Equipotential surfaces — perpendicular to field: Question bank

1.8.10 · D5 · Physics › Electromagnetism › Equipotential surfaces — perpendicular to field

Shuru karne se pehle, ek one-line vocabulary refresh taaki neeche koi word "unearned" na lage:

Koi bhi item padhne se pehle neeche ki picture dekho — isme woh har idea hai jo yeh bank test karta hai.

Figure — Equipotential surfaces — perpendicular to field

Cyan loops equipotentials hain (barabar "electric height"); amber arrows $E$ hain, jo hamesha unhe $9 0^{\circ}$ par cross karte hain aur lower $V$ ki taraf point karte hain. Notice karo jahan loops crowd karte hain wahan arrows lambe hain — yahi hai $E = - d V / d n$ ek nazar mein.

True or false — justify

True or false: Ek single equipotential surface par, ek charge ko zyada lambe raaste se move karna zyada kaam leta hai.

False — electric force conservative hai, isliye

W = q (V_{A} - V_{B})

; dono points ka

V

same hai, toh

W = 0

har path ke liye regardless of uski length. Dekho Work and Conservative Forces.

True or false: Field

E

kisi special point par ek equipotential ka tangent ho sakta hai.

False — ek tangential

E

surface ke saath slide karte charge par kaam karta, lekin equipotential ke along

d V = 0

ka matlab zero work hai; tangential component har jagah vanish karna chahiye.

True or false: Alag

V

ki do equipotentials ek point par touch kar sakti hain.

False — us shared point ko ek saath do values of

V

hold karni hogi, jo impossible hai kyunki

V

position ki single-valued function hai.

True or false: Equipotential surfaces hamesha spheres hoti hain.

False — spheres sirf single point charge ke liye aati hain; uniform field flat planes deta hai, aur ek general charge arrangement blobby shells deta hai jinka form

V = const

se match karta hai.

True or false: Agar do equipotentials barabar

Δ V

par draw ki gayi hain aur perpendicular distance

d n

mein equally spaced hain, toh unke beech ka field uniform hai.

True — barabar

Δ V

barabar perpendicular spacing

d n

(normal gap, koi slanted nahi) ke upar matlab

E = - d V / d n

constant hai, jo exactly uniform-field case hai (parallel plates). Dekho Parallel Plate Capacitor.

True or false: Field lines aur equipotential surfaces parallel families hain.

False — yeh perpendicular hain; field lines equipotentials ko

+

sign ke do strokes ki tarah cross karti hain, hamesha lower

V

ki taraf point karti hain.

True or false: Electrostatic equilibrium mein ek solid conductor ke andar puri volume ek equipotential hai.

True — andar

E = 0

ke saath,

d V = - E \cdot d l = 0

har internal displacement step

d l

ke liye, toh

V

kabhi nahi badalta: poora conductor, surface sahit, ek single equipotential hai. Dekho Conductors in Electrostatic Equilibrium.

True or false: Ek equipotential surface ek closed surface (jaise bubble) honi chahiye.

False — closed sirf isolated charges ke liye; uniform field infinite open planes deta hai, isliye "closed" definition ka part nahi hai.

True or false: Gradient

\nabla V

ek equipotential ke along point karta hai.

False — gradient normal ke along point karta hai (steepest change), yahi reason hai ki yeh constant-

V

surfaces ke perpendicular hota hai; dekho Gradient and Directional Derivative.

Spot the error

Error hunt: "

E = - \nabla V

, isliye

E

higher potential ki taraf point karta hai."

Minus sign usse flip kar deta hai —

\nabla V

higher

V

ki taraf point karta hai, toh

- \nabla V = E

lower

V

ki taraf point karta hai, yaani downhill.

Error hunt: "Equipotentials ek point charge ke around har jagah equally spaced hain."

Galat —

V \propto 1/ r

ke saath, barabar

Δ V

ki surfaces

r

badhne ke saath door hoti jaati hain, jo

1/ r^{2}

ke saath field ke weakening ko mirror karta hai.

Error hunt: "Kyunki

W = q (V_{A} - V_{B}) = 0

ek surface par, electric force kisi bhi path mein space mein zero work karta hai."

Sirf tab sach jab dono endpoints ek hi equipotential par hon; alag

V

ki surfaces ke beech move karo toh kaam nonzero hai.

Error hunt: "Field lines ek hi equipotential par start aur end ho sakti hain."

Woh ek ke andar run nahi kar saktin — ek field line hamesha ek alag (lower) equipotential ko cross karti hai, kyunki

E

ka us surface ke along koi component nahi jis par woh hai.

Error hunt: "Closely packed equipotentials automatically matlab strong field."

Sirf tab agar woh barabar

Δ V

increments par drawn hain aur tum perpendicular gap

d n

compare karo; strength ka claim

E = - d V / d n

padhta hai, toh spacing ko strength maanne se pehle

Δ V

fix karna zaroori hai.

Error hunt: "Ek charged conductor ke bilkul bahar

E

surface ke kisi bhi angle par point kar sakta hai."

