Potential of point charge, potential from field and vice versa: Question bank

1.8.9 · D5 · Physics › Electromagnetism › Potential of point charge, potential from field and vice ver

Shuru karne se pehle, neeche ka anchor picture dhyan se dekho. Black curve ek potential landscape $V (x)$ hai — axis ke upar uski height hi us position $x$ pe $V$ ki value hai. Teen points marked hain: bottom-left pe almost flat spot (choti slope ⇒ almost koi field nahi), hilltop peak pe (zero slope ⇒ $E = 0$ bhalee bhaanti badi $V$ ke bawajood), aur steep right-hand slope pe ek point jahan akela red arrow $E = - d V / d x$ ko slope se neeche point karte hua dikhata hai. Jab bhi koi trap "height" aur "slope" ki baat kare, yeh picture dobara dekho:

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

True or false — justify

A charge feels no force at a point ⇒ the potential is zero there.

False.

E

V

ki local slope hai;

V

height hai. Do equal

+ q

charges ke midpoint pe, symmetry field cancel kar deti hai (

E = 0

) lekin dono charges height badhate hain, toh

V > 0

— ek flat hilltop (jaise anchor figure ka peak) ki bhi altitude hoti hai.

Potential is zero at a point ⇒ the field is zero there.

False.

+ q

aur

- q

ke beech midpoint pe heights cancel ho jaati hain (

V = 0

), lekin dono fields ek hi taraf point karte hain (

- q

ki taraf), toh

E \neq = 0

— flat value flat slope nahi hoti.

For a point charge,

V

aur

E

ek hi rate se fall off hote hain.

False.

V \propto 1/ r

lekin field magnitude

E \propto 1/ r^{2}

; potential dheere fall off hoti hai kyunki yeh field ka accumulated integral hai, jo ek extra power of

r

add karta hai.

Kyunki

V

superpose karta hai aur

E

superpose karta hai, dono ko plain numbers ki tarah add kar sakte hain.

False. Sirf

V

signed numbers ki tarah add hota hai (yeh scalar hai).

E

vector ki tarah superpose karta hai — tumhe components add karney hote hain; magnitudes add karne se galat direction aur magnitude milti hai.

Agar

V

ek path ke along constant hai, toh ussi path pe charge move karne mein koi work nahi hota.

True. Kyunki

V

defined hai work per unit charge ki tarah, jahan

V

change nahi hota wahan move karne ka matlab hai work per charge

= Δ V = 0

, toh

W = q Δ V = 0

. Yahi reason hai kyun Equipotential Surfaces ke along slide karne mein koi work nahi lagta.

Kisi point charge ka potential negative ho sakta hai.

True.

q < 0

ke liye,

V = k q / r < 0

har jagah. Negative

V

ka matlab hai ki ek external agent energy gain karta hai jab positive test charge

q_{0}

ko andar laata hai (field use khींchti hai).

E = - \nabla V

ka matlab hai ki field higher potential ki taraf point karta hai.

False. Minus sign ise flip karta hai:

E

lower potential ki taraf point karta hai (downhill, jaise anchor figure mein red arrow). Minus ke bina, ek positive charge spontaneously higher energy ki taraf chadh jaata, jo energy conservation tod deta.

Tum hamesha

V = 0

at infinity choose kar sakte ho.

Localized charges ke liye True, idealized infinite distributions (infinite line/plane) ke liye false. Wahan

\int

\infty

se diverge ho jaata hai, toh tumhe ek finite reference point choose karna hoga.

Line integral

- \int_{\infty}^{P} E \cdot d l

har path pe same answer deta hai.

Electrostatics mein True. Kyunki

\oint E \cdot d l = 0

(dekho Conservative Fields and Curl), sirf endpoints matter karte hain — yahi cheez

V (P)

ko ek single-valued function banati hai.

Kisi point charge se distance double karne par potential half ho jaata hai.

True.

V \propto 1/ r

, toh

r \to 2 r

V \to V /2

. (Contrast karo field magnitude

E \propto 1/ r^{2}

se, jo quarter tak drop ho jaati hai.)

