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

1. The atom of everything: the point charge

WHY $1/r^2$ ? Picture the charge throwing out field "lines." The number of lines is fixed, but they spread over a sphere of area $4\pi r^2$ . Line density (= field strength) therefore falls as $1/r^2$ . This is literally Gauss's law in disguise.

Define $k \equiv \dfrac{1}{4\pi\varepsilon_0} \approx 8.99\times10^9\ \text{N·m}^2/\text{C}^2$ .

2. Superposition: the master rule

3. The dipole (along the axis & on the bisector)

On the axis (distance $r \gg d$ from center):

Formula Axial field — derive it

WHAT: net of two opposite charges at $r-\tfrac d2$ and $r+\tfrac d2$ . $E_{ax}=k q\left[\frac{1}{(r-\tfrac d2)^2}-\frac{1}{(r+\tfrac d2)^2}\right].$ Why this step? Near charge is closer → stronger, so it wins → net points along $+\vec p$ . Combine the fractions: $E_{ax}=kq\cdot\frac{(r+\tfrac d2)^2-(r-\tfrac d2)^2}{(r^2-\tfrac{d^2}{4})^2}=kq\cdot\frac{2rd}{(r^2-\tfrac{d^2}{4})^2}.$ For $r\gg d$ drop $d^2/4$ : $\boxed{E_{ax}=\frac{2kp}{r^3}}=\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}.$

On the perpendicular bisector:

4. The ring (on its axis)

Ring of radius $R$ , total charge $Q$ , field at point $P$ a distance $x$ along the axis.

Formula Axial field of a ring — derive it

WHY axis only: off-axis is ugly; on the axis symmetry saves us. Take element $dq$ . Distance to $P$ : $r=\sqrt{R^2+x^2}$ . Its field magnitude: $dE=\dfrac{k\,dq}{R^2+x^2}$ . HOW symmetry helps: the radial components from opposite elements cancel; only the axial component survives, weighted by $\cos\theta=\frac{x}{\sqrt{R^2+x^2}}.$ Why this step? Every $dq$ has the same $\cos\theta$ (same $x$ , same $r$ ), so it pulls out of the integral. $E_x=\int \frac{k\,dq}{R^2+x^2}\cdot\frac{x}{\sqrt{R^2+x^2}}=\frac{kx}{(R^2+x^2)^{3/2}}\int dq.$ $\boxed{E_{ring}=\frac{kQx}{(R^2+x^2)^{3/2}}}.$

5. The disk (on its axis) — integrate rings

Uniform surface charge density $\sigma$ , radius $R$ , field at axial distance $x$ .

Formula Disk field — build from rings

WHAT: a disk is nested rings. A ring of radius $s$ , thickness $ds$ has charge $dq=\sigma\,(2\pi s\,ds)$ . Plug into the ring result with $R\to s$ , $Q\to dq$ : $dE=\frac{k\,x\,(\sigma 2\pi s\,ds)}{(s^2+x^2)^{3/2}}.$ Integrate $s:0\to R$ . Let $u=s^2+x^2,\ du=2s\,ds$ : $E=k\,x\,\sigma\,\pi\int_{x^2}^{R^2+x^2}u^{-3/2}\,du= k\pi\sigma x\left[-2u^{-1/2}\right]_{x^2}^{R^2+x^2}.$ Why this step? The substitution turns $\frac{s\,ds}{(\cdots)^{3/2}}$ into a standard power — the $s$ in $dq$ is exactly what we needed. $\boxed{E_{disk}=\frac{\sigma}{2\varepsilon_0}\left[1-\frac{x}{\sqrt{R^2+x^2}}\right]}$ (using $k\pi=\tfrac{1}{4\varepsilon_0}$ , so $2k\pi=\tfrac{1}{2\varepsilon_0}$ ).

6. The infinite line charge (Gauss's law shines)

Linear charge density $\lambda$ (C/m).

Formula Line charge via Gauss

WHY Gauss: cylindrical symmetry → $\vec E$ is radial, magnitude depends only on perpendicular distance $r$ . WHAT surface: a coaxial cylinder, radius $r$ , length $L$ . Flux: only the curved side contributes (ends are parallel to $\vec E$ , zero flux). $\oint\vec E\cdot d\vec A=E\,(2\pi r L)=\frac{Q_{enc}}{\varepsilon_0}=\frac{\lambda L}{\varepsilon_0}.$ Why this step? End caps have $\vec E\perp d\vec A$ ? No — $\vec E\parallel$ ends, so $\vec E\cdot d\vec A=0$ there. Curved wall has $\vec E\parallel d\vec A$ . $\boxed{E_{line}=\frac{\lambda}{2\pi\varepsilon_0 r}=\frac{2k\lambda}{r}}.$ Dies as == $1/r$ == — slower than a point charge, because a line is "spread out infinitely."

