Electric field of point charge, dipole, ring, disk, line charge (Gauss's law): Question bank

1.8.5 · D5 · Physics › Electromagnetism › Electric field of point charge, dipole, ring, disk, line cha

True or false — justify

Ek charged body jiska net charge zero ho, woh har jagah zero electric field produce karta hai.

False. Ek dipole ka net charge zero hota hai phir bhi uska field nonzero hota hai (yeh

1/ r^{3}

ki tarah ghatta hai); sirf far-field average zero hota hai, field khud nahi.

Gauss's law

\oint E \cdot d A = Q_{e n c} / ε_{0}

ek ring of charge ke liye bhi valid hai.

True — yeh hamesha valid hai. Ek ring ke liye jo kaam nahi karta woh hai is se solve karna: koi bhi surface aisi nahi hai jis par

E

constant aur perpendicular ho, isliye integral collapse hokar

E \cdot A

nahi banega.

Ek closed surface se flux ka matlab us surface ke andar charges ki exact position par depend karta hai.

False. Flux sirf total enclosed charge par depend karta hai; andar kisi charge ko idhar-udhar khiskaao, net flux nahi badlega, chahe local field pattern badal jaye. Dekho Flux and field lines.

Ek closed surface ke bahar rakha charge us surface se zero net flux contribute karta hai.

True. Uske field lines ek taraf se andar jaate hain aur doosri taraf se bahar, isliye inflow aur outflow exactly cancel ho jaate hain — us charge ke liye

Q_{e n c} = 0

Uniformly charged ring ke center par field zero hota hai.

True. Har element ka pull uske diametrically opposite element se cancel ho jaata hai; symmetry "closeness" par bhaari padti hai.

Ek infinite sheet ka field

σ / ε_{0}

hota hai.

False. Ek akela isolated sheet

σ /2 ε_{0}

deta hai; flux dono faces se bahar nikalta hai, isliye har face aadha carry karta hai. Poora

σ / ε_{0}

sirf do opposite sheets ke beech milta hai (ek Parallel plate capacitor).

Ek infinite line charge se doori double karne par field aadha ho jaata hai.

True. Line field

\propto 1/ r

hai, isliye

r \to 2 r

karne par

E \to E /2

hota hai — point charge ki tarah nahi jahan woh quarter ho jaata.

Dipole axial field exactly equatorial field ka do guna hota hai, same distance par.

True (far-field

r ≫ d

mein):

E_{a x} = 2 k p / r^{3}

aur

E_{e q} = k p / r^{3}

, aur ye

p

ke relative opposite senses mein point karte hain.

Spot the error

"Ring ke liye, kyunki main center par sabse zyada charge ke paas hoon,

E

wahan maximum hai."

Opposite elements ke pulls center par cancel ho jaate hain, isliye

E = 0

hota hai. Maximum axis ke along

x = \pm R / 2

par hota hai.

"Dipole do point charges hain, har ek

1/ r^{2}

wala, isliye dipole field

1/ r^{2}

hai."

Leading

1/ r^{2}

terms subtract out ho jaate hain kyunki charges opposite hain; difference ko expand karne par

1/ r^{3}

leading term milta hai.

"Gauss's law kisi bhi shape ke liye

E

deta hai kyunki yeh hamesha true hai."

True hai lekin tab tak useless hai jab tak symmetry

E

ko integral se bahar nahi nikaal deti. Ring, disk, aur finite line ke paas aisi koi surface nahi hai — wahan Coulomb's law integrate karna padega.

"Line charge ke Gaussian cylinder ke end caps flux contribute karte hain."

Nahi karte —

E

radial hai, isliye yeh flat end caps ke parallel hai, jisse

E \cdot d A = 0

hota hai wahan. Sirf curved wall contribute karta hai.

"Disk ke liye, jab

x \to \infty

tab field

σ /2 ε_{0}

rehti hai."

Yeh near-field (

x ≪ R

) limit hai. Door jaane par bracket

[1 - x / R^{2} + x^{2}] \to 0

ho jaata hai aur disk point charge jaisi dikhti hai,

E \to k Q / x^{2}

"Ek negative charge field ko outward point karta hai, bas kamzor."

Negative charge field ko inward (apni taraf) point karta hai. Sign ke saath direction flip ho jaati hai; sirf magnitude scale hoti hai.

