Worked examples — Potential of point charge, potential from field and vice versa

1.8.9 · D3 · Physics › Electromagnetism › Potential of point charge, potential from field and vice ver

Do facts pe hum poori tarah rely karte hain (parent me prove hue):

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Scenario matrix

Is topic ke har possible problem in cells mein se ek hogi. Har cell ka kam se kam ek worked example neeche hai.

#	Cell (case class)	Kya tricky hai	Example
C1	Single positive charge, $V$ nikalo ek distance par	baseline, $V$ ka sign	Ex 1
C2	Single negative charge, $V$ nikalo	$V<0$ — ek "valley"	Ex 1
C3	Do charges, scalar superposition (mixed signs)	signed values add karo, vectors nahi	Ex 2
C4	Degenerate zero of $\vec{E}$ but $V\ne 0$	field cancel, potential nahi	Ex 3
C5	Degenerate zero of $V$ but $\vec{E}\ne 0$	potential cancel, field nahi	Ex 3
C6	$\vec{E}$ from $V(x)$ (1-D slope, $E$ ka sign)	gradient ka minus-sign	Ex 4
C7	$\vec{E}$ from $V(x,y,z)$ (partial derivatives)	scalar se vector	Ex 5
C8	$\Delta V$ from a uniform field	integration ki direction, sign	Ex 6
C9	Limiting behaviour $r\to \infty$ aur $r\to 0$	$V,E$ kitni jaldi ghatte/badhte hain	Ex 7
C10	Real-world word problem (energy $U=qV$)	words ko numbers mein translate karo	Ex 8
C11	Exam twist: dipole-like pair ki line par $V=0$ point	ek location ke liye solve karo	Ex 9

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Ex 1 — ek charge, dono signs (cells C1, C2)

Forecast: padhne se pehle dono signs aur rough size guess karo. Positive charge → hill top → $V>0$; negative → valley → $V<0$. Same magnitude.

1. Formula likho. $V=\frac{kq}{r}$. Ye step kyun? Ek point charge hai, aur $V$ ko sirf distance $r$ chahiye, koi direction nahi — yahi reason hai ki $V$ scalar hai.
2. (a) $q=+4\times 10^{-9}$, $r=2$ plug karo. $V=\frac{(9\times 10^9)(4\times 10^{-9})}{2}=\frac{36}{2}=18\mathrm{V}.$ Ye step kyun? Pure arithmetic; units $\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}\cdot C/m=\mathrm{N}m/C=J/C=\mathrm{V}$. ✓
3. (b) $q$ ka sign flip karo. $V=-18\mathrm{V}$. Ye step kyun? Charge ka sign directly $V$ mein aata hai — kabhi bhi absolute values mat lo.

Verify: $|V_a|=|V_b|$ — expected tha; $r$ ko $4\mathrm{m}$ par double karne se dono $\pm 9\mathrm{V}$ ho jaate kyunki $V\propto 1/r$. Sign charge ke saath flip hota hai, magnitude distance se fix hoti hai. ✓

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Ex 2 — scalar superposition, mixed signs (cell C3)

Forecast: point positive charge ke paas hai ($1\mathrm{m}$) aur negative se door ($3\mathrm{m}$), toh guess karo: net positive, lekin bahut bada nahi.

1. Picture se distances padho. $q_1$ se: $r_1=|1-0|=1\mathrm{m}$. $q_2$ se: $r_2=|1-4|=3\mathrm{m}$. Ye step kyun? $V$ direction ignore karta hai — distance ek plain positive length hai. Figure mein do teal brackets dekho.
2. Signed scalar contributions add karo. $V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}=9\times 10^9\left(\frac{2\times 10^{-6}}{1}+\frac{-3\times 10^{-6}}{3}\right).$ Ye step kyun? Potential ke liye superposition algebra hai, trigonometry nahi — signs charges se aate hain, geometry se nahi.
3. Bracket compute karo, phir multiply karo. $=9\times 10^9(2\times 10^{-6}-1\times 10^{-6})=9\times 10^9\times 10^{-6}=9000\mathrm{V}.$

Verify: result positive hai (forecast se match) aur finite hai. Units: $(\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2})(\mathrm{C}{\mathrm{m}}^{-1})=\mathrm{V}$. ✓

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Ex 3 — do famous degeneracies (cells C4, C5)

Forecast: ek case mein field zero ho jaata hai lekin potential bada rehta hai; doosre mein potential zero ho jaata hai lekin field rehta hai. Guess karo kaun kaun sa hai.

Case (A): do like $+q$ (figure ka top).

1. Field by symmetry. Left charge right taraf push karta hai ($+x$), right charge left taraf push karta hai ($-x$), equal magnitude ⇒ $\vec{E}=0$ at origin. Ye step kyun? Field ek vector hai; equal-and-opposite arrows cancel ho jaate hain. Do red arrows head-on milte dekho.
2. Potential by scalar sum. Dono positive hain aur equidistant hain: $V=k\frac{q}{a}+k\frac{q}{a}=\frac{2kq}{a}=\frac{2(9\times 10^9)(10^{-6})}{1}=1.8\times 10^4\mathrm{V}.$ Ye step kyun? Scalars direction se cancel nahi ho sakte — do hills add hoti hain. Toh $\vec{E}=0$ phir bhi $V=18000\mathrm{V}\ne 0$. Cell C4.

