Noesis
1.8.9 · HinglishElectromagnetism

Potential of point charge, potential from field and vice versa

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1.8.9 · Physics › Electromagnetism

1. Potential kya hai? (pehle definition)

Minus sign kyun? External agent ko field ki force ke khilaf push karna padta hai. Agent ka work field ka work . se divide karo.

2. se derive karna — point charge (scratch se)

Hum compute karte hain charge se distance pe, test charge ko se radially andar laate hue.

Step 1 — line integral set up karo. Ye step kyun? Ye potential ki definition hai; hum bas apna jaana-pehchana plug in karte hain.

Step 2 — radial path choose karo. Seedha andar move karo, toh aur . Ye step kyun? Electrostatic path-independent hai (field conservative hai), isliye hum sabse aasaan path choose karte hain — pure radial — aur angular part khatam ho jaata hai.

Step 3 — substitute karo aur integrate karo.

= -\frac{q}{4\pi\varepsilon_0}\left[-\frac{1}{r'}\right]_{\infty}^{r}$$ *Ye step kyun?* $\int r'^{-2}dr' = -r'^{-1}$. Bracket evaluate karo. **Step 4 — limits evaluate karo.** $\infty$ pe, $1/r'\to 0$; $r$ pe hume $1/r$ milta hai. $$\boxed{V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}}$$ > [!formula] Point charge ka potential > $$V(r) = \frac{kq}{r},\qquad k=\frac{1}{4\pi\varepsilon_0}\approx 9\times 10^{9}\ \mathrm{N\,m^2/C^2}$$ > Note: $V\propto 1/r$ (potential), jabki $E\propto 1/r^2$ (field). Potential **dheere** kam hota hai. ![[1.8.09-Potential-of-point-charge,-potential-from-field-and-vice-versa.png]] --- ## 3. Wapas jaana: $V$ se $\vec{E}$ (gradient) Integral ek derivative mein ulat jaata hai. $dV = -\vec E\cdot d\vec l$ se start karke, ek chote radial step ke liye $dV = -E_r\,dr$, toh $$E_r = -\frac{dV}{dr}$$ Poore 3-D mein field hai **negative gradient**: $$\boxed{\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat x + \frac{\partial V}{\partial y}\hat y + \frac{\partial V}{\partial z}\hat z\right)}$$ **Point charge pe check karo:** $V=kq/r$ ke saath, $$E_r = -\frac{d}{dr}\!\left(\frac{kq}{r}\right) = -kq\cdot\left(-\frac{1}{r^2}\right) = \frac{kq}{r^2}\ \checkmark$$ Humne original field wapas paaya — bridge dono taraf kaam karta hai. > [!intuition] Slope picture > $V$ ek "landscape" hai. $\vec E$ **sabse steep descent** ki direction mein point karta hai aur uski magnitude hai **steepness**. Flat region (constant $V$) ⇒ koi field nahi. Jahan contour lines paas paas hain ⇒ strong field. --- ## 4. $V$ path-independent kyun hai (gehra reason) KISI BHI closed loop ke liye, electrostatics mein $\oint \vec E\cdot d\vec l = 0$ (curl-free field). Toh $A\to B$ jaane ka kaam har path pe same hai; sirf endpoints matter karte hain. Yehi cheez hume ek single-valued $V(P)$ *define* karne deti hai. Agar ye sach nahi hota, toh "potential" ek well-defined function nahi hota. --- ## 5. Worked examples > [!example] (a) Do charges ka potential (scalar superposition) > Charges $q_1=+2\ \mu C$ origin pe, $q_2=-3\ \mu C$ at $x=4\ \mathrm m$. $V$ find karo $x=1\ \mathrm m$ pe. > > **Step 1.