Capacitance — parallel plate derivation, cylindrical, spherical: Visual walkthrough

We are going to answer one question: how much charge can two flat metal plates soak up per volt?

Step 1 — What "charge on a plate" even looks like

WHAT: We take two identical flat metal plates, hold them a distance $d$ apart, and move charge from one to the other with a battery.

WHY: Metal lets charge spread out freely and evenly, so the surface ends up uniformly coated — that uniformity is the whole reason the math stays simple. If the coating were lumpy, the field would be a mess.

PICTURE: Look at the figure. The top plate is dusted with pale-yellow $+$ marks, the bottom with blue $-$ marks. They are spread evenly — same crowd everywhere.

Figure — Capacitance — parallel plate derivation, cylindrical, spherical

Because the charge is spread evenly, we describe it not by the total $Q$ but by how crowded the surface is:

Step 2 — Every charge makes an arrow: the electric field

WHAT: The $+ Q$ plate pushes test charges away; the $- Q$ plate pulls them toward it. In the gap, both effects point the same way — from $+$ plate straight down to the $-$ plate.

WHY: We need the field because the field is the bridge between "charge" and "voltage." No field, no push, no voltage difference. Field is the middle-man in the whole recipe.

PICTURE: In the figure the arrows in the gap are all the same length and all point straight down — a perfectly uniform stack of parallel arrows. Outside the plates the arrows cancel and there's essentially nothing.

Notice: we claimed the field is uniform. We haven't proven it yet. That's the next step, and the tool that proves it is Gauss's law.

Step 3 — WHY Gauss's law, and the pillbox trick

WHAT: We slide a tiny imaginary sealed box — a pillbox — halfway into the positive plate, so one flat face sits inside the metal and the other floats in the gap.

WHY the pillbox: Inside solid metal the field is zero (charges would keep moving until the push cancels), so no field pokes through the inside face. The field slides along the four thin side walls, so nothing pokes through them either. Only the gap-face lets field through. One surface, one term.

PICTURE: The figure shows the pillbox: a short blue cylinder. The top cap (pink) is buried in the plate — zero field there. The side walls have arrows sliding along them — zero flux. Only the bottom cap (yellow) in the gap has arrows stabbing straight through.

So the entire left-hand side of Gauss's law collapses to one term: $field \times area of the one gap-face E \cdot A_{box} = \frac{σ A _{box} charge trapped = crowdedness \times that same area}{ε _{0}}$

$A_{box}$ = area of the pillbox face (a number we chose; it must cancel).
$σ A_{box}$ = charge sealed inside = crowdedness × the face's slice of plate.

Step 4 — The area cancels: a uniform field is born

WHAT: The made-up area $A_{box}$ appears on both sides. Divide it out.

WHY this matters: If our answer depended on the size of an imaginary box we invented, the physics would be nonsense. The cancellation is nature telling us the field does not care where you look — it's the same everywhere in the gap. That is exactly what "uniform" means.

$E = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$

Term by term:

$E$ — the field strength in the gap, and there's no $r$ , no $x$ in the answer → it's constant everywhere.
$σ = Q / A$ — we just substituted the crowdedness from Step 1.
$ε_{0}$ — the conversion constant.

PICTURE: The figure shows two different pillboxes, fat and thin. Both give the same $E$ — the arrow length is identical. Different box, same field: proof of uniformity.

Step 5 — From field to voltage: climbing the hill

WHAT: March from the bottom plate to the top plate, a distance $d$ , feeling the field the whole way.

WHY it's easy here: The field is constant (Step 4), so "add up $E$ over the distance" is just $E$ times $d$ — no calculus gymnastics needed. A constant integrand makes the integral a plain multiplication.

$V = \int_{0}^{d} E d x = E d = \frac{Q d}{ε _{0} A}$

$\int_{0}^{d} \dots d x$ — sum over every sliver of the gap from $0$ to $d$ .
$E d$ — constant field × gap width = total voltage climbed.
$d$ — plate separation (m). Wider gap → longer climb → bigger $V$ .