Case (B): $+q$ left, $-q$ right (figure ka bottom).

3. Potential by scalar sum. $V=k\frac{+q}{a}+k\frac{-q}{a}=0.$ Ye step kyun? Ek hill aur equal valley equal distance par as numbers cancel ho jaate hain → $V=0$.
4. Field by vectors. Left $+q$ ek test $+$ charge ko right dhakelta hai; right $-q$ usse right kheenchta hai. Dono $+x$ direction mein hain, ye add ho jaate hain: $E=2\cdot \frac{kq}{a^2}=2\frac{(9\times 10^9)(10^{-6})}{1^2}=1.8\times 10^{4}V/m(\text{along }+x)\ne 0.$ Ye step kyun? Same direction mein point karne wale vectors reinforce karte hain. Toh $V=0$ phir bhi $\vec{E}\ne 0$. Cell C5.

Verify: dono cases mirror images hain — (A) mein vectors cancel aur scalars add; (B) mein scalars cancel aur vectors add. Ye definitive counter-example hai "$V=0\iff E=0$" ke liye. ✓

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Ex 4 — 1-D potential se field, sign dhyan se (cell C6)

Forecast: $x=2$ par parabola steeply upar ja rahi hai, toh ball ko chota $x$ ki taraf roll karna chahiye — expect karo $E_x<0$.

1. Differentiate karo aur negate karo. $E_x=-\frac{dV}{dx}=-(6x-2)$. Ye step kyun? Field minus the slope of potential hota hai — minus arrow ko downhill point karata hai. Iske bina energy conservation toot jaati hai.
2. $x=2$ par evaluate karo. $E_x=-(6\cdot 2-2)=-(10)=-10V/m$. Ye step kyun? Bas plug in karo. Negative ⇒ $-x$ mein point karta hai, exactly forecast ke anusaar (rising $V$ se downhill).

Verify: vertex par slope zero hai: $6x-2=0\Rightarrow x=\frac{1}{3}$, toh $E_x=0$ wahan — flat spot mein koi field nahi, slope picture se consistent hai. ✓ Units: $V/m$. ✓

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Ex 5 — 3-D potential se vector field (cell C7)

Forecast: teen partial derivatives → teen components. Genuinely 3-D arrow milega.

1. Har partial derivative lo (baaki variables ko constants maano). $\frac{\partial V}{\partial x}=4xy,\frac{\partial V}{\partial y}=2x^2,\frac{\partial V}{\partial z}=-1.$ Ye step kyun? Gradient $\nabla V$ in teen slopes ko ek vector mein stack karta hai; har ek measure karta hai ki $V$ us axis ke along kitna chadh raha hai.
2. Negate karke field nikalo. $\vec{E}=-\nabla V=(-4xy,-2x^2,+1)$. Ye step kyun? Same minus sign jaisa 1-D mein — field wahan point karta hai jahan $V$ sabse tezi se ghatta hai.
3. $(1,1,3)$ par evaluate karo. $\vec{E}=(-4,-2,1)V/m$. Ye step kyun? Coordinates plug in karo; note karo ki $z$ drop out ho gaya kyunki $\partial V/\partial z$ constant tha.

Verify: magnitude $|\vec{E}|=\sqrt{(-4{)}^2+(-2{)}^2+1^2}=\sqrt{21}\approx 4.58V/m$. Teen components independent hain — yahi vector nature hai jo $V$ chupaata hai lekin $\vec{E}$ reveal karta hai. ✓

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Ex 6 — uniform field se potential difference (cell C8)

Forecast: field $+x$ direction mein point karta hai (downhill $+x$ hai), toh chota $x$ = zyada upar. $A$ ($x=1$ par) $B$ ($x=4$ par) se upar hai: expect karo $V_A-V_B>0$.

1. $V_A-V_B=-{\int }_B^A\vec{E}\cdot d\vec{l}$ use karo. $\vec{E}$ constant hai $x$ ke along: $V_A-V_B=-E(x_A-x_B).$ Ye step kyun? Uniform field integral se seedha bahar aa jaata hai; sirf endpoint displacement $x_A-x_B$ bachta hai.
2. Numbers plug karo. $V_A-V_B=-200(1-4)=-200(-3)=600\mathrm{V}$. Ye step kyun? Arithmetic. Positive ⇒ $A$ higher potential par hai — forecast se match (field ke against jaane par tum chadh rahe ho).

