Electric potential — definition V = −∫E·dl: Worked examples

1.8.8 · D3 · Physics › Electromagnetism › Electric potential — definition V = −∫E·dl

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The scenario matrix

Har potential problem in cells mein se ek hota hai. Neeche har worked example ko us cell ke saath tag kiya gaya hai jise woh cover karta hai.

#	Cell (case class)	Kya cheez tricky banati hai	Example
A	Single positive point charge, radial path	$V$ ka sign; $V$ kahaan bada hai	Ex 1
B	Single negative point charge	$V < 0$ har jagah; "downhill" flip ho jaata hai	Ex 2
C	Kai charges ka Superposition	$V$ plain numbers ki tarah add hota hai (no vectors)	Ex 3
D	Uniform field (parallel plates), walk ka sign	walk ki direction vs. $E$	Ex 4
E	Gradient se $E$ recover karna $V$ ke through	kaun sa sign, kaun sa partial derivative	Ex 5
F	Work / energy of a moving charge (dono signs of $q$ )	$q$ ka sign times $Δ V$ ka sign	Ex 6
G	Degenerate / limiting: infinite line of charge	$\infty$ par reference fail ho jaati hai (diverge ho jaati hai)	Ex 7
H	Zero case: point on a perpendicular bisector, $V \neq = 0$ but $E = 0$ -component subtlety	$V$ bada ho sakta hai jahaan $E$ vanish kare	Ex 8
I	Real-world word problem	words ko formula mein translate karo	Ex 9
J	Exam twist: path independence — ek curved path	path ki shape kyun matter nahi karti	Ex 10

Ex 1 — Positive point charge (Cell A)

Worked example Statement

Ek charge $q = + 3 nC$ origin par rakha hai. $r_{1} = 0.20 m$ aur $r_{2} = 0.10 m$ par $V$ nikalo. Kaun sa point zyada potential par hai?

Forecast: Compute karne se pehle guess karo — jab tum ek positive charge ki taraf jaate ho, toh electrical "height" upar jaati hai ya neeche? (Sochो ek pahaad chadh rahe ho jiska peak woh charge hai.)

Steps.

Parent note ka master result use karo, $V (r) = \frac{k q}{r}$ . Yeh step kyun? Point charge ka field radial hai, isliye radial path $E \cdot d l = E d r$ banata hai; integral $k q / r$ par collapse ho jaata hai. Ise reuse karo, re-derive mat karo.
$V (r_{1}) = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{0.20} = 134.85 V$ . Yeh step kyun? Dur wali radius $r_{1}$ ko $k q / r$ mein plug karo taaki move karne se pehle ka baseline height mile — humein dono points compare karne ke liye chahiye.
$V (r_{2}) = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{0.10} = 269.7 V$ . Yeh step kyun? Nazdiki radius $r_{2}$ ke saath repeat karo; $r$ ko aadha karne se $V$ double ho jaata hai, kyunki $1/ r$ ki shape ka matlab hai "closer = taller."

Verify: $V > 0$ har jagah (jaise ki $+$ charge ke paas hona chahiye), aur nazdiki point $r_{2}$ ka potential zyada hai. Yeh forecast se match karta hai: $+ q$ ki taraf jaana upar chadna hai. Units: $N\cdotm^{2} / C^{2} \cdot C / m = N\cdotm/C = J/C = V$ . ✓

Figure — Electric potential — definition V = −∫E·dl

Ex 2 — Negative point charge (Cell B)

Worked example Statement

Ab $q = - 3 nC$ origin par hai. $r = 0.20 m$ par $V$ nikalo. Jaise tum paas aate ho, $V$ upar jaata hai ya neeche?

Forecast: Formula $V = \frac{k q}{r}$ abhi bhi bilkul waise hi kaam karta hai — lekin ab $q$ khud ek minus sign carry karta hai. $V$ ka sign guess karo aur kya paas aane par $V$ bada hoga ya zyada negative hoga.

Steps.

