Potential of point charge, potential from field and vice versa: Visual walkthrough

1.8.9 · D2 · Physics › Electromagnetism › Potential of point charge, potential from field and vice ver

Step 1 — Khiladi: ek charge, ek test charge, ek doori

KYA. Hamare paas ek fixed charge $q$ hai jo ek jagah baitha hai. Uske aas-paas ek invisible influence rehti hai. Us influence ko feel karne ke liye hum ek doosra, bahut chota sa positive charge — jise $q_{0}$ kehte hain — lekar kisi jagah jaate hain jo $q$ se $r$ doori par hai.

KYUN. "Potential" define hota hai ek baahri haath ko $q_{0}$ ko idhar-udhar move karne mein jo kaam karna padta hai, us se. Toh $q_{0}$ picture mein hona zaroori hai, aur yeh naapne ka tariqa bhi ki hum kitne door hain: woh doori hai $r$ (metres mein).

DIKHTA KAISA HAI. Figure dekhiye. Purple dot source charge $q$ hai. Coral dot hamara test charge $q_{0}$ hai. Unke beech ki dashed line $r$ hai — sirf yehi ek geometric quantity hai jis par answer depend karega, kyunki yahan sab kuch $q$ ke aas-paas round (symmetric) hai.

Step 2 — Charge per unit force: field $E$

KYA. Doori $r$ par, source charge hamare test charge ko ek force se dhakelta hai. Us force ko $q_{0}$ se divide karo toh milta hai force per unit charge — electric field $E$ . Coulomb's Law se yeh hai:

$field E = strength constant k \frac{1}{4 π ε _{0}} shrinks like 1/ r^{2} \frac{q}{r ^{2}} points straight out \overset{r}{^}$

Left se right padhein: $E$ woh arrow hai jo chahiye; $\frac{1}{4 π ε _{0}} = k \approx 9 \times 1 0^{9}$ sirf ek units constant hai; $\frac{q}{r ^{2}}$ kehta hai ki push doori ke square ke saath kamzor hoti jaati hai; $\overset{r}{^}$ (length-one arrow) kehta hai ki push radially bahar ki taraf point karti hai ek positive $q$ ke liye.

KYUN. Potential field ke through define hota hai, isliye pehle field jaanna zaroori hai. Yahi hamara ek starting fact hai — dekhein Electric Field of Point Charge.

DIKHTA KAISA HAI. Figure mein arrows bahar ki taraf fan karte hue dikhte hain. Dhyan do ki $r$ badhne ke saath woh chote hote jaate hain — woh chhotaapan hi $1/ r^{2}$ law ka drawn version hai.

Step 3 — Potential hai work per charge, toh ek path ke saath push add karna hoga

KYA. $q_{0}$ ko bahut door (infinity) se andar doori $r$ tak laane ke liye, ek baahri haath poore safar mein field se ladhta rehta hai. Field jo total work karta hai, $q_{0}$ se divide karne par potential milta hai:

$V (r) = - \int_{\infty}^{r} E \cdot d l$

Term by term: $V (r)$ woh number hai jo chahiye (volts mein); bada $\int$ matlab hai "poore safar mein add up karo"; $d l$ safar ka ek chota sa kadam hai; $E \cdot d l$ hai ki us kadam mein field kitni madad karta hai ya rokta hai; minus sign field ka kaam palat ke haath ka kaam bana deta hai.

YEH TOOL KYUN — integral? Ek akeli push $E$ sirf ek doori par sach hai; safar infinitely many distances cross karta hai, har ek alag push ke saath. Integral exactly woh tool hai jo infinitely many choti-choti continuously changing contributions ko sum karta hai. Seedha multiplication tab hi kaam karta jab push constant hoti — woh hai nahi.

MINUS SIGN KYUN? Haath field ke khilaaf push karta hai. Haath ka kaam $= -$ (field ka kaam). $q_{0}$ se divide karne par woh minus preserve rehta hai.

DIKHTA KAISA HAI. Figure incoming path ko chote segments $d l$ mein kaata hai; har segment kaam ka ek chota sa slice hai. Hum slices stack karke $V$ lete hain.

Step 4 — Sabse smart path chuno: $\overset{r}{^}$ ke saath seedha andar

KYA. Hum $\infty$ se $r$ tak koi bhi raasta le sakte hain, kyunki electrostatic field conservative hai (path matter nahi karta — dekhein Conservative Fields and Curl). Toh sabse aasaan raasta chuno: radial line ke saath seedha andar chalo. Tab kadam hai $d l = d r^{'} \overset{r}{^}$ aur dot product simple ho jaata hai:

$E \cdot d l = (E \overset{r}{^}) \cdot (d r^{'} \overset{r}{^}) = E d r^{'} (\overset{r}{^} \cdot \overset{r}{^} = 1)$

Yahan $r^{'}$ ek moving doori hai (ek dummy variable) jo $\infty$ se slide hoti hui final $r$ tak jaati hai; hum ise rename karte hain taaki endpoint $r$ se clash na ho.