Har jagah

V

mein same constant add karne se field change ho jaati hai.

False.

E = - \nabla V

sirf

V

ke variation pe depend karta hai, uski absolute value pe nahi;

\nabla (V + C) = \nabla V

, toh field untouched rehti hai — yahi reason hai ki

V = 0

reference ki choice free hai.

Spot the error

V = k q / r

, toh

r = 0

pe potential infinite hai — yahi stored energy hai."

Yeh formula ek point ka mathematical idealization hai; real charges ki finite size hoti hai, toh

r \to 0

kabhi nahi hota. Divergence model ki breakdown dikhata hai, koi physical infinite energy nahi.

E = - \nabla V

, aur

V = k q / r

positive hai, toh

E

inward point karta hoga."

Galat sign logic. Kyunki

E_{r} = - d V / d r

, minus sign slope ko flip karta hai:

\frac{d}{d r} (1/ r) = - 1/ r^{2}

, toh

E_{r} = - k q \cdot (- 1/ r^{2}) = + k q / r^{2} > 0

q > 0

ke liye — field outward point karta hai.

E_{r} = - d V / d r

mein sign-flip exactly wahi hai jo ek falling potential ko outward field mein badalta hai.

"Do plates ke beech,

E = 200

V/m hai aur plates 3 m apart hain, toh

V = 200 \times 3

beech har jagah hai."

V

gap mein koi single number nahi — yeh linearly vary karta hai. Sirf 3 m across difference

600

V hai; potential khud depend karta hai ki tum kis point pe ho.

E

paane ke liye

V

integrate karta hoon;

V

paane ke liye

E

differentiate karta hoon."

Ulta hai.

V

field ko integrate karne se milta hai (

V = - \int E \cdot d l

);

E

potential ko differentiate karne se milta hai (

E = - \nabla V

V_{A} - V_{B} = - \int_{A}^{B} E d x

Limits flipped hain.

V_{A} - V_{B} = - \int_{B}^{A} E \cdot d l = + \int_{A}^{B} E \cdot d l

. Endpoints galat karne se answer ka sign flip ho jaata hai.

"Field yahan strong hai, toh potential bhi yahan high hoga."

Field magnitude steepness (slope) naapti hai, height (value) nahi. Steep hillside kisi bhi altitude pe ho sakti hai; strong field low, high, ya zero potential pe ho sakti hai.

V

ek fixed absolute quantity hai, toh kisi point pe uska numerical value apne aap meaningful hai."

Sirf

V

ke differences physical hain. Tum har jagah koi bhi constant add kar sakte ho bina koi field ya force badle; raw number poori tarah depend karta hai ki tumne

V = 0

kahan set kiya.

Why questions

E

ke mukable

V

ko energy calculations ke liye kyun prefer kiya jaata hai?

V

ek scalar hai, toh contributions signed numbers ki tarah add hoti hain bina trigonometry ke; energy bookkeeping vector addition ki jagah algebra ban jaata hai.

V = - \int E \cdot d l

mein minus sign kyun aata hai?

External agent ko test charge

q_{0}

pe field ki force

q_{0} E

ke against push karna hota hai; uski work field ki work ka negative hai, aur

q_{0}

se divide karne par woh minus

V

mein aa jaata hai.

E

hamesha equipotential surface ke perpendicular kyun hota hai?

Surface ke along

V

constant hai, toh us direction mein uski slope zero hai — koi field component surface mein nahi rehta;

E = - \nabla V

ka poora hissa iske across point karta hai (steepest descent). Dekho Equipotential Surfaces.

Point charge ka

V

derive karte waqt "radial path" kyun choose karte hain?

Kyunki field conservative hai (

\oint E \cdot d l = 0

), answer path-independent hai, toh hum woh path choose karte hain jahan

E ∥ d l

ho aur integral sabse asaan ho.

E = - \nabla V

exactly

1/ r^{2}

Coulomb field ko

V = k q / r

se kyun recover karta hai?