Ek hi fundamental formula $V (r) = \frac{k q}{r}$ rakho aur $q = - 3 \times 1 0^{- 9} C$ substitute karo: $V (0.20) = \frac{( 8.99 \times 1 0 ^{9} ) ( - 3 \times 1 0 ^{- 9} )}{0.20} = - 134.85 V .$ Yeh step kyun? Hum ek naya " $- k q / r$ " formula nahi banate; wahi $V = k q / r$ kaam karta hai, aur sirf $q$ ka sign $V$ ko negative drive karta hai. Isse har charge ke liye ek hi rule rehta hai.
Nazdiki radius $r = 0.10 m$ ko same formula mein substitute karo: $V (0.10) = \frac{( 8.99 \times 1 0 ^{9} ) ( - 3 \times 1 0 ^{- 9} )}{0.10} = - 269.7 V .$ Yeh step kyun? Hum trend dekhne ke liye doosra point compute karte hain: paas aane par $V$ $- 134.85$ se $- 269.7$ tak jaata hai — ek ghaati mein gehre jaana, pahaad nahi chadna.

Verify: $V < 0$ har jagah $- q$ ke paas. Field $E$ charge ki taraf point karta hai (inward), jo decreasing $V$ ki direction hai ( $- 134.85$ se $- 269.7$ tak). " $E$ downhill point karta hai" satisfy ho gaya. ✓

Ex 3 — Superposition of charges (Cell C)

Worked example Statement

Do charges: $q_{1} = + 4 nC$ at $(0, 0)$ aur $q_{2} = - 2 nC$ at $(0.30, 0) m$ . Point $P = (0.30, 0.40) m$ par $V$ nikalo.

Forecast: Potential ek scalar hai, isliye hum bas do numbers add karte hain — koi arrows nahi, koi components nahi. Guess karo ki total $+$ aayega ya $-$ .

Steps.

$q_{1}$ se $P$ tak distance: $r_{1} = 0.3 0^{2} + 0.4 0^{2} = 0.50 m$ . Yeh step kyun? $V$ sirf distance par depend karta hai, isliye pehle har separation chahiye.
$q_{2}$ se $P$ tak distance: yeh seedha $P$ ke neeche baitha hai, isliye $r_{2} = 0.40 m$ . Yeh step kyun? $q_{2}$ aur $P$ dono ka $x = 0.30$ same hai, isliye unka separation sirf vertical gap $0.40$ hai — koi square root ki zarurat nahi. Doosre $k q / r$ term ko feed karne ke liye yeh doosri distance chahiye.
$V = \frac{k q _{1}}{r _{1}} + \frac{k q _{2}}{r _{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 9} )}{0.50} + \frac{( 8.99 \times 1 0 ^{9} ) ( - 2 \times 1 0 ^{- 9} )}{0.40}$ . Yeh step kyun? Superposition: total $V$ har charge ke $k q / r$ ka plain algebraic sum hai. Yahi " $V$ is scalar" ka payoff hai — koi angles resolve nahi karne.
$= 71.92 + (- 44.95) = 26.97 V$ . Yeh step kyun? Do divisions karo aur signed numbers add karo; negative charge ka term simply subtract ho jaata hai, koi vector geometry involve nahi.

Verify: Bada, kaafi-kaafi-nazdik positive charge jeet jaata hai, isliye $V > 0$ — forecast se match. Agar do charges equal aur opposite hote equal distance par, toh $V = 0$ cancel ho jaata; yahan nahi hota, aur hume ek moderate positive number milta hai. ✓

Ex 4 — Uniform field, walk ka sign (Cell D)

Worked example Statement

Ek parallel-plate region mein uniform field $E = 500 V/m$ hai jo $+ \overset{x}{^}$ direction mein point kar raha hai. Plates $d = 0.04 m$ door hain. $V_{b} - V_{a}$ nikalo jab tum $a$ (at $x = 0$ ) se $b$ (at $x = 0.04$ ) tak chalte ho.

Forecast: Tum $E$ ke along chal rahe ho. Potential upar jaayega ya neeche? (Yaad karo $E$ downhill point karta hai.)

Steps.

$E \cdot d l = (500) (d x)$ kyunki dono $+ \overset{x}{^}$ along point karte hain; dot product sirf magnitudes ka product hai. Yeh step kyun? Jab do arrows parallel hote hain, dot product mein angle khatam ho jaata hai — yeh plain multiplication hai. Isliye humne seedha cross walk choose kiya.
$V_{b} - V_{a} = - \int_{0}^{0.04} 500 d x = - 500 \times 0.04 = - 20 V$ . Yeh step kyun? Constant $E$ integral se bahar aa jaata hai, sirf length $d = 0.04$ integrate karne ke liye reh jaata hai — isliye answer simply $- E d$ hai.
Toh $b$ $20 V$ neeche hai $a$ se; equivalently $V_{a} - V_{b} = + 20 V$ , aur $E = \frac{Δ V}{d} = \frac{20}{0.04} = 500 V/m$ . Yeh step kyun? $E = Δ V / d$ par wapas rearrange karne se given field recover hoti hai, ek consistency check aur reason ki field units $V/m$ hain.