Nahi — surface ek equipotential hai, isliye bahar

E

purely normal (perpendicular) hona chahiye; koi bhi tangential part free charges ko push karta jab tak self-cancel na ho jaaye.

Error hunt: "

d V = - E \cdot d l

dikhata hai ki

V

E

ki direction mein rise karta hai."

Dot product minus sign ke saath ulta dikhata hai: ek displacement

d l

$E$ ke along step lena

d V < 0

banata hai, toh

V

field ki direction mein girta hai.

Why questions

Kyun

E

ek equipotential ke perpendicular hona chahiye, ek clean sentence mein?

Kyunki

E \cdot d l = 0

har in-surface (tangent) displacement step

d l

ke liye, aur ek nonzero vector jo sab tangent directions ke perpendicular hai woh by definition surface ka normal hai.

"Surface ke along koi kaam nahi" wala fact perpendicularity ko kyun force karta hai?

Kisi bhi tangent ke along zero kaam ka matlab

E \cdot d l_{tangent} = 0

; kyunki yeh sab tangent directions ke liye hold karna chahiye,

E

ka koi tangential component nahi bachta aur woh purely normal point karta hai.

Equipotentials ek doosre ko kyun cross nahi karti?

Ek crossing point single-valued function

V

ke do alag values ek saath carry karega, jo logically impossible hai.

Ek parallel-plate capacitor ke andar equipotentials flat, evenly spaced planes kyun hain?

Field uniform hai,

E = E \overset{x}{^}

, toh

V = - E x + C

x

mein linear hai;

V = const

set karna

x

fix karta hai, barabar

Δ V

per barabar perpendicular spacing ke saath parallel planes deta hai.

Ek hiking-map contour picture ise itna achhi tarah kyun capture karta hai?

Map par height

V

ka role play karti hai; ek ball seedha steepest slope ke neeche roll karta hai, jo equal-height contours ke right angles par hota hai — exactly jaise

E

equipotentials ko cross karta hai.

Edge cases

Edge case: Ek region ka "equipotential" kya hai jahan

E = 0

ho (jaise conductor ke andar deep)?

Kyunki

d V = - E \cdot d l = 0

sab displacement steps

d l

ke liye, poora region ek equipotential volume mein collapse ho jaata hai, sirf ek surface nahi —

V

poore mein ek single constant hai.

Edge case: Ek point par jahan

E = 0

ho (do equal charges ke beech ek field null), kya hum phir bhi bol sakte hain

E ⊥

surface?

Zero vector ki koi direction nahi, toh "perpendicular" vacuously satisfied hai; surface

V = const

se abhi bhi well-defined hai, field ka wahan bas koi arrow nahi hai.

Edge case: Ek point charge par bilkul right (

r \to 0

), equipotentials ka kya hota hai?

Woh infinitely close crowd karte hain (

V \propto 1/ r \to \infty

), toh

E = - d V / d n \to \infty

— ek genuine singularity jahan surfaces bina bound ke pile up karti hain.

Edge case: Kisi bhi charge se bahut door (

r \to \infty

), barabar-

Δ V

equipotentials kaisi behave karti hain?

Woh infinitely door spread hoti hain kyunki

V \to 0

itni gently se,

E \to 0

se match karti hain — field fade hota hai aur "hill" flat ho jaati hai.

Edge case: Agar do conductors same potential par held hain, kya unke beech ka gap ek equipotential hai?

Sirf unki surfaces us value ko share karti hain; unke beech ka region generally doosri

V

values rakhta hai, toh woh ek equipotential nahi hai jab tak ki poora gap field-free na ho.

Edge case: Kya ek equipotential surface ka sharp corner ya kink ho sakta hai?

Haan — ek sharp conductor tip ke paas surface conductor ki geometric singularity inherit karti hai, toh equipotential apna corner ya kink develop kar sakti hai; equipotentials wahan tightly bunch bhi karti hain aur

E

spike karta hai (the lightning-rod effect), toh aisi points par smoothness guaranteed nahi hai.

Recall One-line self-test

Agar tum yeh answer kar sako toh topic tumhara hai: " $E$ kahan point karta hai, aur kis angle par, ek given equipotential ke relative — aur surface spacing tumhe kya batata hai?" Answer ::: $E$ surface ke normal (at $9 0^{\circ}$ ) point karta hai, lower $V$ ki taraf aim karta hai; barabar $Δ V$ par drawn equipotentials perpendicular distance $d n$ mein close hain jahan $E$ strong hai aur door hain jahan $E$ weak hai.

Connections

Electric Potential — woh scalar $V$ jiske level sets yeh surfaces hain.
Electric Field Lines — equipotentials ko $9 0^{\circ}$ par cross karti hain.
Gradient and Directional Derivative — kyun $\nabla V$ level sets ka normal hai.
Conductors in Electrostatic Equilibrium — killer boundary case.
Work and Conservative Forces — kyun path length ek surface par irrelevant hai.
Parallel Plate Capacitor — uniform-field, flat-plane example.