Equipotential surfaces — perpendicular to field: Worked examples

1.8.10 · D3 · Physics › Electromagnetism › Equipotential surfaces — perpendicular to field

Yeh page ek workout hai. Parent note ne ek bada fact establish kiya tha — electric field $E$ hamesha equipotential surface ke perpendicular hota hai, aur $E = - \nabla V$ . Yahan hum us fact ko har tarah ki situation mein le jaate hain, taaki jab exam ya real problem koi case pheke, tumne pehle se uska "twin" dekha hua ho.

Shuru karne se pehle, teen symbols jo hum baar baar use karenge, plain words mein:

Scenario matrix

Is topic ke har problem ka answer inhi cells mein se ek mein hai. Neeche ke examples cell(s) label ke saath hain, aur sab milke inhe cover karte hain.

#	Case class	Isme kya special hai	Covered by
A	Positive point charge	Spheres, $E$ baahir point karta hai	Ex 1
B	Negative point charge	Spheres, $E$ andar point karta hai (sign flip)	Ex 2
C	Uniform field (parallel plates)	Flat, equally spaced planes	Ex 3
D	Do equal charges (dipole-ish geometry)	Curved surfaces, ek zero-field saddle point	Ex 4
E	Degenerate: surface pe hi	Motion along — $W = 0$ dena hi chahiye	Ex 5
F	Limiting behaviour: $r \to \infty$ aur $r \to 0$	Spacing spreads / crowds without bound	Ex 6
G	Conductor surface	Poora body ek equipotential, $E ⊥$ surface	Ex 7
H	Real-world word problem	Units ke saath numbers, capacitor-style	Ex 8
I	Exam twist: sign / direction trap	Jahan naive answer galat hota hai	Ex 9

Cells A aur B source ki sign split karte hain. C, D shapes cover karte hain. E, F degenerate/limiting inputs cover karte hain. G killer application hai. H, I word-problem aur trap cells hain.

Ex 1 — Positive point charge (Cell A)

Worked example Ek charge

$+ Q = + 2 nC$ origin pe baitha hai. $r = 0.30 m$ pe $V$ aur $E$ nikaalo, aur check karo ke woh wahan equipotential ke perpendicular hain.

Forecast: Equipotential ki shape guess karo aur yeh bhi ke $E$ charge ki taraf point karega ya usse door — aage padhne se pehle.

Step 1 — Potential likho. Point charge ke liye, $V (r) = k \frac{Q}{r}, k = 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2} .$ Yeh step kyun? $V$ sirf $r$ par depend karta hai (charge se distance). " $V = const$ " ka matlab isliye " $r = const$ " hai — ek sphere. Isse hi equipotential ki shape pata chal jaati hai.

Step 2 — Numbers plug karo. $V = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.30} = 59.9 V .$ Yeh step kyun? Hum apne "landscape" pe ek concrete height chahte hain taaki picture anchor ho sake.

Step 3 — Field magnitude. $E = k \frac{Q}{r ^{2}} = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.3 0 ^{2}} = 199.8 V/m .$ Yeh step kyun? $E = - d V / d n$ aur yahan normal step $d n$ bas radial step $d r$ hai (spheres ka normal radial hota hai); $Q / r$ differentiate karne se $- Q / r^{2}$ aata hai, toh field $V$ se zyada fast fall off karta hai.

Step 4 — Direction & perpendicularity. $E = E \overset{r}{^}$ radially outward point karta hai ( $+ Q$ se door). Outward radial direction $\overset{r}{^}$ exactly sphere $r = const$ ka normal hai. Toh $E ⊥$ equipotential. ✓

Neeche wali figure in teeno spheres (blue) aur field arrows (orange) ko dikhati hai jo unse seedhe baahir nikal rahe hain. Woh chhota green right-angle mark dekho jahan ek arrow ek sphere cross karta hai: woh $9 0^{\circ}$ crossing Cell A ka poora point hai — aur notice karo ke charge ke paas arrows zyada tight packed hain, jo wahan stronger field ko foreshadow karta hai.

Verify: Units: $\frac{N\cdotm ^{2} / C ^{2} \cdot C}{m ^{2}} = N/C = V/m$ ✓. Sanity: $E / V = 1/ r = 1/0.30 = 3.33 m^{- 1}$ , aur sach mein $199.8/59.9 = 3.34$ ✓.

Figure — Equipotential surfaces — perpendicular to field

Ex 2 — Negative point charge (Cell B — sign flip)

Worked example Source ko

$- Q = - 2 nC$ origin pe replace karo. $r = 0.30 m$ pe kya badlega?