Verify: $E$ ke along chalna potential giraata hai (negative $Δ$ ), exactly jaise "downhill" rule predict karta hai. $Δ V / d$ se $E$ re-derive karna original $500 V/m$ deta hai, self-consistency confirm karta hai. Capacitance and parallel-plate fields (E = ΔV/d) dekho. ✓

Ex 5 — $V$ se $E$ recover karna (Cell E)

Worked example Statement

Ek region mein $V (x, y, z) = 2 x^{2} - 3 y z + 4 z (volts, with x, y, z in metres)$ hai. Point $(1, 2, 1)$ par $E$ nikalo.

Forecast: $E = - \nabla V$ . Compute karne se pehle notice karo ki $V$ is point ke paas $x$ mein sabse tezi se grow karta hai — toh guess karo $E$ ka kaun sa component magnitude mein sabse bada hoga.

Steps.

Partial derivatives (baaki variables ko constants treat karke differentiate karo): $\frac{\partial V}{\partial x} = 4 x, \frac{\partial V}{\partial y} = - 3 z, \frac{\partial V}{\partial z} = - 3 y + 4.$ Yeh step kyun? $\nabla V$ exactly slopes ki yeh list hai; Gradient operator ∇ and directional derivatives dekho.
Sign flip karo: $E = - (4 x) \overset{x}{^} - (- 3 z) \overset{y}{^} - (- 3 y + 4) \overset{z}{^} = - 4 x \overset{x}{^} + 3 z \overset{y}{^} + (3 y - 4) \overset{z}{^} .$ Yeh step kyun? $E = - \nabla V$ ; minus "uphill slope" ko "downhill field" mein badhalta hai.
$(1, 2, 1)$ par: $E = (- 4) \overset{x}{^} + (3) \overset{y}{^} + (2) \overset{z}{^} V/m .$ Yeh step kyun? Slope ko interest ke point par evaluate karte hain — gradient position ka function hai, isliye woh definite arrow tab hi deta hai jab hum $(1, 2, 1)$ plug in karte hain.

Verify: Units: $V$ volts mein, metres ke saath derivative $V/m$ deta hai — correct field units. $\overset{x}{^}$ component ki magnitude $4$ hai, teeno mein sabse badi, forecast se match karta hai ki $V$ yahan $x$ mein sabse tezi se steep hai. ✓

Ex 6 — Moving charge par work, dono signs of $q$ (Cell F)

Worked example Statement

Point $a$ par $V_{a} = 80 V$ hai, point $b$ par $V_{b} = 30 V$ . (i) Ek $+ 5 μ C$ charge $a \to b$ move karta hai. (ii) Ek $- 5 μ C$ charge $a \to b$ move karta hai. Har case mein field ke dwara ki gayi work nikalo.

Forecast: $W_{field} = q (V_{a} - V_{b})$ use karo. Arithmetic se pehle har case mein work ka sign guess karo — ek positive hoga, ek negative.

Steps.

$V_{a} - V_{b} = 80 - 30 = 50 V$ . Yeh step kyun? Work formula ko motion ke along potential drop, $V_{a} - V_{b}$ chahiye, isliye ise ek baar compute karke dono charges ke liye reuse karte hain.
(i) $W = (+ 5 \times 1 0^{- 6}) (50) = + 2.5 \times 1 0^{- 4} J$ . Yeh step kyun? Ek positive charge jo high $V$ se low $V$ par jaata hai, field ke saath push hota hai — woh "downhill roll" karta hai, isliye field positive work karta hai.
(ii) $W = (- 5 \times 1 0^{- 6}) (50) = - 2.5 \times 1 0^{- 4} J$ . Yeh step kyun? Ek negative charge $E$ ke opposite direction mein force feel karta hai; space mein same direction mein jaana ab energy mein uphill jaana hai, isliye field negative work karta hai.