E

inside a Gaussian sphere constant hai, isliye main ise bahar nikal sakta hoon — yeh point-charge law prove karta hai."

Yeh statement sirf point charge ke around spherical symmetry ki wajah se sahi hai. Agar charge off-center ya non-spherical hota, toh

E

sphere par vary karta aur bahar nahi nikala ja sakta.

Why questions

Point charge ka field

1/ r^{2}

kyun ghatta hai?

Uske fixed number of field lines ek sphere ke area

4 π r^{2}

par spread ho jaate hain; line density (= field strength) isliye

1/ r^{2}

ki tarah ghatti hai. Yahi Gauss's law picture form mein hai.

Dipole field point charge se zyada tezi (

1/ r^{3}

) se kyun ghatta hai?

Do opposite charges zyada door par nearly cancel ho jaate hain; jo bachta hai woh sirf unka chhota difference hai, jo ek extra power of

r

zyada tezi se ghatta hai.

Infinite line

1/ r

se point charge se dheere kyun ghatta hai?

Door jaate waqt tum line ka proportionally zyada lamba hissa dekhte ho; woh extra charge

1/ r^{2}

single-point falloff ko partially compensate karta hai, net result

1/ r

hota hai.

Infinite sheet ka field distance se independent kyun hai?

Door jaao tum proportionally zyada sheet area lete ho; extra charge exactly

1/ r^{2}

weakening ko cancel karta hai — ek perfect trade, isliye

E

flat rehta hai.

Ring ke axis par sirf axial components kyun bachte hain?

Har element ke liye axis ke across ek mirror element hota hai; unke radial components opposite direction mein point karte hain aur cancel ho jaate hain, sirf common axial component bachta hai (har ek ke saath same

cos θ = x / R^{2} + x^{2}

Point charge ke liye sphere kyun chunte hain lekin line ke liye cylinder?

Tum Gaussian surface ko source ki symmetry se match karte ho taaki

E

us par constant aur perpendicular ho — point ke liye spherical symmetry, line ke liye cylindrical.

Compact dipole formulas

2 k p / r^{3}

aur

k p / r^{3}

ke liye

r ≫ d

kyun assume karna padta hai?

Tabhi hum

d^{2} /4

terms drop kar sakte hain aur leading behaviour rakh sakte hain; paas mein exact two-charge expression use karna padega.

Edge cases

Ring ke axis par jab

x \to \infty

tab

E

kya hota hai?

Yeh

k Q / x^{2}

approach karta hai — ek point charge, kyunki ring ka size bahut door se negligible ho jaata hai.

Charged ring ke exact center par

E

kya hota hai?

Exactly zero, symmetry se — har pull opposite element se cancel ho jaati hai.

Disk field

x = 0

par (surface par, axis par) kya hota hai?

σ /2 ε_{0}

— bracket

1 - 0 = 1

ban jaata hai; yeh finite disk ki central value hai, jo infinite-sheet result se match karti hai.

Line-charge field ka

r \to 0

par kya hota hai?

Yeh diverge karta hai (

E \to \infty

), kyunki idealized line ki thickness zero hai — paas mein infinite charge density. Real wires ki finite radius hoti hai, isliye yeh ek model artifact hai.

Equatorial dipole ke liye

E

kis direction mein point karta hai

p

ke relative?

p

ke opposite. Axis par yeh

+ p

ke saath point karta hai; yeh reversal ek classic sign trap hai.

Dipole axis par

+ q

aur

- q

ke exact midpoint par field ki direction kya hai?

Dono charges wahan same taraf push karte hain (+ se door, - ki taraf), isliye field strong hai aur $+$ se $-$ ki taraf point karta hai, yaani gap ke andar

- p

ke along — far-axial direction ke opposite.

Ek hi sign ki do identical parallel sheets ke beech

E

kya hota hai?

Zero — do

σ /2 ε_{0}

fields gap mein opposite direction mein point karte hain aur cancel ho jaate hain. (Opposite signs ke liye ye add hokar

σ / ε_{0}

dete hain; compare karo Parallel plate capacitor se.)

Recall Quick self-test

Dipole field falloff exponent ::: $1/ r^{3}$ Isolated infinite sheet field ::: $σ /2 ε_{0}$ , constant Field at center of a ring ::: zero, by symmetry Does Gauss's law hold for a ring? ::: yes, but it can't be solved by it — no symmetric surface