Forecast: Ab $E$ kis taraf point karega — aur kya perpendicularity fact survive karega?

Step 1 — Potential negative ho jaata hai. $V = (8.99 \times 1 0^{9}) \frac{- 2 \times 1 0 ^{- 9}}{0.30} = - 59.9 V .$ Yeh step kyun? $Q$ ki sign seedha $V$ mein chali jaati hai. "Landscape" ab ek hill nahi, ek valley hai.

Step 2 — Equipotentials phir bhi spheres hain. " $V = const$ " abhi bhi " $r = const$ " force karta hai. Shape unchanged. Yeh step kyun? Geometry $r$ par dependence pe depend karti hai, sign pe nahi — yahi cell ka point hai.

Step 3 — Field direction flip hoti hai. $E = k \frac{( - Q )}{r ^{2}} \overset{r}{^} = - 199.8 \overset{r}{^} V/m,$ yaani magnitude $199.8 V/m$ inward point karta hai (charge ki taraf). Yeh step kyun? $E = - \nabla V$ : kyunki jab tum negative charge se door jaate ho toh $V$ increase karta hai ( $- 59.9$ se $0$ ki taraf), field opposite direction mein point karta hai — andar.

Step 4 — Perpendicularity. Inward radial abhi bhi $\overset{r}{^}$ ke along hai (bas $- \overset{r}{^}$ ), abhi bhi sphere ka normal hai. $E ⊥$ equipotential hold karta hai. ✓

Verify: $∣ E ∣$ Ex 1 ke identical hai ( $199.8 V/m$ ) kyunki $E \propto ∣ Q ∣/ r^{2}$ . Direction reverse hai. Consistent hai " $E$ higher $V$ ( $0$ dur) se lower $V$ ( $- 59.9$ charge ke paas) ki taraf point karta hai" ke saath — dekho Work and Conservative Forces yeh jaanne ke liye ke downhill lower $V$ ki taraf kyun hai.

Ex 3 — Uniform field / parallel plates (Cell C)

Worked example Do plates

$d = 5.0 mm$ apart hain aur $Δ V = 100 V$ potential difference hold karti hain. $E$ nikalo, $Δ V = 20 V$ steps ke liye equipotential spacing nikalo, aur confirm karo ke planes flat aur evenly spaced hain. Dekho Parallel Plate Capacitor.

Forecast: $20 V$ sheets evenly spaced hongi ya unevenly?

Step 1 — Spacing se field. $E = \frac{Δ V}{d} = \frac{100}{5.0 \times 1 0 ^{- 3}} = 2.0 \times 1 0^{4} V/m .$ Yeh step kyun? Plates ke andar $E$ uniform hai, toh $E = - d V / d n$ ek plain ratio ban jaata hai (calculus nahi chahiye): total drop over total gap, kyunki normal step $d n$ plate se plate seedha cross karta hai.

Step 2 — $V$ gap mein linear hai. $E = E \overset{x}{^}$ ke saath, integrate karne se $V = - E x + C$ aata hai. Toh " $V = const$ " $\Rightarrow$ " $x = const$ ": flat planes $\overset{x}{^}$ ke perpendicular. Yeh step kyun? Linear $V$ hi sheets ko flat banata hai aur (next step) evenly spaced bhi.

Step 3 — $20 V$ step ke liye spacing. Har equal $Δ V$ ko equal $Δ x$ chahiye: $Δ x = \frac{Δ V _{step}}{E} = \frac{20}{2.0 \times 1 0 ^{4}} = 1.0 \times 1 0^{- 3} m = 1.0 mm .$ $5.0 mm$ gap mein paanch equal sheets. ✓ Yeh step kyun? Equal $Δ V$ aur equal spacing $\Rightarrow$ uniform field — parallel-plate cell ka signature.

Neeche wali figure do plates (red $+ 100 V$ , blue $0 V$ ), $20 V$ steps pe chaar green dashed equipotential planes, aur high se low jaate orange field arrows dikhati hai. Notice karo ke arrows sab same length ke hain aur dashed planes sab same distance pe hain — woh even spacing hi uniform field ka visual signature hai.

Verify: $5$ steps $\times 1.0 mm = 5.0 mm = d$ ✓. Units: $V / m$ from $V / m$ ✓.

Ex 4 — Do equal positive charges: saddle / zero-field point (Cell D)

Worked example Do charges

$+ Q = + 2 nC$ $x$ -axis pe $x = \pm a$ pe baithey hain jahan $a = 0.10 m$ hai. Dikhao ke midpoint (origin) pe field zero hai, phir bhi woh point ek perfectly good equipotential pe baitha hai.

