^{tot} = 0$ .
Verticals add hote hain. Dono charges $P$ ko upar push karte hain (dono $P$ ke neeche hain aur isse upar repel karte hain). Same size, same direction → yeh double ho jaate hain.

$E_{y}^{tot} = 2 E_{1 y} = 2 E_{1} cos θ .$

Cancellation guaranteed KYUN hai (luck nahi). Poori picture $P$ se guzarne waali vertical line ke baare mein ek mirror reflection hai. Reflection left↔right flip karta hai, toh har horizontal component ka sign flip karta hai lekin verticals ko untouched chhod deta hai. Ek quantity jo apne aap ke negative ke equal honi chahiye sirf zero ho sakti hai. Yahi reason hai $E_{x} = 0$ — pure symmetry, koi arithmetic zaroori nahi.

PICTURE. Do diagonal arrows $P$ par tail-to-tail rakhe hue; unke yellow horizontals opposite ways point karte hain aur ek doosre ko erase karte hain (greyed dikhaya gaya), jabki green verticals ek tall arrow mein stack ho jaate hain.

Step 6 — Final formula assemble karo

KYA. $E_{1}$ aur $cos θ$ ko $E_{y}^{tot} = 2 E_{1} cos θ$ mein substitute karo:

=\frac{2kqy}{(a^2+y^2)^{3/2}}.$$ Track karo ki exponent $3/2$ kahan paida hota hai: - ek full power $(a^2+y^2)^{1}$ Coulomb ke $r^2$ se $E_1$ mein, - aadha power $(a^2+y^2)^{1/2}$ $\cos\theta$ ke $\sqrt{\,}$ se, - milake $(a^2+y^2)^{1+1/2}=(a^2+y^2)^{3/2}$. **Woh mixed power KYUN matter karta hai.** Yeh ek pure inverse-square *nahi* hai. Extra half-power slant ko vertical par project karne ki geometry se aati hai — ek signature jo tumhe "symmetric arrangement ke on-axis" wale har problem mein phir milegi. **PICTURE.** Clean result: $P$ par ek single upward green arrow, uski length boxed formula se labelled, neeche do charges. ![[deepdives/dd-physics-1.8.04-d2-s06.png]] > [!formula] Is walkthrough ka central result > $$\boxed{\,E_y=\frac{2kqy}{(a^2+y^2)^{3/2}}\,}\qquad(\text{seedha upar point karta hai, }E_x=0)$$ --- ## Step 7 — Edge aur degenerate cases (reader ko kabhi stranded mat chhodo) **Case A — bahut door, $y\gg a$.** Jab $y$, $a$ ko dwarf kare, root ke andar $a^2$ term $y^2$ ke saamne ek crumb hai; isse drop karo: $a^2+y^2\approx y^2$. Tab $$E_y\approx\frac{2kqy}{(y^2)^{3/2}}=\frac{2kqy}{y^3}=\frac{2kq}{y^2}=\frac{k(2q)}{y^2}.$$ Door se dono charges **ek lump of charge $2q$** mein blur ho jaate hain — ek ordinary inverse-square point-charge field. *Forecast confirmed:* door se, sirf **total** charge matter karta hai, uska spread nahi. Isliye bahut door ke charge clusters [[Continuous Charge Distributions]] ki tarah ek point par collapsed lagte hain. **Case B — bilkul centre par, $y=0$.** $y=0$ plug karo: $$E_y=\frac{2kq\cdot 0}{(a^2)^{3/2}}=0.$$ Do equal positive charges ke exact midpoint par, left charge daayein push karta hai aur right charge baayein equal strength se push karta hai — total field **zero**. Yeh ek *balance point* hai (ek unstable wala: test charge ko sideways nudge karo aur woh bhaag jaata hai). [[Electric Field|field-line]] saddle dekho. **Case C — opposite arrangement ([[Electric Dipole]] ka ek hint).** Agar right charge $-q$ hota $+q$ ki jagah, toh **verticals** cancel ho jaate aur **horizontals** bachte — upar sab kuch ka mirror image. Woh flipped problem dipole ka $1/y^3$ field deta hai. Same triangle, $\sin$ aur $\cos$ ke roles swapped. **PICTURE.** Teen mini-panels: (A) far field arrow shrinking as one point charge, (B) origin par zero arrow with two opposing pushes, (C) flipped dipole case with sideways arrow. ![[deepdives/dd-physics-1.8.04-d2-s07.png]] > [!mistake] "Bas magnitudes add karo: $E=E_1+E_1=2E_1$." > Woh naked "2" ignore karta hai ki arrows **slanted** hain. Real total hai $2E_1\cos\theta$, hamesha $2E_1$ se *chhota* kyunki $\cos\theta\le 1$. Factor $\cos\theta$ horizontal pieces ke khud ko throw away karne ki price hai. --- ## Ek-picture summary Ek canvas par sab kuch: do triangles, do slanted $E_1$ arrows, horizontals ka annihilate hona, verticals ka stack hona, aur final boxed formula jisme har factor colour-tagged hai is jagah ki taraf jahan woh paida hua. ![[deepdives/dd-physics-1.8.04-d2-s08.png]] > [!recall]- Feynman retelling — poora walk plain words mein > Do identical positive balls zameen par baithe hain, ek baayein aur ek daayein ek centre spot ke. Hum ek "field-feeler" seedha centre ke upar hover karte hain. **Step 1–2:** har ball same slanted distance par hai — right triangle draw karo, uski hypotenuse $\sqrt{a^2+y^2}$ hai. **Step 3:** har ball ka push us distance ke *square* ke saath kamzor hota hai (ek sphere par phailta hua). **Step 4:** har push diagonally apni ball se door point karta hai, toh hum isse ek sideways part aur ek up part mein split karte hain; up part poora times $\cos\theta=y/r$ hai. **Step 5:** do sideways parts opposite ways point karte hain aur cancel ho jaate hain (picture ek mirror hai), jabki do up-parts same way point karte hain aur double ho jaate hain. **Step 6:** up-part ko double karne par milta hai $\dfrac{2kqy}{(a^2+y^2)^{3/2}}$ — ajeeb $3/2$ power Coulomb se ek power plus projection se aadha power hai. **Step 7:** door khade ho jaao aur pair ek charge $2q$ jaisa lagta hai; exact middle mein khado aur pushes tie ho jaate hain, toh field zero hai. Ek charge ka sign flip karo aur poori story sideways ho jaati hai — woh dipole hai. > [!recall]- Quick self-test > Centre-line par $E_x=0$ KYUN hai? ::: Mirror symmetry horizontal components flip karti hai; ek quantity jo apne aap ke negative ke equal ho woh zero honi chahiye. > $3/2$ exponent kahan se aata hai? ::: $(a^2+y^2)$ ka $1$ power Coulomb ke $r^2$ se plus $\cos\theta$ ke square root se $1/2$ power. > $y\gg a$ ke liye $E_y$ kya ban jaata hai? ::: $k(2q)/y^2$ — combined charge ka ek single point charge. > Exactly $y=0$ par $E_y$ kya hai? ::: Zero — do equal pushes midpoint par cancel ho jaate hain.