Electric field of point charge, dipole, ring, disk, line charge (Gauss's law): Visual walkthrough

1.8.5 · D2 · Physics › Electromagnetism › Electric field of point charge, dipole, ring, disk, line cha

Prerequisites jinpe hum rely karenge (waise yahan bhi build kiye gaye hain): Coulomb's law ek point charge ke liye, Flux and field lines "arrows pointing away" ki picture ke liye, aur parent ka superposition ka rule.

Step 1 — Woh ek cheez draw karo jis par hum already trust karte hain: ek single point charge

KYA HAI. Ring se pehle, ek akela positive charge ka speck space mein akele baitha hua dekho. Yeh ek tiny test speck ko apne se door dhakelta hai. Is dhakele ki ek strength (kitna zor) aur ek direction (seedha door) hoti hai. Hum isse ek arrow ki tarah draw karte hain.

KYUN. Har complicated field inhi single-speck arrows ka ek sum hai. Agar hum ek arrow sahi se draw nahi kar sakte, toh hazaar arrows add nahi kar sakte. Isliye hum pehle one-speck rule ko pin down karte hain.

PICTURE. Figure mein, purple dot hamara charge $q$ hai. Coral arrow woh field hai jo yeh ek point par banata hai jo $r$ distance par hai.

Figure — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

Woh word "direction" hi pura game hai. Isse nazar mein rakho.

Step 2 — Stage set karo: ek ring, uske axis par ek point, ek chosen piece

KYA HAI. Ab bahut saara charge ek circular hoop mein bend karo jiska radius $R$ hai aur total charge $Q$ hai. Axis par — woh seedhi line jo center se hoke ring ke plane ke perpendicular jaati hai — center se $x$ distance par ek point $P$ par khade ho jao. Hoop ka ek tiny sa piece chuno aur uske charge ko $d q$ bolo.

KYUN. Axis ek jaadui vantage point hai: wahan se ring ka har piece symmetrically placed dikhta hai. Yahi symmetry hame ek monstrous integral se bachayegi. Hum ek piece $d q$ choose karte hain kyunki Step 1 pehle se uska field bata deta hai — bas itne sare pieces ko add karna hai.

PICTURE. Mint circle ring hai. Purple line axis hai. Chosen speck $d q$ rim par baitha hai; dashed slate line woh straight-line distance $r$ hai us speck se $P$ tak.

Step 3 — Har speck ek slanted arrow fire karta hai. Ise do honest pieces mein split karo

KYA HAI. Step 1 se, hamara speck $d q$ $P$ par $d E = \frac{k d q}{r ^{2}} = \frac{k d q}{R ^{2} + x ^{2}}$ magnitude ka field banata hai, woh dashed line ke saath, slant karte hue $P$ ki taraf. Ek slanted arrow add karna awkward hai. Toh hum isse do perpendicular pieces mein split karte hain: ek axis ke saath ( $d E_{x}$ ) aur ek axis ke perpendicular ( $d E_{⊥}$ ), sideways point karte hue.

KYUN. Arrows add karna tab hi aasaan hota hai jab woh fixed directions mein point karein. Har slanted arrow ko "axial part" + "sideways part" mein split karne se hum saare axial parts ko alag aur saare sideways parts ko alag add kar sakte hain. Yeh har superposition problem ki core move hai.

PICTURE. Coral arrow poora slanted field $d E$ hai. Lavender arrow uska axial share $d E_{x}$ hai; butter arrow uska sideways share $d E_{⊥}$ hai. Dashed line $r$ aur axis ke beech angle $θ$ split ko control karta hai.

Step 4 — Sideways arrows cancel ho jaate hain. Dekho unhe pairs mein khatam hote

KYA HAI. Hamare chosen speck ke liye, ring par diametrically opposite ek speck hai. Uska axial arrow same direction mein point karta hai (forward, $P$ ki side ki taraf), lekin uska sideways arrow bilkul opposite sideways direction mein point karta hai. Do sideways arrows equal aur opposite hain — inka sum zero ho jaata hai.

KYUN. Isliye axis magic hai. Har speck ko uske ring ke paar mirror ke saath pair karo; har pair ke sideways contributions annihilate ho jaate hain. Puri ring ko pair karne ke baad, ek bhi scrap of sideways field nahi bachta. Sirf axial parts rehte hain — aur woh sab same direction mein point karte hain, toh reinforce karte hain. Humne symmetry use karke half problem delete kar di.

PICTURE. Do opposite specks (mint dots), unke do slanted arrows, aur butter sideways components left/right point karte hue — clearly equal aur opposite — jabki lavender axial components dono right point karte hain.

Recall Sideways field kyun vanish ho jaata hai?

Kyunki har speck ka ring ke paar ek partner hota hai jiska sideways push exactly oppose karta hai ::: unke perpendicular components pairs mein cancel ho jaate hain, sirf axial component bachta hai.

Step 5 — Saare axial parts add karo. Woh lucky constant jo bahar nikal aata hai

KYA HAI. Sirf axial component bachta hai, toh total field hai $E_{x} = \int d E_{x} = \int d E cos θ = \int \frac{k d q}{R ^{2} + x ^{2}} \cdot \frac{x}{R ^{2} + x ^{2}} .$ Ab miracle: $R$ , $x$ , aur isliye poora factor $\frac{k x}{( R ^{2} + x ^{2} ) ^{3/2}}$ har speck ke liye same hai — yeh is baat par depend nahi karta ki humne kaunsa speck choose kiya. Constants integral se bahar nikal jaate hain: $E_{x} = \frac{k x}{( R ^{2} + x ^{2} ) ^{3/2}} \int d q .$

KYUN. Ek integral $\int (stuff) d q$ tab hi trivial banta hai jab "stuff" constant ho. Yahan hai, Steps 2–3 ke equal-distance aur equal-angle facts ki wajah se. Bacha hua $\int d q$ sirf matlab hai "charge ka har pinch add karo," jo ki poora charge $Q$ hai.