Forecast: Agar kisi point pe $E = 0$ ho, toh "equipotential ke perpendicular" ka matlab kya hoga?

Step 1 — Origin pe har charge ka field. Har charge $a$ distance pe hai: $E_{each} = k \frac{Q}{a ^{2}} = (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.1 0 ^{2}} = 1798 V/m .$ Yeh step kyun? Hum total field superposition se build karte hain — do arrows add karo.

Step 2 — Woh cancel ho jaate hain. Left charge ek test $+$ charge ko right ( $+ \overset{x}{^}$ ) push karta hai; right charge use left ( $- \overset{x}{^}$ ) push karta hai. Equal magnitudes, opposite directions: $E_{total} = 1798 \overset{x}{^} - 1798 \overset{x}{^} = 0.$ Yeh step kyun? Yahi woh degenerate zero-field input hai jo matrix demand karta hai.

Step 3 — Potential wahan zero nahi hai. Potential ek scalar hai — koi cancellation nahi: $V_{origin} = 2 \times k \frac{Q}{a} = 2 \times (8.99 \times 1 0^{9}) \frac{2 \times 1 0 ^{- 9}}{0.10} = 359.6 V .$ Yeh step kyun? Yeh dikhata hai ke $V$ aur $E$ alag creatures hain: numbers add karna vs. arrows add karna.

Step 4 — " $⊥$ " ka matlab jab $E = 0$ ho. Jahan $E = 0$ , "downhill direction" undefined hai — steepest descent ki koi direction nahi, toh perpendicularity vacuously satisfied hoti hai. Origin potential landscape ka ek saddle hai: $x$ -axis ke along har charge ki taraf step lena uphill hai, lekin axis se hat ke yeh doosri taraf curve karta hai. Saddle ke paas equipotentials mountain pass ki crossed hyperbolas jaisi dikhti hain — lekin do branches same $V$ value ki hain, toh yeh do alag equipotentials ka cross karna nahi hai (woh abhi bhi forbidden hai).

Verify: $E_{total} = 0$ exactly by symmetry ✓. $V_{origin} = 2 \times 179.8 = 359.6 V$ ✓ ( $0.10 m$ pe single-charge value ka double).

Ex 5 — Degenerate move: equipotential ke along slide karna (Cell E)

Worked example Origin pe

$+ Q = + 2 nC$ lo (Ex 1). Ek charge $q = + 5 nC$ ko point $A = (0.30, 0) m$ se point $B = (0, 0.30) m$ tak circular arc $r = 0.30 m$ ke along drag kiya jaata hai. Electric force dwara kiya gaya work nikalo.

Forecast: Compute karne se pehle work guess karo. Kya curvy path matter karta hai?

Step 1 — Kya $A$ aur $B$ same equipotential pe hain? Dono $r = 0.30 m$ pe hain, toh dono ka $V = 59.9 V$ hai (Ex 1 se). Haan — same sphere. Yeh step kyun? Equipotential ka poora point yahi hai ke "same $V$ " hi matter karta hai.

Step 2 — Work formula apply karo. $W_{A B} = q (V_{A} - V_{B}) = q (59.9 - 59.9) = 0 J .$ Yeh step kyun? Electric force conservative hai (Work and Conservative Forces): work sirf endpoints pe depend karta hai, aur yahan endpoints tie karte hain.

Step 3 — Geometric echo. Arc ke along, har step $d l$ sphere ke tangent hai, aur $E$ radial (normal) hai. Unka dot product $E \cdot d l = 0$ har point pe hai, toh koi work accumulate nahi hota — Step 2 se consistent. Yeh step kyun? Yeh algebra ( $V_{A} = V_{B}$ ) ko geometry (perpendicularity) se connect karta hai — dono same fact hain.

Verify: $W = 5 \times 1 0^{- 9} \times (59.9 - 59.9) = 0$ ✓. Same sphere pe koi bhi doosra path bhi $0$ dega — path length irrelevant hai.

Ex 6 — Spacing ka limiting behaviour (Cell F)

Worked example

$+ Q = + 2 nC$ ke liye, $V = 60, 30, 15 V$ pe equipotentials draw karo (har ek pehle wale ka half). Unke radii nikalo aur $r \to \infty$ aur $r \to 0$ pe spacing ke baare mein comment karo.