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

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

Source	$E \propto$	Key formula
Point charge	$1/r^2$	$kq/r^2$
Dipole	$1/r^3$	$2kp/r^3$ (axial)
Infinite line	$1/r$	$2k\lambda/r$
Infinite sheet	$r^0$	$\sigma/2\varepsilon_0$
Ring (axis)	—	$kQx/(R^2+x^2)^{3/2}$

8. Common mistakes (Steel-manned)

Flashcards

Field of a point charge at distance

r

E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}

, radial.

Why does point-charge field go as

1/r^2

Fixed flux spreads over sphere area

4\pi r^2

(Gauss's law).

Axial dipole field for

r\gg d

E=\dfrac{2kp}{r^3}

, along

\vec p

Equatorial dipole field

E=\dfrac{kp}{r^3}

, opposite to

\vec p

(half the axial value).

Why does a dipole field fall as

1/r^3

Leading

1/r^2

terms of the two charges cancel.

Axial field of a ring (charge

Q

, radius

R

, distance

x

)

E=\dfrac{kQx}{(R^2+x^2)^{3/2}}

Field at the center of a charged ring

Zero, by symmetry.

Where is a ring's axial field maximum?

x=\pm R/\sqrt2

Axial field of a uniformly charged disk

E=\dfrac{\sigma}{2\varepsilon_0}\left[1-\dfrac{x}{\sqrt{R^2+x^2}}\right]

Field of an infinite charged sheet

E=\dfrac{\sigma}{2\varepsilon_0}

, constant.

Field of an infinite line charge

E=\dfrac{\lambda}{2\pi\varepsilon_0 r}=\dfrac{2k\lambda}{r}

Why do end caps contribute zero flux for a line's Gaussian cylinder?

\vec E

is parallel to the end caps, so

\vec E\cdot d\vec A=0

When CAN you use Gauss's law to find

E

When symmetry (sphere/cylinder/plane) makes

E

constant & perpendicular on a chosen surface.

Recall Feynman: explain to a 12-year-old

Imagine each charge is a sprinkler shooting invisible "spray" outward. Close up the spray is dense (strong field); far away it spreads thin (weak). A single drop-charge thins as you go in every direction → $1/r^2$ . A long hose-line of charge only spreads sideways, not along its length → thins slower, $1/r$ . A huge wall of charge spreads nowhere new — the spray stays the same thickness everywhere → constant. And a $+$ and $-$ glued together almost cancel each other's spray, so far away there's barely anything left → it fades super fast, $1/r^3$ . To add up a ring of sprinklers: the sideways sprays push against each other and cancel, only the along-axis spray survives.

Connections

Gauss's law — the symmetry shortcut behind point/line/sheet results
Coulomb's law — the $1/r^2$ atom of all superposition
Electric potential — $\vec E=-\nabla V$ ; often easier to integrate scalar $V$ first
Electric dipole in a uniform field — torque $\vec\tau=\vec p\times\vec E$ , energy $U=-\vec p\cdot\vec E$
Parallel plate capacitor — two sheets give $\sigma/\varepsilon_0$
Flux and field lines — visual basis of the falloff exponents

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, saare electric field problems ek hi basic idea pe khade hain: har charge apne aas-paas field banata hai jo positive charge se bahar ki taraf aur negative ki taraf andar point karta hai. Point charge ka field $kq/r^2$ hota hai — yeh $1/r^2$ isliye girta hai kyunki field lines ek sphere (area $4\pi r^2$ ) pe phailti hain. Yahi cheez Gauss's law kehti hai: flux $\oint\vec E\cdot d\vec A = Q_{enc}/\varepsilon_0$ .

Jab charge ek shape me phaila ho (ring, disk, line), toh hum usko chhote-chhote $dq$ tukdo me todte hain, har tukde ka field nikaalte hain, aur vector me jod dete hain — isi ko superposition kehte hain. Trick yeh hai ki symmetry se kuch components cancel ho jaate hain. Jaise ring ke axis pe sirf axial component bachta hai (radial cancel ho jaata hai), aur center pe to field bilkul zero hota hai — kyunki sab taraf se pull barabar hai.

Falloff yaad rakho: Point $1/r^2$ , Dipole $1/r^3$ (kyunki $+$ aur $-$ lagभग cancel ho jaate hain, isliye fast girta hai), Line $1/r$ (slow, kyunki line infinitely lambi hai), aur Sheet me to field constant $\sigma/2\varepsilon_0$ — distance se matlab hi nahi! Disk ka formula $\frac{\sigma}{2\varepsilon_0}[1-\frac{x}{\sqrt{R^2+x^2}}]$ hai, aur $R\to

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

1. The atom of everything: the point charge

2. Superposition: the master rule

3. The dipole (along the axis & on the bisector)

4. The ring (on its axis)

5. The disk (on its axis) — integrate rings

6. The infinite line charge (Gauss's law shines)

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

8. Common mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Electromagnetism

Connections