KYUN. Radial path par, field aur motion perfectly parallel hain, isliye messy angle gayab ho jaata hai aur $E \cdot d l$ ek ordinary 1-D integral ban jaata hai jo hum actually compute kar sakte hain.

DIKHTA KAISA HAI. Figure mein path (coral) ek field arrow (lavender) ke bilkul upar lie karta hai — woh overlap karte hain. Greyed-out curvy path compare karo: woh same answer deta hai lekin compute karna mushkil hai. Yeh equality conservative field ka poora gift hai.

Step 5 — Integral karo (calculus ka punchline)

KYA. Field substitute karo aur integrate karo:

\;=\; -\frac{q}{4\pi\varepsilon_0}\int_{\infty}^{r} r'^{-2}\,dr'$$ Constants bahar nikaal lo. Bacha hua piece ek hi fact use karta hai ki $r'^{-2}$ ka antiderivative $-\,r'^{-1}$ hai: $$\int r'^{-2}\,dr' = -\frac{1}{r'}\quad\Longrightarrow\quad V(r) = -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r}$$ **YEH TOOL KYUN — $1/r'^2$ ka antiderivative?** Integral poochh raha hai "kaunsa function, differentiate karne par $1/r'^2$ deta hai?" Woh hai $-1/r'$. Yeh Step 7 mein lete jaane wale derivative ka *ulta* hai — bridge pehle se dikh raha hai. **DIKHTA KAISA HAI.** Figure $1/r'^2$ plot karta hai; $\infty$ se $r$ tak uske neeche shaded area **hi** integral hai. Area finite hai even though tail infinity tak jaata hai — kyunki $1/r'^2$ itni tezi se shrink karta hai. Isliye potential ek finite number hota hai. --- ## Step 6 — Limits plug karo: infinity par reference gayab ho jaata hai **KYA.** Bracket ko dono endpoints par evaluate karo: $$V(r) = -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r} - \Big(-\underbrace{\frac{1}{\infty}}_{\to\,0}\Big)\right) = -\frac{q}{4\pi\varepsilon_0}\left(-\frac{1}{r}\right) = \boxed{\;\frac{1}{4\pi\varepsilon_0}\frac{q}{r} = \frac{kq}{r}\;}$$ $r'=\infty$ par term $1/r'\to 0$ hoti hai — yahi poora reason hai ki humne infinity ko apna starting point kyun chuna: yeh far end ka contribution **kuch nahi** banata, toh $V=0$ bahut door. Do minus signs cancel hoke positive $q$ ke liye clean positive answer dete hain. **KYUN.** Potential sirf ek reference ke *relative* define hota hai. $V(\infty)=0$ choose karna woh choice hai jo ek term ko khatam kar deti hai aur tidy formula deti hai. **DIKHTA KAISA HAI.** Figure $V(r)=kq/r$ draw karta hai: ek curve jo charge ke paas steeply dive karta hai aur door jaake zero ki taraf flat hota hai. Faintly overlaid hai $E\propto 1/r^2$ — note karo $V$ **slower** girta hai (ek power of $r$, do nahi). > [!formula] Result > $$V(r)=\frac{kq}{r},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx 9\times10^{9}\ \mathrm{N\,m^2/C^2}$$ --- ## Step 7 — Wapas jaake cross-check karo: $V$ se $\vec E$ **KYA.** Bridge dono directions mein kaam karna chahiye. $dV=-\vec E\cdot d\vec l$ se, ek chota radial step deta hai $E_r=-\dfrac{dV}{dr}$. Apna fresh result daalo: $$E_r = -\frac{d}{dr}\!\left(\frac{kq}{r}\right) = -kq\cdot\left(-\frac{1}{r^2}\right) = \frac{kq}{r^2}\quad\checkmark$$ Humein **original** field wapas mil gaya. Step 7 ka derivative exactly Step 5 ke integral ko undo karta hai. **YEH TOOL KYUN — derivative $-dV/dr$?** Potential ek "height landscape" hai. Field uski steepest slope hai. Derivative *hi* slope hai, aur minus sign ise **neeche ki taraf** — lower potential ki taraf — point karta hai. Puri 3-D mein yeh hai $\vec E=-\nabla V$ (dekhein [[Gradient Operator]]). **DIKHTA KAISA HAI.** Figure $V$ curve ko ek hill ki tarah show karta hai; ek jagah ek tangent line draw ki gayi hai, aur ek red arrow field ko us slope ke **neeche** point karte hua mark karta hai. Hill ka steep hissa ⇒ lamba arrow ⇒ strong field. --- ## Step 8 — Sign aur degenerate cases (koi gap mat chhodna) **KYA.