Electric potential — definition V = −∫E·dl: Question bank

1.8.8 · D5 · Physics › Electromagnetism › Electric potential — definition V = −∫E·dl

Shuru karne se pehle, kuch symbols apni jagah earn karte hain taaki neeche koi mark aisa na ho jo tumne dekha na ho:

Definition Characters ki cast

Potential $V$ — har point se attached ek number: us jagah ki electrical "height", volts mein.
Field $E$ — har point par ek arrow: yeh $V$ mein neeche ki taraf point karta hai aur iski length steepness hai.
Test charge $q_{0}$ — ek tiny positive charge jise hum kisi point par imagine karte hain use probe karne ke liye, itna chhota ki woh khud us field ko disturb na kare jise woh measure kar raha hai. Jab koi problem kisi real, arbitrary charge ki baat kare toh use $q$ kehte hain.
Potential difference $Δ V = V_{b} - V_{a}$ — do points ke beech ka height gap; yahi ek cheez physically measurable hai.
Coulomb constant $k$ — $V = \frac{k q}{r}$ aur Coulomb's law mein fixed proportionality number, $k = \frac{1}{4 π ε _{0}} \approx 9 \times 1 0^{9} N\cdotm^{2} / C^{2}$ . Yeh sirf yeh set karta hai ki electric effects kitni strong hain; ise ek positive constant maano. Dekho Electric field E — definition and Coulomb's law.
Plate separation $d$ — do parallel charged plates ke liye, $d$ simply unke beech ka gap (distance) hai, field ke along measure kiya gaya. Yeh $E = Δ V / d$ mein aata hai. Dekho Capacitance and parallel-plate fields (E = ΔV/d).

Figure 1 neeche tumhara reference "hill" hai har height-vs-slope question ke liye (True/False items 2, 3, 7 aur scalars par Why questions). Figure 2 reference equipotential-and-perpendicular-field picture hai equipotential items aur shell edge case ke liye — jab bhi koi prompt "downhill", "perpendicular", ya "same height" mention kare toh inhe dobara dekho.

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

Ek teesri picture "divergent reference" aur "same potential, nonzero field" ke do traps ek nazar mein settle kar deti hai — ise Edge cases section ke saath rakhna:

True or false — justify karo

Har item ek claim hai. Reveal verdict aur reason deta hai — sirf "true/false" bolna kuch kaam ka nahi.

Potential

V

ek vector hai kyunki yeh vector

E

se bana hai.

False.

V

ek scalar hai — dot product

E \cdot d l

arrow ko integrate karne se pehle ek number mein collapse kar deta hai. Direction

E = - \nabla V

mein hoti hai,

V

mein nahi.

Agar kisi point par

E = 0

hai, toh wahan

V = 0

bhi hai.

False.

E

V

ki slope hai (Figure 1); slope zero ka matlab hai

V

flat hai (hilltop ya valley), yeh nahi ki uski height zero hai. Do equal

+

charges ke beech mein

E = 0

hota hai lekin

V

large aur positive hota hai.

Agar kisi point par

V = 0

hai, toh wahan

E = 0

bhi hai.

False. Height zero steepness ke baare mein kuch nahi kehti (Figure 1: ek hill zero height cross karte hue bhi slope kar sakti hai).

+ q

aur

- q

ke beech mein

V = 0

hota hai phir bhi

E

strong hota hai (dono charges same direction mein push karte hain).

Equipotential surface ke along move karne par, field zero work karta hai.

True. Equipotential par

Δ V = 0

hai, aur field ka kaam

= q (V_{a} - V_{b}) = 0

. Geometrically

E ⊥

surface hai (Figure 2), toh har step ke liye

E \cdot d l = 0

— dekho Equipotential surfaces and why E ⟂ them.

Rest se release hua positive charge hamesha higher potential ki taraf move karta hai.

False. Field positive work karta hai jab charge lower

V

ki taraf move kare, toh ek free positive charge

V

-hill ke neeche accelerate karta hai (Figure 1), lower potential ki taraf. (Negative charge higher

V

ki taraf jaata hai.)

V = 0

ko infinity par set karna hamesha ek valid choice hai.

False. Infinite line ya plane of charge ke liye integral infinity tak diverge karta hai (Edge cases mein

\int d r / r

aur

\int d r

arguments dekho); tumhe koi finite reference chunna hoga. Sirf

Δ V

physical hai, toh reference ek free choice hai — lekin exist zaroor karna chahiye.

Do points ka potential same ho sakta hai phir bhi unke beech field nonzero ho.