PICTURE: The figure draws voltage as a staircase / ramp: because $E$ is constant, the potential rises as a straight slanted line from $0$ at the bottom plate to $V$ at the top. The slope of that line is $E$ .

Step 6 — Divide, and watch $Q$ vanish

WHAT: Plug $V$ into the definition $C = Q / V$ .

WHY the magic happens: $E \propto Q$ , therefore $V \propto Q$ . When we divide $Q$ by something proportional to $Q$ , the $Q$ cancels. What's left cannot depend on how much charge you loaded — only on shape and material. That's the deep truth of capacitance.

$C = \frac{Q}{V} = \frac{Q}{\frac{Q d}{ε _{0} A}} = \frac{ε _{0} A}{d}$

PICTURE: The figure shows the cancellation visually — a $Q$ in a numerator box and a $Q$ inside the $V$ box, with a chalk stroke crossing both out, leaving pure geometry glowing.

Step 7 — The edge cases (never leave the reader stranded)

Case $A \to 0$ (shrink the plates to nothing): $C = ε_{0} A / d \to 0$ . No plate → nowhere to store charge → zero capacitance. Sanity ✅.

Case $d \to 0$ (plates touch): $C \to \infty$ . Zero climb means any charge makes zero voltage, so $Q / V$ blows up. In reality the plates short out or the dielectric breaks down first — the formula honestly warns you the design is unphysical.

Case $d \to \infty$ (pull plates far apart): $C \to 0$ . The climb becomes enormous, so the same charge makes a huge voltage → tiny capacitance. The plates stop "seeing" each other.

Fixed charge, then squeeze ( $d$ smaller): $C$ grows, and the stored energy $U = Q^{2} /2 C$ (see Energy Stored in a Capacitor) falls. So squeezing releases energy — which is exactly why the two plates attract each other. The formula secretly contains the force.

PICTURE: The figure plots $C$ versus $d$ — a $1/ d$ curve. It spikes toward infinity as $d \to 0$ and flattens toward zero as $d$ grows. Three chalk dots mark the three limits above.

The one-picture summary

The whole derivation on one board: charge → (Gauss + pillbox) → uniform field → (integrate over $d$ ) → voltage → (divide, $Q$ cancels) → $C = ε_{0} A / d$ .

Recall Feynman retelling — the whole walk in plain words

We laid two metal sheets face to face and pumped electrons off one onto the other, so one sheet is crowded with plus and the other with minus — and evenly, because metal lets charge spread out. That crowding makes a push (a field) filling the gap, and because both sheets push the same way, the push is the same everywhere — an even stack of arrows. We proved the "same everywhere" part with a clever imaginary box (a pillbox): field only pokes through its gap-face, and the box's made-up area cancels, which is nature's way of saying the field doesn't depend on where you look. Then we walked a test charge across the gap; since the push is constant, the voltage is just push times gap-width. Finally we divided charge by voltage — and the charge on top cancelled the charge hiding inside the voltage, leaving only shape: area over gap, times nature's constant. Bigger plates and a thinner gap make a more "generous" capacitor. And when the gap goes to zero the number blows up, when the plates vanish it drops to zero — the formula tells the whole truth, including its own breaking points.

Recall Quick self-check

Why is the field between the plates uniform? ::: A pillbox Gauss surface gives $E = σ / ε_{0}$ with the box area cancelling — no position dependence, so $E$ is the same everywhere. Why does $Q$ disappear from the final formula? ::: $E \propto Q$ so $V \propto Q$ ; dividing $Q$ by something $\propto Q$ cancels it, leaving pure geometry. What happens to $C$ as $d \to 0$ ? ::: $C = ε_{0} A / d \to \infty$ — but physically the dielectric breaks down first. What is $C$ for $A = 2 m^{2}$ , $d = 0.5 mm$ ? ::: $C = ε_{0} (2) / (5 \times 1 0^{- 4}) \approx 3.54 \times 1 0^{- 8} F = 35.4 nF$ .

Related deep-uses of this same field: Coaxial Cable Transmission Lines, Capacitors in Series and Parallel.