Applications — sphere, cylinder, infinite plane: Visual walkthrough

What we are allowed to assume

Before symbol one, here is our entire toolkit. If you have not met these, they are built in Gauss's Law, Electric Flux and Symmetry in Physics.

Formula The one law everything rests on

$arrows out of the bag \oint_{S} E \cdot d A = charge in the bag \frac{Q _{e n c}}{ε _{0}}$

$\oint_{S}$ — the little circle means "add up over a closed surface $S$ " (a surface with an inside and an outside, like a balloon).
$d A$ — a tiny patch of that surface, drawn as an arrow pointing straight out of the bag, its length equal to the patch's area.
$E \cdot d A$ — how much of the field actually pierces that patch (from Electric Flux).
$Q_{e n c}$ — only the charge trapped inside the surface. Charge outside contributes nothing.
$ε_{0}$ — a fixed constant of nature, $8.85 \times 1 0^{- 12}$ , that sets the "exchange rate" between charge and flux.

WHY this law and not Coulomb's Law? Coulomb gives the field of one point and forces you to add up infinitely many pieces. Gauss lets us skip all that summation — if we are clever about the surface. That cleverness is symmetry, and the sphere is where it shines brightest.

Step 1 — Draw the source and ask "which way can the arrows point?"

WHAT. We place a ball of radius $R$ carrying total charge $Q$ spread perfectly evenly through its volume, and we ask: at some outside point, which direction is $E$ ?

WHY. We cannot use Gauss's law until we know the direction and pattern of $E$ . Symmetry answers this for free — before any math.

PICTURE. Look at the figure. Imagine you are a field-arrow standing at a point away from the ball. Could you tilt left? If left were allowed, then by turning the whole picture (which changes nothing — the ball looks identical from every angle) you'd get a contradiction: the arrow must point somewhere that no rotation can prefer. The only such direction is straight away from the centre — radially outward.

Figure — Applications — sphere, cylinder, infinite plane

Step 2 — Choose the bag that matches the symmetry

WHAT. We wrap an imaginary sphere (the Gaussian surface) of radius $r$ around the charge, sharing the same centre.

WHY. We just learned $E$ points radially and has one strength at each radius. On a concentric sphere, every patch $d A$ also points radially outward — so $E$ and $d A$ are parallel everywhere, and $∣ E ∣$ is the same everywhere on it. That is exactly the double-jackpot that makes the flux integral trivial.

PICTURE. The red dashed sphere is our bag. At every point on it, the black field arrow and the outward patch arrow lie on the same line. No grazing, no variation.

Step 3 — Collapse the flux integral

WHAT. We compute the left side $\oint E \cdot d A$ on our chosen sphere.

WHY. Because $E ∥ d A$ everywhere, the dot product becomes plain multiplication $E d A$ . Because $E$ is constant on the surface, it slides out of the sum.

PICTURE. Every arrow pierces its patch head-on (angle $0$ , so $cos 0 = 1$ ). Adding up all patches just gives the sphere's total area.

$\oint E \cdot d A = \oint E d A = E total area \oint d A = E (4 π r^{2})$

$E \cdot d A \to E d A$ — the dot product's $cos θ$ became $cos 0 = 1$ because field and patch are parallel.
$E$ pulled outside the $\oint$ — legal only because $E$ is the same on every patch (Step 2).
$\oint d A = 4 π r^{2}$ — the surface area of a sphere of radius $r$ ; this is pure geometry, no physics.

Step 4 — Count the charge inside (outside case, $r > R$ )

WHAT. For a bag bigger than the ball, all of $Q$ is inside: $Q_{e n c} = Q$ .

WHY. Gauss's law only cares what's trapped inside our chosen surface. When $r > R$ , the whole ball is swallowed.

PICTURE. The red ball sits entirely within the dashed Gaussian sphere — everything is enclosed.