** Walk through karo ki formula har extreme par kya karta hai: - **$q>0$ (positive charge):** $V=kq/r>0$ everywhere. Landscape ek **hill** hai jo charge par $+\infty$ tak uthti hai. - **$q<0$ (negative charge):** $V=kq/r<0$ everywhere. Landscape ek **funnel/well** hai jo $-\infty$ tak girta hai. - **$r\to 0$ (charge ke bilkul paas):** $V\to\pm\infty$. Yeh ek *idealisation* hai — real charges ka finite size hota hai, isliye yeh blow-up physical nahi hai. - **$r\to\infty$:** $V\to 0$, hamare reference choice se match karta hai. - **Field zero lekin $V\neq 0$:** do equal positive charges ke midpoint par dono $\vec E$ arrows cancel hote hain, phir bhi dono potentials **add** hote hain (scalars) ek large positive $V$ banane ke liye. Potential poore safar par depend karta hai, local push par nahi. **KYUN.** Ek reader ko kabhi koi sign ya limiting case nahi milna chahiye jo tumne skip kar diya ho. **DIKHTA KAISA HAI.** Figure positive-charge hill (zero ke upar) aur negative-charge well (zero ke neeche) ko ek axis par stack karta hai, beech mein dur ki flat $V=0$ line mark ki gayi hai. > [!mistake] "$V=0$ jahan bhi $\vec E=0$ hota hai." > Dono ek integral se linked hain, equality se nahi. Cancelling arrows ($\vec E=0$) phir bhi ek badi height ($V\neq0$) par baith sakte hain. Flip side ke liye dekhein [[Equipotential Surfaces]] — flat height ka matlab no field. --- ## Ek-picture summary Poora safar ek canvas par: ek source charge, infinity se andar ki taraf ek radial path, shrinking $1/r^2$ field arrows jo humne sum kiye, unka enclosed finite area, aur resulting $V=kq/r$ hill — uske downhill slope ke roop mein field ke saath. Left→right padhein derivation ke liye; right→left check ke liye. > [!recall]- Feynman retelling — ek dost ko batao > Ek chota glowing dot (charge) imagine karo aur tum ek tiny bead (test charge) lekar bahut, bahut door ho. Jaise tum bead ko andar laate ho, dot wapas push karta hai — door se halka, paas se zyada (yahi hai $1/r^2$ arrows distance ke saath chote hote hue). "Potential" bas *total effort per unit bead* hai jo woh walk karne mein lagi — tum raaste mein har choti push ko add karte ho, yahi integral karta hai. Kyunki koi bhi path le sako, hum seedha andar chale jaate hain sum aasaan rakhne ke liye. $1/r'^2$ ko infinity se $r$ tak sum karne par neat $1/r$ milta hai — wahi hai $V=kq/r$, ek hill jo dot ke paas oonchi aur door flat hai. Aur agar koi tumhe hill de de, tum field uss hill ki slope padhke neeche ki taraf pointing karke dhundh lete ho — wahi hai $\vec E=-dV/dr$. Ek derivation, dono taraf se walk karne layak. > [!recall]- Two-line self-test > Integral ka $\infty$ end kyun drop out ho jaata hai? ::: Kyunki wahan $1/r'\to 0$ — yahi poora point hai $V(\infty)=0$ choose karne ka. > $1/r^2$ integrate karne par $1/r$ potential kyun milta hai? ::: $r^{-2}$ ka antiderivative $-r^{-1}$ hai, isliye denominator mein $r$ ki ek power survive karti hai. --- ## Connections - [[Coulomb's Law]] — woh $1/r^2$ field deta hai jo humne integrate kiya. - [[Electric Field of Point Charge]] — Step 7 mein exactly recover kiya gaya. - [[Gradient Operator]] — $\vec E=-\nabla V$ ke peeche ki 3-D machinery. - [[Conservative Fields and Curl]] — kyun hum Step 4 mein radial shortcut le sake. - [[Equipotential Surfaces]] — "no field" ka flat-height view. - [[Potential Energy of Charge System]] — $V$ ko ek charge se multiply karo toh energy milti hai. --- #flashcards/physics Why can we choose a radial path when computing $V$ of a point charge? ::: Field conservative hai, isliye line integral path-independent hota hai — sabse aasaan (radial) raasta chuno. What makes the infinity limit vanish in the integral? ::: $1/r'\to 0$ as $r'\to\infty$, isliye choose kiya taaki $V(\infty)=0$ ho. Why is $V$ finite even though the path starts at infinity? ::: $1/r'^2$ itni tezi se shrink karta hai ki uske neeche ka area finite hota hai.