Verify: Same $Δ V$ , opposite $q$ , opposite work signs — $W \propto q$ ke consistent. Energy check: field dwara positive work = $+$ charge ki kinetic energy gain, Work–energy theorem in electrostatics se match karta hai. ✓

Ex 7 — Degenerate case: infinite line of charge (Cell G)

Worked example Statement

Ek infinite straight line linear charge density $λ$ carry karti hai. Distance $s$ par uska field $E = \frac{λ}{2 π ε _{0} s}$ radially out point karta hai. $V (\infty) = 0$ try karo; phir problem fix karo.

Forecast: Hum $1/ s$ ko infinity tak integrate karte hain. Calculus se yaad karo $\int^{\infty} \frac{d s}{s}$ kya karta hai — guess karo kya $V$ finite rehta hai.

Steps.

Naive attempt: $V (s) = - \int_{\infty}^{s} \frac{λ}{2 π ε _{0} s ^{'}} d s^{'} = - \frac{λ}{2 π ε _{0}} [ln s^{'}]_{\infty}^{s} .$ Yeh step kyun? Radial path phir se $E \cdot d l = E d s$ banata hai.
$ln s^{'}$ upper limit $\infty$ par blow up karta hai — infinity par reference ek infinite number deta hai. Naive choice fail ho jaati hai. Yeh step kyun? Charge localized nahi hai; woh infinity tak stretch karta hai, isliye "infinitely far" charge se infinitely far nahi hai. Parent note ke mistake box ne exactly isi ka warning diya tha.
Fix: ek finite reference distance $s_{0}$ choose karo. Tab $V (s) - V (s_{0}) = - \frac{λ}{2 π ε _{0}} ln \frac{s}{s _{0}} = \frac{λ}{2 π ε _{0}} ln \frac{s _{0}}{s} .$ Yeh step kyun? Sirf potential ke differences physical hain; finite reference har number ko finite banata hai.

Verify: Jaise $s \to \infty$ fixed $s_{0}$ ke saath, $V \to - \infty$ (for $λ > 0$ ) — woh divergence real hai, isliye $\infty$ zero nahi ho sakta. $s < s_{0}$ ke liye (ek $+$ line ke nazdik) hume $ln (s_{0} / s) > 0$ milta hai, isliye line ke paas $V$ zyada hai — correct "hill toward positive charge." ✓

Ex 8 — Zero-field spot mein nonzero potential (Cell H)

Worked example Statement

Do equal charges $+ q = + 6 nC$ $(\pm 0.30, 0) m$ par rakhe hain. Midpoint $M = (0, 0)$ par, fields cancel ho jaate hain isliye $E = 0$ . Kya wahan $V = 0$ bhi hai?

Forecast: Fields $M$ par opposite directions mein point karte hain aur cancel ho jaate hain. Lekin potentials numbers hain — kya numbers yahan cancel hote hain? Compute karne se pehle guess karo.

Steps.

Har charge $M$ se $r = 0.30 m$ door hai. Yeh step kyun? $V$ ko distances chahiye, aur dono yahan equal hain.
$V = \frac{k q}{r} + \frac{k q}{r} = 2 \cdot \frac{( 8.99 \times 1 0 ^{9} ) ( 6 \times 1 0 ^{- 9} )}{0.30} = 2 (179.8) = 359.6 V .$ Yeh step kyun? Dono contributions positive hain (same-sign charges) — woh add hote hain, cancel nahi hote. Sirf vector $E$ cancel hota hai, kyunki uske arrows opposite directions mein point karte hain jabki do numbers $k q / r$ dono "upar" point karte hain.

Verify: $E = 0$ lekin $V = 359.6 V \neq = 0$ . Yahi crucial lesson hai: $E = - \nabla V$ slope measure karta hai, aur ek pahaad ka flat top ho sakta hai (zero slope) jabki abhi bhi upar ho (nonzero height). Ise yaad rakhne ka ek smart tarika. ✓

Ex 9 — Real-world word problem (Cell I)

Worked example Statement

Ek cloud-to-ground lightning setup roughly ek capacitor ki tarah behave karta hai: cloud base $500 m$ ground se upar hai, aur strike se pehle (near-uniform) field approximately $3 \times 1 0^{6} V/m$ hai (air ka breakdown field). Cloud aur ground ke beech potential difference estimate karo.

Forecast: Distance $d$ par uniform field: kaun sa relation $E$ , $Δ V$ , $d$ connect karta hai? Order of magnitude guess karo — millions? billions of volts?

Steps.