Verify: endpoints flip karo: $V_B-V_A=-200(4-1)=-600\mathrm{V}$, hamara answer ka negative — jaisa kisi bhi well-defined potential difference ke liye hona chahiye. Units $(V/m)\cdot \mathrm{m}=\mathrm{V}$. ✓

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Ex 7 — limiting behaviour (cell C9)

Forecast: $r$ double karne par $V$ half ho jaana chahiye lekin $E$ quarter ho jaana chahiye, kyunki $V\propto 1/r$ aur $E\propto 1/r^2$.

1. $r=1\mathrm{m}$ par values. $V=\frac{(9\times 10^9)(10^{-6})}{1}=9000\mathrm{V},E=\frac{(9\times 10^9)(10^{-6})}{1^2}=9000V/m.$ Ye step kyun? Compare karne ke liye baseline.
2. $r=2\mathrm{m}$ par values. $V=\frac{9000}{2}=4500\mathrm{V}(\times \frac{1}{2}),E=\frac{9000}{2^2}=2250V/m(\times \frac{1}{4}).$ Ye step kyun? Different power laws confirm karta hai — $V$ $E$ se dhire marta hai.
3. Limits. $r\to \infty$ par: dono $\to 0$ (hamara chosen reference). $r\to 0$ par: dono blow up ho jaate hain ($V\to \infty$, $E\to \infty$) — idealised point charge ek mathematical singularity hai, real object nahi. Ye step kyun? Har acchi analysis apne extreme cases name karti hai taaki koi reader surprised na ho.

Verify: ratio check $V(1)/V(2)=2$ aur $E(1)/E(2)=4$ — exactly $1/r$ vs $1/r^2$ signatures. ✓

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Ex 8 — real-world word problem, $U=qV$ (cell C10)

Forecast: ek positive charge high $V$ se low $V$ ki taraf roll karta hai (downhill), toh use kinetic energy gain honi chahiye.

1. Energy charge times potential hai: $U=qV$. Potential energy mein change: $\Delta U=q(V_{\text{final}}-V_{\text{initial}})=q(20-50)=q(-30).$ Ye step kyun? Potential energy of a charge in a field $U=qV$ hoti hai; motion ke liye sirf change matter karta hai.
2. Energy conservation: kinetic gained $=-\Delta U$. $\Delta K=-\Delta U=-q(-30)=30q=30(1.6\times 10^{-19})=4.8\times 10^{-18}\mathrm{J}.$ Ye step kyun? Total energy conserved hai, toh PE lost, KE gain hoti hai.

Verify: positive $\Delta K$ ⇒ speed badhti hai, "positive charge rolls downhill" forecast se match. Electron-volts mein, $\Delta K=q\cdot 30\mathrm{V}=30\mathrm{eV}$ — ek clean sanity check, kyunki $1\mathrm{eV}$ exactly ek electronic charge ka $1\mathrm{V}$ across energy hai. ✓

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Ex 9 — exam twist: woh point dhundho jahan $V=0$ ho (cell C11)

Forecast: positive charge stronger hai, toh uski hill ko (weaker) valley se zero karne ke liye zyada door hona padega — expect karo ki null negative charge ke paas ho, yani $x>3$.

1. Total potential zero set karo position $x$ par ($0<x<6$, toh $r_1=x$, $r_2=6-x$): $k\frac{q_1}{x}+k\frac{q_2}{6-x}=0\Rightarrow \frac{2}{x}+\frac{-1}{6-x}=0.$ Ye step kyun? $V=0$ ek scalar equation hai — koi vectors nahi, toh hum sirf signed terms add karte hain aur sum zero set karte hain ($k$ aur $10^{-6}$ cancel ho jaate hain).
2. Solve karo. $\frac{2}{x}=\frac{1}{6-x}\Rightarrow 2(6-x)=x\Rightarrow 12-2x=x\Rightarrow 3x=12\Rightarrow x=4\mathrm{m}$. Ye step kyun? Cross-multiply karo; ek linear equation, ek root in range.

Verify: $x=4$ par: $V\propto \frac{2}{4}+\frac{-1}{2}=0.5-0.5=0$ ✓. Aur $x=4>3$ hai, negative charge ke paas — exactly forecast. Note karo ki $\vec{E}\ne 0$ yahan hai (Ex 3 ka lesson: $V=0$ spot field-free spot nahi hota). ✓

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Recall

Recall Kaun sa cell kaun sa trap hai?

Field zero lekin potential bada ::: do like charges ka midpoint (Ex 3A). Potential zero lekin field nonzero ::: do opposite charges ka midpoint (Ex 3B). Potentials kaise add karte hain? ::: signed scalars, no components (Ex 2). $V$ se $\vec{E}$ kaise nikalte hain? ::: minus the gradient, $E_x=-dV/dx$ (Ex 4, 5). $r$ double karo: $V$ aur $E$ ka kya hoga? ::: $V$ half hoga, $E$ quarter hoga (Ex 7).

Connections

• Coulomb's Law · Electric Field of Point Charge · Potential Energy of Charge System · Equipotential Surfaces · Conservative Fields and Curl · Gradient Operator