True. Do jagahon par equal height beech mein hill hone se nahi rokti (Figure 3);

\int_{a}^{b} E \cdot d l

cancellation se zero ho sakta hai jabki raaste mein

E \neq = 0

raha ho.

V = - \int E \cdot d l

mein minus sign sirf ek convention hai jise hum drop kar sakte hain.

False (physically). Yeh sign encode karta hai ki potential us direction mein decrease hota hai jis taraf

E

point karta hai; drop karo toh tum predict karoge ki positive charges spontaneously hills chadhte hain — jo Work–energy theorem in electrostatics ka violation hai.

Test charge

q_{0}

ko double karne par uski location par potential

V

double ho jaata hai.

False.

V

work per unit charge hai —

q_{0}

precisely isliye divide kiya jaata hai taaki

V

sirf location ki property rahe. Sirf potential energy

U = q_{0} V

double hoti hai.

Error dhundho

Har line mein ek flawed statement ya step hai. Reveal bug ka naam batata hai aur use repair karta hai.

"Plate ka field

E = Δ V / d

hai, toh badi plate spacing bada field deti hai."

Backwards causation. Fixed

Δ V

ke liye, bada gap

d

chhota

E = Δ V / d

deta hai. Dekho Capacitance and parallel-plate fields (E = ΔV/d).

V = \frac{k q}{r}

, toh

r \to 0

par potential zero hai."

Wrong limit. Jab

r \to 0

V = k q / r \to \infty

(blow up hota hai), zero nahi.

r \to \infty

hone par

V \to 0

hota hai (yahan

k

Coulomb constant hai, ek positive scale factor).

"Field ka work

W = q (V_{b} - V_{a})

hai."

Sign flipped. Yeh

W_{field} = q (V_{a} - V_{b})

hai: field positive work karta hai jab charge high

V_{a}

se low

V_{b}

tak gire.

E = + \nabla V

, kyunki field aur potential saath-saath badhte hain."

Missing minus.

E = - \nabla V

: field

V

ki steepest increase ke opposite point karta hai — neeche ki taraf, upar nahi. Dekho Gradient operator ∇ and directional derivatives.

"Test charge ko quasi-statically laane ke liye agent

F_{ext} = + q_{0} E

apply karta hai."

Wrong direction. Agent ko field oppose karna chahiye:

F_{ext} = - q_{0} E

, taaki net force zero rahe aur kinetic energy fixed rahe.

"Kyunki sirf force ka displacement ke along component work karta hai, hum

W = \int F d l

likhte hain (no dot)."

Dot product hi yeh statement hai "component along displacement"; ise drop karna silently assume karta hai ki force har step par already parallel hai, jo curved path par galat hai.

\int F \cdot d l

rakho.

"Point charge ke liye maine curved path par integrate kiya, toh angle matter karta hai aur

E \cdot d l \neq = E d r

Overcomplication.

V

electrostatic fields ke liye path-independent hai, toh ek radial path choose karo jahan

E ∥ d l

ho aur

E \cdot d l = E d r

ho. Result

k q / r

kisi bhi path par same hoga.

Why questions

V

ko sirf work ki jagah work per unit charge kyun define kiya jaata hai?

Kyunki force, aur isliye work, test charge

q_{0}

ke saath scale karta hai. Use divide karne par ek aisi quantity bachti hai jo sirf field/location describe karti hai, reusable hai kisi bhi charge ke liye jo tum baad mein wahan drop karo.

Test charge

q_{0}

ko kinetic energy change kiye bina kyun move karna chahiye?

Taaki saara agent-work potential energy ban jaaye (kuch bhi speed mein leak na ho). Isse

W_{ext} = Δ U

exact hota hai, jo ise

q_{0} Δ V

ke barabar rakhne deta hai.

Calculation ke liye scalar

V

, vector

E

se zyada useful kyun hai?

Scalars simply add ho jaate hain (koi components nahi, koi angles nahi); tum

V

contributions sum karo, phir poora vector

E = - \nabla V

recover karne ke liye ek gradient lo. Yeh "easy solve karo, hard recover karo" shortcut hai.

E

hamesha equipotential surfaces ke perpendicular kyun point karta hai?

Surface ke along

V

constant hota hai, toh us direction mein uski slope zero hai. Saari steepness — aur isliye

E

— ek remaining perpendicular direction mein honi chahiye (Figure 2). Dekho Equipotential surfaces and why E ⟂ them.

Sirf potential differences kyun measure ki ja sakti hain, absolute

V

nahi?