$Q_{e n c} = Q (r > R)$

Now equate the two sides of Gauss's law:

$flux, Step 3 E (4 π r^{2}) = charge/ ε_{0} \frac{Q}{ε _{0}} ⟹ E_{o u t} = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$

Divide both sides by $4 π r^{2}$ → the $4 π$ regroups into the familiar Coulomb constant $\frac{1}{4 π ε _{0}}$ .
The result depends on $r^{2}$ in the denominator → an inverse-square law.

Step 5 — Shrink the bag inside the ball ( $r < R$ )

WHAT. Now let the Gaussian sphere be smaller than the ball. Only the charge within radius $r$ is enclosed — the outer shell of charge is outside our bag and ignored.

WHY. Gauss again: charge beyond radius $r$ contributes nothing to this surface's flux. We must recompute $Q_{e n c}$ .

PICTURE. The dashed Gaussian sphere sits inside the red ball. Only the shaded inner core counts; the outer rind is excluded.

First, the charge density $ρ$ — charge per unit volume, constant because the ball is uniform:

$ρ = \frac{Q}{\frac{4}{3} π R ^{3}}$

Numerator $Q$ — the total charge.
Denominator $\frac{4}{3} π R^{3}$ — the ball's full volume. So $ρ$ = "how densely packed" the charge is.

The charge inside radius $r$ is density times the small volume:

$Q_{e n c} = ρ \cdot \frac{4}{3} π r^{3}$

$\frac{4}{3} π r^{3}$ — volume of the inner sphere only (radius $r$ , not $R$ ).

Feed this into Gauss's law with the same flux $E (4 π r^{2})$ :