** Distances: $q_1$ se, $r_1=1\,\mathrm m$; $q_2$ se, $r_2=3\,\mathrm m$. > *Kyun?* $V$ sirf distance pe depend karta hai, isliye bas har ek measure karo. > > **Step 2.** Scalars jodhno (signs ke saath): > $$V = k\frac{q_1}{r_1}+k\frac{q_2}{r_2}=9\times10^9\!\left(\frac{2\times10^{-6}}{1}+\frac{-3\times10^{-6}}{3}\right)$$ > *Kyun?* Potential algebraically superpose hota hai — koi components nahi, charge ka sign seedha carry hota hai. > > **Step 3.** Compute karo: $9\times10^9(2\times10^{-6}-1\times10^{-6})=9\times10^9\times10^{-6}=9000\ \mathrm V$. > [!example] (b) Diye gaye potential se field > $V(x)=3x^2-2x$ (volts, $x$ metres mein). $E_x$ find karo $x=2$ pe. > > **Step 1.** $E_x=-\dfrac{dV}{dx}=-(6x-2)$. > *Kyun?* Field potential ka minus slope hai. > > **Step 2.** $x=2$ pe: $E_x=-(12-2)=-10\ \mathrm{V/m}$ ($-x$ direction mein point karta hai). > *Sign kyun?* $V$ yahan $x$ ke saath badh raha hai, isliye field doosri taraf point karta hai — downhill. > [!example] (c) Field se potential difference > Uniform field $E=200\ \mathrm{V/m}$ $+x$ direction mein. $V_A-V_B$ find karo $x_A=1$ aur $x_B=4\,\mathrm m$ ke beech. > > **Step 1.** $V_A-V_B=-\int_B^A E\,dx = -E\,(x_A-x_B)=-200(1-4)=600\ \mathrm V$. > *Kyun?* Field ke khilaf move karna (high $x$ se low $x$ ki taraf) potential badhata hai; $A$ zyada $V$ pe hai. --- ## 6. Common mistakes (steel-manned) > [!mistake] "$V$ zero hai jahan $\vec E$ zero hai." > **Kyun sahi lagta hai:** kai simple cases mein (jaise door se) dono saath khatam hote hain. > **Fix:** $V$ depend karta hai $E$ ke *integral* pe, uski local value pe nahi. **Do equal positive charges ke midpoint pe**, symmetry se $\vec E=0$ hai lekin $V$ large aur positive hai. Waise hi, $+q$ aur $-q$ ke beech midpoint pe, $V=0$ hai lekin $\vec E\neq 0$. > [!mistake] "Fields ko scalars ki tarah jodhno jaise potentials." > **Kyun sahi lagta hai:** superposition dono ke liye kaam karta hai. > **Fix:** $\vec E$ ek **vector** hai — components jodhno. $V$ ek **scalar** hai — (signed) magnitudes jodhno. Inhe mix karne se galat angles aur galat magnitudes milte hain. > [!mistake] "$\vec E=-\nabla V$ mein minus sign chhodh do." > **Kyun sahi lagta hai:** sign sirf bookkeeping jaisa lagta hai. > **Fix:** uske bina, field *uphill* point karega aur energy conservation toot jaayegi. Minus encode karta hai "force lower potential energy ki taraf point karta hai." --- ## 7. Active recall > [!recall]- Quick self-test (answers chhupao) > - $V$ scalar kyun hai lekin $\vec E$ vector kyun? → $V$ energy/charge hai (koi direction nahi); $\vec E$ force/charge hai. > - Point charge ke liye $V\propto ?$ aur $E\propto ?$? → $V\propto 1/r$, $E\propto 1/r^2$. > - $V$ se $\vec E$ kaise nikaalte hain? → $\vec E=-\nabla V$. > - Kya $V\neq0$ ho sakta hai jahan $\vec E=0$ ho? → Haan (do same charges ka midpoint). > [!recall]- Feynman: 12-saal ke bachche ko explain karo > Ek pahadi khet ka socho. Har jagah zameen ki **unchai** potential $V$ hai — bas ek number jo batata hai tum kitne "oopar" ho. Wahan rakhi ball ko ek push milta hai — wo push electric field $\vec E$ hai, aur ye hamesha **sabse steep slope ke neeche** point karta hai. Agar zameen flat hai, ball ko koi push nahi (koi field nahi). Steep cliff ⇒ strong push. Toh agar tum mujhe pahaadiyon ki shape bata do, main jaanta hoon har ball kahan jaayegi aur kitni tez — yahi hai potential se field nikaalana. Aur agar tum mujhe bata do har jagah balls kitni tez push ho rahe hain, main chadhaaiyon ko jodh ke pahaadiyon ki unchai rebuild kar sakta hoon — yahi hai field se potential nikaalana. > [!mnemonic] > **"Volts neeche jaate hain, Fields slope ke neeche jaate hain."** > $V$ use karta hai $1/r$ (neeche ek $r$ = "one"); $E$ use karta hai $1/r^2$ ("squared, stronger drop"). Aur **E = minus slope of V**: *Eat Minus the Slope*. --- ## 8. Connections - [[Coulomb's Law]] — $1/r^2$ field jise humne integrate kiya. - [[Electric Field of Point Charge]] — jo hum $-\nabla V$ se recover karte hain. - [[Potential Energy of Charge System]] — $U=qV$. - [[Equipotential Surfaces]] — constant $V$ ki surfaces, hamesha $\perp \vec E$. - [[Conservative Fields and Curl]] — kyun $\oint\vec E\cdot d\vec l=0$. - [[Gradient Operator]] — $-\nabla V$ ka math machinery. --- #flashcards/physics Ek point pe electric potential define karo ::: External agent dwara infinity se us point tak unit positive test charge laane mein kiya gaya work per unit charge: $V=-\int_\infty^P \vec E\cdot d\vec l$. Unit: volt (J/C). Point charge ka potential formula ::: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}=\dfrac{kq}{r}$. $V$ aur $\vec E$ kaise related hain (integral form mein)? ::: $V=-\int \vec E\cdot d\vec l$ (field ka negative line integral). $V$ se $\vec E$ kaise nikaalte hain? ::: $\vec E=-\nabla V$; 1-D mein $E_x=-dV/dx$. $\vec E=-\nabla V$ mein minus sign kyun? ::: Field decreasing potential ki taraf point karta hai (downhill), ye encode karta hai ki force potential energy kam karta hai. Point charge ke liye $V$ vs $r$ aur $E$ vs $r$ scaling ::: $V\propto 1/r$, $E\propto 1/r^2$; potential dheere kam hota hai. $V$ scalar hai ya vector, aur ise kaise superpose karte hain? ::: Scalar; signs ke saath algebraically jodhno: $V=\sum kq_i/r_i$. Kya field zero ho sakta hai jahan potential nonzero ho? ::: Haan — jaise do equal positive charges ka midpoint: $\vec E=0$ lekin $V>0$. Electrostatic potential path-independent kyun hai? ::: Kyunki $\oint\vec E\cdot d\vec l=0$ (field conservative/curl-free hai); kaam sirf endpoints pe depend karta hai. $V=kq/r$ se $E$ recover karo ::: $E_r=-dV/dr=-kq(-1/r^2)=kq/r^2$. ✓ ## 🖼️ Concept Map ```mermaid flowchart TD E["Field E vector"] V["Potential V scalar"] DEF["V = work per unit charge"] LINT["V = -∫ E·dl"] GRAD["E = -∇V"] PC["Point charge field E = kq/r²"] VPC["V = kq/r"] RAD["Radial path choice"] CONS["Conservative field"] SLOPE["Field points downhill"] DEF -->|leads to| LINT E -->|line integral| LINT LINT -->|defines| V V -->|negative gradient| GRAD GRAD -->|recovers| E PC -->|integrate radially| VPC CONS -->|allows| RAD RAD -->|simplifies| LINT LINT -->|applied to point charge| VPC VPC -->|falls off as 1 over r| V GRAD -->|explains| SLOPE ```