Physics

E = - \nabla V

par depend karti hai, aur

V

mein har jagah ek constant add karne par gradient unchanged rehta hai. Constant (reference height) unobservable hai, toh sirf

Δ V

kisi bhi experiment mein show hoti hai.

Potential energy

U

aur potential

V

alag kyun hain jabke dono similar lagte hain?

U = q_{0} V

: potential "height per kilogram" hai, energy actual mass times height hai. Dekho Potential energy of a charge configuration —

U

charge-plus-field ki property hai,

V

location ki property hai.

Field volts per metre mein kyun measure hota hai?

E = Δ V / d

se, field potential-drop per length hai — literally volts per metre, jo dikhata hai ki

E

woh rate hai jis par

V

-hill utarta hai jab tum chalte ho.

Edge cases

Har degenerate ya limiting scenario mein batao kya hota hai (aur kyun).

Ek charge exactly ek equipotential par baitha hai aur tum use us surface par ek closed loop mein move karte ho.

Net work zero hai — kyunki

Δ V = 0

(start=end height) aur kyunki

E ⊥

har step. Electrostatic fields conservative hote hain, toh koi bhi closed loopaise bhi zero deta hai.

Charge distribution ek infinite charged plane hai; reference kahan rakhoge?

Infinity par nahi. Ek plane constant field

E \sim const

deta hai, toh

\int^{\infty} E d r \sim \int^{\infty} d r

diverge karta hai (bina bound ke badhta hai). Koi convenient finite point chuno (aksar plane khud,

V = 0

wahan) aur

Δ V

usse measure karo.

Charge distribution ek infinite line hai; infinity yahan bhi kyun fail karta hai?

Line

E \sim 1/ r

deta hai, toh

\int^{\infty} \frac{d r}{r} = ln r^{\infty}

logarithmically diverge karta hai. Phir se finite reference chuno; physical answer hamesha

Δ V = ln (r_{1} / r_{2})

-type ratio hoga.

Hollow conducting shell ke andar (cavity mein koi charge nahi) —

V

kya hai?

V

andar har jagah constant hota hai, surface value ke barabar. Reason:

E = 0

interior mein poori jagah, toh

V_{b} - V_{a} = - \int_{a}^{b} E \cdot d l = 0

kisi bhi do interior points ke liye — koi slope nahi matlab uniform height, ek bilkul flat plateau.

Do identical charges

+ q

ek doosre se

2 a

door hain; tum midpoint par ho, toh har charge

r = a

door hai.

V

aur

E

kya hain?

V = k q / a + k q / a = 2 k q / a

large aur positive hai (scalars add hote hain);

E = 0

kyunki dono equal arrows opposite directions mein point karte hain aur cancel ho jaate hain. Ek hilltop: zyada height, zero slope.

Ek dipole

+ q, - q

ek doosre se

2 a

door; tum midpoint par ho, har charge

r = a

door.

V

aur

E

kya hain?

V = k q / a - k q / a = 0

(equal-and-opposite scalars cancel) lekin

E \neq = 0

(dono arrows same direction mein point karte hain,

+

-

ki taraf). Zero height, nonzero slope — yeh Figure 3 wali situation hai.

Field uniform hai (

E

constant) — field ke along

V

kaise vary karta hai?

Linearly:

V

har metre par

E

se girta hai, ek seedha, tilted ramp deta hai. Equipotentials equally spaced parallel planes hain

E

ke perpendicular.

Tum ek test charge infinity se ek point tak laate ho negative charge ke paas;

V

ka sign kya hai?

Negative. Field inward point karta hai (

- q

ki taraf), incoming positive charge par positive work karta hai, toh uski potential energy — aur

V = U / q_{0}

— zero se neeche aati hai:

V = - k ∣ q ∣/ r

Ek ideal point charge ki exact location (

r = 0

) par

V

kya hai?

Yeh

\pm \infty

tak diverge karta hai (

V = k q / r

). Yeh idealised point ka ek artefact hai; real charges finite size ke hote hain, toh physically koi true infinity nahi hoti.

Recall Ek-line self-test

Agar ek positive charge rest se release hone par speed up karta hai, toh uska potential kis taraf gaya? Uski potential energy kis taraf gayi? Answer ::: Potential decrease hua ( $V$ -hill se neeche lower $V$ ki taraf rola); potential energy $U = q_{0} V$ bhi decrease hui, kinetic energy mein convert hote hue — exactly Work–energy theorem in electrostatics.

Connections

Parent: Electric potential — definition
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)