Differentiation explicitly karo:

E_{r} = - \frac{d}{d r} (\frac{k q}{r}) = - k q \cdot (- \frac{1}{r ^{2}}) = \frac{k q}{r ^{2}}

1/ r

potential ko differentiate karne se automatic extra power of

r

aa jaati hai, jo Coulomb ka

1/ r^{2}

field magnitude return karta hai — bridge genuinely two-way hai (dekho Gradient Operator).

V

ke genuine maximum ya minimum pe

E = 0

kyun hota hai?

Ek true extremum pe slope

d V / d x = 0

har direction mein hota hai, aur kyunki

E = - \nabla V

, field wahan vanish ho jaata hai. Yeh

V = 0

se same nahi hai: anchor figure ka hilltop ka slope zero hai (toh

E = 0

) lekin badi height hai (

V \neq = 0

). Zero slope field ko khatam karta hai; zero value nahi.

V = 0

kahan set karo yeh choice arbitrary kyun hai?

Kyunki physics sirf

E = - \nabla V

aur energy differences

q Δ V

use karta hai; dono

V

mein add kiye gaye kisi bhi constant ko ignore karte hain. Reference point ek bookkeeping convenience hai, koi physical fact nahi.

Edge cases

Do equal

+ q

charges ke exact midpoint pe (maano

d

dono charges ke beech separation hai, toh har charge midpoint se

d /2

dur hai):

V

aur

E

kya hain?

E = 0

symmetry se (fields cancel ho jaate hain), lekin

V = 4 k q / d > 0

— har charge distance

d /2

k q / (d /2) = 2 k q / d

contribute karta hai, aur dono milke

4 k q / d

dete hain. Zero field ke saath nonzero potential.

+ q

aur

- q

ke exact midpoint pe:

V

aur

E

kya hain?

V = 0

(heights cancel ho jaate hain), lekin

E \neq = 0

— dono fields

+ q

- q

ki taraf point karte hain aur add ho jaate hain.

Kisi bhi finite charge distribution se infinitely dur

V

kya hota hai?

V \to 0

, hamara chosen reference. Yahi boundary condition hai jo

V = k q / r

ko koi extra constant ke bina clean banata hai.

Negative charge ke liye,

E

kis taraf point karta hai aur

V

ka sign kya hoga?

E

radially inward point karta hai (charge ki taraf) aur

V < 0

har jagah; field phir bhi downhill point karta hai,

V = 0

infinity se negative well mein neeche.

Agar kisi region mein

V

constant hai, toh wahan

E

kya hai?

E = - \nabla V = 0

us region mein har jagah — flat ground matlab koi push nahi, jaise ek charged conductor ka interior.

Jab ek test charge

q_{0}

ko ek closed loop mein move karke starting point pe wapas laaya jaata hai, toh net work per charge kya hai?

Zero: electrostatic fields ke liye

\oint E \cdot d l = 0

, toh kisi bhi closed path ke around

Δ V = 0

— yahi cheez

V

ko well-defined banati hai.

Connections

Coulomb's Law — har field-vs-potential trap ke peeche $1/ r^{2}$ force supply karta hai; " $r = 0$ pe infinite hai" aur " $E \propto 1/ r^{2}$ vs $V \propto 1/ r$ " wale traps yahan rehte hain.
Electric Field of Point Charge — woh vector $E$ jo hum $V$ ko differentiate karke recover karte hain; Spot the error mein "sign of $E_{r} = - d V / d r$ " trap ko underpin karta hai.
Potential Energy of Charge System — $U = q V$ sign-of- $V$ aur negative-charge edge cases ko energy statements mein convert karta hai ("negative $V$ ⇒ agent energy gain karta hai").
Equipotential Surfaces — flat-value cases: kyun "constant $V$ ke along koi work nahi" True hai aur kyun $E ⊥$ surface Why questions mein.
Conservative Fields and Curl — reason ki $\oint E \cdot d l = 0$ , jo path-independence True/False aur closed-loop edge case ko justify karta hai.
Gradient Operator — $E = - \nabla V$ ki machinery, gauge-freedom ("add a constant") trap aur hilltop $E = 0$ reasoning mein invoke hoti hai.