Forecast: Jab voltage half karo, toh spheres ke beech gap barhta hai, ghatta hai, ya equal rehta hai?

Step 1 — Radii pane ke liye $V (r)$ invert karo. $V = \frac{k Q}{r}$ se (jahan $k$ Coulomb's constant hai, toh $k Q = 17.98 V\cdotm$ ): $r = \frac{k Q}{V} .$ Yeh step kyun? Hum surfaces ki positions chahte hain taaki unki spacing measure kar sakein.

Step 2 — Compute karo. $r_{60} = \frac{17.98}{60} = 0.300 m, r_{30} = \frac{17.98}{30} = 0.599 m, r_{15} = \frac{17.98}{15} = 1.199 m .$ Gaps: $r_{30} - r_{60} = 0.299 m$ , phir $r_{15} - r_{30} = 0.599 m$ — har baar double ho raha hai. Yeh step kyun? Yeh spreading ko concrete banata hai.

Step 3 — Do limits.

Jab $r \to \infty$ ( $V \to 0^{+}$ ): equal $Δ V$ ke liye spheres zyada aur zyada door ho jaati hain → field $0$ ki taraf weak hota hai ( $E \sim 1/ r^{2}$ ).
Jab $r \to 0$ ( $V \to \infty$ ): spheres infinitely close crowd ho jaate hain → field blow up karta hai. Point charge ek singular ("infinitely steep") spot hai.

Yeh step kyun? Yeh woh boundary inputs hain jo matrix insist karta hai — reader ab jaanta hai extremes pe kya hota hai, sirf comfortable middle mein nahi.

Verify: Ratio check — $V$ half karne se $r$ double hota hai: $r_{30} / r_{60} = 0.599/0.300 = 2.00$ ✓, $r_{15} / r_{30} = 1.199/0.599 = 2.00$ ✓.

Ex 7 — Conductor surface (Cell G — killer application)

Worked example Radius

$R = 0.05 m$ wala ek solid metal sphere $Q = + 2 nC$ carry karta hai. Andar aur surface pe $V$ nikalo, aur baahir field direction explain karo. Dekho Conductors in Electrostatic Equilibrium.

Forecast: Metal ka inside surface ke compare mein lower, higher, ya equal potential pe hai?

Step 1 — Conductor ke andar $E = 0$ . Free electrons rearrange ho jaate hain jab tak andar koi field nahi rehta (electrostatic equilibrium). Yeh step kyun? $E = 0$ conductors ke liye starting axiom hai.

Step 2 — Zero field ⇒ constant $V$ . Kyunki $E = - d V / d n = 0$ andar har jagah hai, $V$ change nahi ho sakta: poora sphere, surface samete, ek equipotential hai. $V_{surface} = \frac{k Q}{R} = \frac{17.98}{0.05} = 359.6 V .$ Yeh step kyun? Charged sphere ke baahir field point charge jaisa hota hai, toh $V = k Q / r$ ; $r = R$ pe evaluate karne se constant interior value milti hai.

Step 3 — Baahir field perpendicular hai. Surface ek equipotential hai, toh $E$ (jo har equipotential ke $⊥$ hai) metal se exactly $9 0^{\circ}$ pe milna chahiye. Koi bhi tangential component surface charges ko push karta jab tak woh cancel nahi ho jaata. Yeh step kyun? Isliye field lines metal se right angles pe strike karti hain — yahi reason hai ke yeh topic real devices mein matter karta hai.

Verify: $V_{surface} = 17.98/0.05 = 359.6 V$ ✓. Note karo yeh Ex 4 ki origin value se coincidentally match karta hai kyunki dono ne $k Q$ aur $0.10$ / $0.05$ -type ratio use ki; yahan $r = R = 0.05$ .

Ex 8 — Real-world word problem (Cell H)

Worked example Ek smoke detector mein air-gap capacitor ki plates

$2.0 mm$ apart hain aur $9.0 V$ across hain. (a) Plates ke beech field kya hai? (b) Ek dust particle jo $q = + 1.6 \times 1 0^{- 19} C$ carry karta hai low plate se high plate tak poora move kiya jaata hai — external agent field ke against kitna work karta hai? (c) $3.0 V$ equipotential sheets kitni door hain?

Forecast: Kya work positive ya negative hoga ek external agent ke liye jo $+$ charge ko high plate ki taraf push kar raha hai?

Step 1 — Field (uniform). $E = \frac{V}{d} = \frac{9.0}{2.0 \times 1 0 ^{- 3}} = 4500 V/m .$ Yeh step kyun? Parallel plates ⇒ uniform field ⇒ simple ratio $E = - d V / d n$ total drop over gap ban jaata hai.