Ise parallel-plate gap ki tarah model karo: $∣Δ V ∣ = E d$ . Yeh step kyun? Strike channel par field almost constant hai, isliye $\int E d l$ $E \times d$ ban jaata hai (exactly Ex 4 ki logic) — cloud aur ground hamare do "plates" hain.
$∣Δ V ∣ = (3 \times 1 0^{6} V/m) (500 m) = 1.5 \times 1 0^{9} V .$ Yeh step kyun? Air-breakdown field aur $500 m$ gap ko $E d$ mein substitute karo; multiplication hi bacha hai kyunki $E$ constant treat hai.

Verify: $1.5 \times 1 0^{9} V$ (1.5 billion volts) — lightning ke liye correct real-world ballpark. Units: $(V/m) (m) = V$ . ✓ Itna bada number isliye hai ki air break down ho jaati hai aur ek spark banta hai.

Ex 10 — Exam twist: path independence (Cell J)

Worked example Statement

Origin par charge $+ q = + 2 nC$ hai. $a$ at $r_{a} = 0.10 m$ aur $b$ at $r_{b} = 0.25 m$ ke beech $V_{b} - V_{a}$ compute karo (i) ek straight radial line ke along, aur (ii) ek weird path ke along jo pehle sideways swing karta hai phir spiral out karta hai. Dikhao ki answer same hai.

Forecast: Electrostatic field conservative hai. Guess karo ki tedha path answer badalta hai ya nahi.

Steps.

Radial path: $V_{b} - V_{a} = \frac{k q}{r _{b}} - \frac{k q}{r _{a}} = (8.99 \times 1 0^{9}) (2 \times 1 0^{- 9}) (\frac{1}{0.25} - \frac{1}{0.10}) .$ Yeh step kyun? $V (r) = k q / r$ already radial integral se aaya hai, isliye do radii ke beech difference sirf do $k q / r$ values ka difference hai.
$= 17.98 (4 - 10) = 17.98 \times (- 6) = - 107.88 V .$ Yeh step kyun? $1/0.25 = 4$ aur $1/0.10 = 10$ compute karo, subtract karo, aur $k q = 17.98$ se multiply karo; result negative hai kyunki $b$ zyada door hai (pahaad par neeche).
Crooked path: kisi bhi $d l$ ko radial part $d r \overset{r}{^}$ aur sideways part mein split karo. Sideways part $⊥ E$ hai, isliye $E \cdot d l$ sirf radial piece $E d r$ pick karta hai — step 1 ke identical. Yeh step kyun? $E \cdot d l$ $E$ ke perpendicular motion discard kar deta hai; kyunki $E$ purely radial hai, sirf tumhara radius mein change matter karta hai, wandering nahi.

Verify: Dono paths $V_{b} - V_{a} = - 107.88 V$ dete hain. Potential difference sirf endpoints par depend karta hai — conservative field ki pehchaan, aur reason ki $V$ bilkul well-defined hai. Equipotential surfaces and why E ⟂ them dekho: equipotential ke along sideways motion se koi potential change nahi hota. ✓

Recall

Recall Matrix par quick self-test

$+$ charge ke paas aate hue, $V$ upar jaata hai ya neeche? ::: Upar — yeh ek pahaad hai; $V = k q / r$ badata hai jaise $r$ kam hota hai. $-$ charge ke paas, $V$ ka sign? ::: Har jagah negative; same $V = k q / r$ use karo $q < 0$ ke saath. Kai charges ke potentials kaise combine hote hain? ::: Har $k q / r$ ka plain algebraic sum (scalar). $E$ ke along chalne par, $V$ upar jaata hai ya neeche? ::: Neeche — $E$ lower $V$ ki taraf point karta hai. Kya $E = 0$ ho sakta hai jabki $V \neq = 0$ ho? ::: Haan — do equal charges ke beech flat hilltop. Infinite line ke liye $\infty$ reference kyun fail hoti hai? ::: $\int^{\infty} d s / s = ln \infty$ diverge karta hai; finite reference use karo. Kya tedha path $V_{b} - V_{a}$ badalta hai? ::: Nahi — field conservative hai; sirf endpoints matter karte hain. Potential se ek formula mein field? ::: $E = - \nabla V$ .

Connections

1.8.08 Electric potential — definition V = −∫E·dl (Hinglish)
Electric field E — definition and Coulomb's law
Potential energy of a charge configuration
Equipotential surfaces and why E ⟂ them
Gradient operator ∇ and directional derivatives
Work–energy theorem in electrostatics
Capacitance and parallel-plate fields (E = ΔV/d)