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

1.8.5 · Physics › Electromagnetism

1. Har cheez ka atom: point charge

$1/ r^{2}$ kyun? Socho charge field "lines" bahar phek raha hai. Lines ki sankhya fixed hai, lekin wo $4 π r^{2}$ area wale ek sphere par failti hain. Line density (= field strength) isliye $1/ r^{2}$ ki tarah girta hai. Ye literally Gauss's law disguise mein hai.

Define karo $k \equiv \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ .

2. Superposition: master rule

3. Dipole (axis par aur bisector par)

Axis par (center se distance $r ≫ d$ ):

Perpendicular bisector par:

4. Ring (apne axis par)

Radius $R$ ki ring, total charge $Q$ , field point $P$ par jo axis ke along distance $x$ par hai.

5. Disk (apne axis par) — rings integrate karke

Uniform surface charge density $σ$ , radius $R$ , axial distance $x$ par field.

6. Infinite line charge (Gauss's law chamakta hai)

Linear charge density $λ$ (C/m).

Figure — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

7. Field-falloff summary (80/20 table)

Source	$E \propto$	Key formula
Point charge	$1/ r^{2}$	$k q / r^{2}$
Dipole	$1/ r^{3}$	$2 k p / r^{3}$ (axial)
Infinite line	$1/ r$	$2 k λ / r$
Infinite sheet	$r^{0}$	$σ /2 ε_{0}$
Ring (axis)	—	$k Q x / (R^{2} + x^{2})^{3/2}$

8. Common mistakes (Steel-manned)

Flashcards

Distance

r

par point charge ka field

E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}

, radial.

Point-charge field

1/ r^{2}

ki tarah kyun jaata hai?

Fixed flux sphere area

4 π r^{2}

par failta hai (Gauss's law).

r ≫ d

ke liye axial dipole field

E = \frac{2 k p}{r ^{3}}

p

ke along.

Equatorial dipole field

E = \frac{k p}{r ^{3}}

p

ke opposite (axial value ka aadha).

Dipole field

1/ r^{3}

ki tarah kyun girta hai?

Dono charges ke leading

1/ r^{2}

terms cancel ho jaate hain.

Ring (charge

Q

, radius

R

, distance

x

) ka axial field

E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}

Charged ring ke center par field

Zero, symmetry se.

Ring ka axial field maximum kahan hai?

x = \pm R / 2

par.

Uniformly charged disk ka axial field

E = \frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]

Infinite charged sheet ka field

E = \frac{σ}{2 ε _{0}}

, constant.

Infinite line charge ka field

E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}

Line ke Gaussian cylinder ke end caps zero flux kyun contribute karte hain?

E

end caps ke parallel hai, isliye

E \cdot d A = 0

Gauss's law se

E

kab find kar SAKTE HO?

Jab symmetry (sphere/cylinder/plane) kisi chosen surface par

E

ko constant aur perpendicular banaye.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho har charge ek sprinkler hai jo invisible "spray" bahar phenkta hai. Nazdik spray dense hai (strong field); door se wo phatl jaata hai (weak). Ek single drop-charge har direction mein thinta hai → $1/ r^{2}$ . Charge ki ek lambi hose-line sirf sideways failti hai, apni length ke along nahi → slowly thinta hai, $1/ r$ . Charge ki ek badi wall kahi naya nahi failti — spray har jagah same thickness rehti hai → constant. Aur ek $+$ aur $-$ jo saath chipke hain, almost ek doosre ki spray cancel karte hain, isliye door se bahut kuch bachta hi nahi → bahut tezi se fade hota hai, $1/ r^{3}$ . Ek ring of sprinklers ko add karne ke liye: sideways sprays ek doosre ke against push karte hain aur cancel ho jaate hain, sirf along-axis spray bachti hai.

Connections

Gauss's law — point/line/sheet results ke peeche symmetry shortcut
Coulomb's law — saari superposition ka $1/ r^{2}$ atom
Electric potential — $E = - \nabla V$ ; aksar pehle scalar $V$ integrate karna aasaan hota hai
Electric dipole in a uniform field — torque $τ = p \times E$ , energy $U = - p \cdot E$
Parallel plate capacitor — do sheets milke $σ / ε_{0}$ dete hain
Flux and field lines — falloff exponents ka visual basis