Step 2 — External agent ka work. $+ q$ ko low plate ( $V = 0$ ) se high plate ( $V = 9.0 V$ ) move karne par, field karta hai $W_{field} = q (V_{A} - V_{B}) = q (0 - 9.0)$ . Agent opposite karta hai: $W_{agent} = q (V_{B} - V_{A}) = (1.6 \times 1 0^{- 19}) (9.0) = 1.44 \times 1 0^{- 18} J .$ Positive — agent ko $+$ charge ko potential hill upar push karna padta hai. Yeh step kyun? Sign discipline: agent work $= + q Δ V$ field ke against.

Step 3 — $3.0 V$ sheets ki spacing. $Δ x = \frac{3.0}{4500} = 6.67 \times 1 0^{- 4} m = 0.667 mm .$ Teen sheets $2.0 mm$ gap fill karti hain. Yeh step kyun? Equal- $Δ V$ spacing phir se — everyday capacitor picture.

Verify: $3 \times 0.667 mm = 2.0 mm = d$ ✓. $W_{agent} = 1.6 \times 1 0^{- 19} \times 9.0 = 1.44 \times 1 0^{- 18} J$ ✓. Units: $C \cdot V = J$ ✓.

Ex 9 — Exam twist: direction trap (Cell I)

Worked example Do plates ke beech, left plate

$V = + 40 V$ pe hai aur right $V = + 10 V$ pe, $d = 0.020 m$ separated. Ek student likhta hai " $E$ left se right point karta hai kyunki left zyada positive hai... ruko, ya doosra taraf?" $E$ ka magnitude aur correct direction nikalo.

Forecast: Left plate higher potential pe hai. $E$ usse door point karta hai ya uski taraf?

Step 1 — Magnitude. $E = \frac{∣ V _{left} - V _{right} ∣}{d} = \frac{∣40 - 10∣}{0.020} = \frac{30}{0.020} = 1500 V/m .$ Yeh step kyun? Sirf potential difference aur gap strength set karte hain; absolute values ( $40$ vs $10$ ) matter nahi karte.

Step 2 — Direction $E = - \nabla V$ se. Field $V$ mein downhill point karta hai, yaani higher-potential left plate ( $40 V$ ) se lower-potential right plate ( $10 V$ ) ki taraf — toh left → right. Dekho Gradient and Directional Derivative: $- \nabla V$ decreasing $V$ ki taraf point karta hai. Yeh step kyun? Trap yeh sochna hai " $+$ plate repel karta hai, toh field jaata hai...?" Clean rule hamesha yeh hai: "field high $V$ se low $V$ ki taraf point karta hai", koi repulsion story nahi chahiye.

Step 3 — Perpendicularity check. Yahan equipotentials vertical planes hain ( $x = const$ ); $E$ horizontal hai ( $x$ ke along) — unse $9 0^{\circ}$ pe milta hai. ✓

Verify: $E = 30/0.020 = 1500 V/m$ ✓. Direction: high $\to$ low = left $\to$ right ✓. (Agar dono plates negative hote, maano $- 40$ aur $- 10$ , field tab bhi $- 40$ se $- 10$ ki taraf point karta — yaani less negative plate ki taraf — absolute sign kabhi nahi aata, sirf difference aata hai.)

Recall Matrix ke across quick self-test

Sign flip point charge ke equipotentials ke baare mein kya change karta hai? ::: Sirf field direction (aur $V$ ki sign); spherical shape unchanged rehti hai. Zero-field saddle point pe, kya $E ⊥$ equipotential hai? ::: Vacuously haan — koi field direction nahi hai, toh koi violation nahi; aur crossing hyperbolas same $V$ ki hain, do alag surfaces nahi. Ek equipotential ke along charge slide karne ka work? ::: Zero, kisi bhi path ke liye — force conservative hai aur $V_{A} = V_{B}$ . Point charge ke liye $r \to \infty$ jaane par, equal- $Δ V$ spheres... ::: zyada door ho jaati hain (field $1/ r^{2}$ ke saath weak hota hai). $+ 40 V$ aur $+ 10 V$ wali plates ke beech $E$ ki direction? ::: $40 V$ plate se $10 V$ plate ki taraf (high $V$ → low $V$ ).

Connections

Parent topic
Electric Potential · Electric Field Lines · Gradient and Directional Derivative
Conductors in Electrostatic Equilibrium · Parallel Plate Capacitor · Work and Conservative Forces