PICTURE. Bar-chart-style stack jo bahut sare tiny equal $d q$ pieces dikhata hai jo $Q$ mein sum ho rahe hain, har ek identical geometric factor se multiply hua — visually, ek shared factor times badhta hua pile.

Step 6 — Edge case A: bilkul center par khade ho ( $x = 0$ )

KYA HAI. $P$ ko ring ke center par rakho: $x = 0$ . Formula deta hai $E_{r in g} = \frac{k Q \cdot 0}{( R ^{2} + 0 ) ^{3/2}} = 0.$

KYUN. Center par tum sab charge se equidistant ho aur har speck ka arrow ring ke plane mein radially inward-ya-outward point karta hai; woh sab directions mein cancel ho jaate hain, sirf sideways nahi. Numerator mein $x$ exactly yahi encode karta hai: koi axial preference nahi matlab koi field nahi.

PICTURE. $P$ center par with arrows chaaro taraf se symmetrically outward point karte hue — ek starburst jo kuch bhi sum nahi karta.

Step 7 — Edge case B: bahut door khade ho ( $x ≫ R$ )

KYA HAI. Jab $x$ , $R$ ke comparison mein huge ho, toh bracket ke andar $R^{2}$ negligible ho jaata hai, toh $(R^{2} + x^{2})^{3/2} \to x^{3}$ : $E_{r in g} \to \frac{k Q x}{x ^{3}} = \frac{k Q}{x ^{2}} .$

KYUN. Itni door se, ring ki shape matter karna band kar deti hai — yeh ek single dot ki tarah dikhti hai jo charge $Q$ hold kar rahi hai. Point-charge law $k Q / x^{2}$ recover karna woh sanity check hai jo confirm karta hai ki hamari poori derivation honest hai. Koi bhi correct extended-charge formula zaroor large distance par point charge reduce hona chahiye.

PICTURE. Ring ek dot mein shrink hoti hui jab $P$ door jaata hai, uska field arrow Step 1 ke plain Coulomb arrow se match karta hua.

Step 8 — Edge case C: field sabse strong kahan hai? ( $x = \pm R / 2$ )

KYA HAI. Field center par zero hai aur bahut door bhi zero hai, toh beeche mein kahin peak hogi. $\frac{d E _{r in g}}{d x} = 0$ set karne se peak milti hai $x = \pm \frac{R}{2} .$

KYUN. Hum derivative use karte hain kyunki "maximum" matlab hai " $E$ versus $x$ ka slope momentarily flat hai." Do zeros ke beech, ek smooth positive curve ka ek highest point zaroor hoga, aur calculus use locate karta hai. Do signs $\pm$ keh rahe hain ki ring ka field ring ke plane ke dono taraf mirror-symmetric hai.

PICTURE. Poora curve $E$ vs $x$ : center par $0$ se uthta hai, $x = R / 2$ par peak karta hai, phir $1/ x^{2}$ ki tarah decay karta hai — peak marked ke saath.

Recall Kaun si axial distance par ring ka field maximal hai?

$d E / d x = 0$ solve karo ::: $x = \pm R / 2$ par, jahan field center-zero aur far-field decay ke beech apni peak reach karta hai.

Ek-picture summary

Har arrow slant karta hai; har ek ko axial (bachta hai) aur sideways (pairs mein cancel hota hai) mein split karo; saare survivors ek geometric factor share karte hain, toh woh bahar nikal jaate hain aur total charge $Q$ se multiply ho jaate hain. Result zero se uthta hai, $R / 2$ par peak karta hai, aur ek point charge mein fade ho jaata hai.

Recall Feynman retelling — plain words mein bolo

Imagine karo ek hula-hoop of charge aur tum us pole par float kar rahe ho jo iske middle se guzarti hai. Hoop par har chhota bead tumhare test charge ko ek slanted line ke saath tumhari taraf dhakelta hai. Har dhakele ko "forward-along-the-pole" aur "sideways" mein split karo. Ab yeh trick hai: har bead ka ek twin hai hoop ke doosri side par, aur unke sideways dhakele equal aur opposite hain — woh ek doosre ko wipe out kar dete hain. Sirf forward dhakele bachte hain, aur kyunki har bead tumse same distance aur same slant par hai, in sab ko same number se multiply kiya jaata hai. Toh tum bas saara charge ek bade $Q$ mein add kar lo aur ek baar multiply karo. Isse $E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}$ milta hai. Dead center par khado aur forward part zero hai (koi forward direction special nahi hai) toh field vanish ho jaata hai. Bahut door float karo aur hoop ek dot ki tarah dikhta hai, toh plain old $k Q / x^{2}$ wapas mil jaata hai. Beeche mein, field $x = R / 2$ par strongest hai. Yahi poori kahani hai — split, cancel, add, check the ends.

Is exact machinery use karke related builds: disk rings stack karta hai, dipole do point charges pair karta hai, aur jahan symmetry aur bhi zyada kind hai hum Gauss's law ke saath integration skip kar dete hain. Energy view ke liye, Electric potential dekho.