\;\;\Longrightarrow\;\; \boxed{\,E_{in} = \frac{\rho\,r}{3\varepsilon_0} = \frac{Q\,r}{4\pi\varepsilon_0 R^3}\,}$$ - Left side has $r^2$; enclosed charge has $r^3$; dividing leaves **one power of $r$ on top**. - So inside, $E$ **grows linearly** with $r$ — the deeper you go, the *weaker* the field, reaching zero at the very centre. --- ## Step 6 — The degenerate case: the exact centre, and the conductor **WHAT.** Two edge cases the formula must survive: > (a) $r=0$ (dead centre), and (b) a hollow conducting shell where all charge lives on the surface. **WHY.** A derivation is only trustworthy if it behaves sensibly at its extremes. Skip these and a reader hits a wall we never showed. **PICTURE.** At the centre, shrink the Gaussian sphere to a dot: it encloses no charge, so $Q_{enc}\to 0$ and flux $\to 0$. For the conductor, the charge sits as a thin red skin; any bag *inside* the skin encloses nothing. ![[deepdives/dd-physics-1.8.07-d2-s06.png]] **(a) Centre of the solid ball.** Put $r=0$ into $E_{in}=\dfrac{Qr}{4\pi\varepsilon_0R^3}$ → $E=0$. It matches the linear-growth picture: the field vanishes at the middle. Physically, charge pulls equally in every direction, cancelling. **(b) Conducting shell.** All charge sits on the outer surface. For any $r<R$, $Q_{enc}=0$, so $E_{in}=0$ **everywhere inside**, not just at the centre. See [[Conductors in Electrostatics]]. $$E_{in}^{\text{conductor}} = 0 \quad (r<R), \qquad E_{out} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\quad (r>R)$$ > [!mistake] Confusing the two "insides" > **Solid uniform ball:** field inside grows like $r$ (not zero except at the centre). **Hollow conductor:** field inside is zero *everywhere*. **Fix:** ask "is charge spread through the volume, or stuck on the surface?" — that decides which inside-formula you use. --- ## Step 7 — Stitch the pieces into one graph **WHAT.** Assemble $E(r)$ across all radii for the uniform ball. **WHY.** Seeing the two regimes joined proves they *agree at the boundary* $r=R$ — a final consistency check. **PICTURE.** Rising red line ($E\propto r$) inside, falling black curve ($E\propto 1/r^2$) outside. They meet exactly at $r=R$: the field peaks at the surface. $$E(R)_{\text{inside}} = \frac{Q\,R}{4\pi\varepsilon_0 R^3} = \frac{Q}{4\pi\varepsilon_0 R^2} = E(R)_{\text{outside}} \checkmark$$ - Plug $r=R$ into both boxed formulas → identical. The graph has **no jump** at the surface; the two branches shake hands. ![[deepdives/dd-physics-1.8.07-d2-s07.png]] > [!example] Numbers to feel the shape > Take $Q = 2\times10^{-6}$ C, $R = 0.1$ m, and $\frac{1}{4\pi\varepsilon_0}=9\times10^9$. > - **At the surface** $r=R=0.1$: $E = \dfrac{(9\times10^9)(2\times10^{-6})}{0.1^2} = 1.8\times10^6$ N/C — the **peak**. > - **Outside**, $r=0.2$: $E = \dfrac{(9\times10^9)(2\times10^{-6})}{0.2^2} = 4.5\times10^5$ N/C — a quarter of the peak (double the distance → quarter the field, inverse-square). > - **Inside** (uniform ball), $r=0.05$: $E = \dfrac{(9\times10^9)(2\times10^{-6})(0.05)}{0.1^3} = 9\times10^5$ N/C — exactly **half** the peak (half the radius → half the field, linear). *Why:* used $E_{in}=\frac{Qr}{4\pi\varepsilon_0R^3}$. --- ## The one-picture summary ![[deepdives/dd-physics-1.8.07-d2-s08.png]] Everything compressed: pick the radial-symmetry surface → flux collapses to $E\cdot4\pi r^2$ → equate to $Q_{enc}/\varepsilon_0$ → read off $E$. The enclosed charge grows like $r^3$ inside (giving $E\propto r$) and freezes at $Q$ outside (giving $E\propto 1/r^2$), so the field climbs to a peak at the surface and then fades. > [!recall]- Feynman: the whole walkthrough in plain words > Picture a fuzzy red ball of electric "stuff." Field-arrows want to shoot out of it, but which way? The ball looks the same spun any direction, so the arrows have no reason to lean sideways — they fly straight outward, and their strength can only depend on how far out you are. > > Now imagine slipping a stretchy bubble around the ball. Gauss's rule says the number of arrows poking through the bubble equals the charge trapped inside it (times a fixed constant). Because we picked a *round* bubble, every arrow pierces it head-on and the counting is just "field strength times bubble area = $E\times 4\pi r^2$." > > If the bubble is bigger than the ball, it traps *all* the charge — so as you move out and the bubble grows, the fixed pile of arrows spreads thin over more area, fading like $1/r^2$. That's why the ball, from outside, looks exactly like a single dot at its centre. > > If the bubble is *smaller*, buried inside the ball, it only traps the charge in the little core — and a small core holds very little charge. So near the centre the field is tiny; it grows straight-line as you move outward, hitting its strongest right at the surface. At the very middle: nothing, pulled equally every way. And if the ball is a hollow metal shell instead, all the charge hides on the skin, so *anywhere inside* is dead calm — zero field. > [!mnemonic] The sphere in one line > **Outside: point-charge, $1/r^2$. Inside solid: linear, $\propto r$. Inside hollow: zero. Peak at the skin.** --- ## #flashcards/physics Why must $\vec E$ be radial for a charged sphere ::: Rotational symmetry — the ball looks identical from every angle, so no sideways direction is special Flux through a concentric Gaussian sphere ::: $E\cdot 4\pi r^2$ (field constant and parallel to every patch) Field outside a charged sphere ($r>R$) ::: $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ Field inside a uniform solid ball ($r<R$) ::: $E=\dfrac{Qr}{4\pi\varepsilon_0R^3}\propto r$ Field inside a hollow conductor ::: $0$ everywhere Where is the sphere's field strongest ::: At the surface, $r=R$