Capacitance — parallel plate derivation, cylindrical, spherical: Worked examples

This page is the drill hall for the parent capacitance note. There we built three formulas from the same four-step recipe. Here we make sure you never meet a case you haven't already beaten: every sign, every degenerate limit, every real-world twist.

Everything rests on the three boxed results (all derived by Gauss's Law + Electric Potential):

The scenario matrix

Before working anything, let's list every kind of question this topic can throw. Each worked example below is tagged with the cell(s) it covers.

Cell	What makes it different	Example
A. Plain plug-in	Give geometry, want $C$	Ex 1 (plate), Ex 3 (cyl), Ex 5 (sph)
B. Degenerate / zero input	$d \to 0$ , $b \to a$ , $b \to \infty$	Ex 2, Ex 6
C. Limiting behaviour	thin-gap sphere ≈ plate	Ex 6
D. Dielectric ( $κ$ ) inserted	material multiplies $C$	Ex 4
E. Solve backwards	given $C$ , find a dimension	Ex 7
F. Real-world word problem	translate a device to geometry	Ex 8 (coax cable), Ex 5
G. Exam twist / combination	half-filled or series idea	Ex 9

We deliberately hit all seven letters. Follow the "Forecast" line each time — guess before you compute; that is how the number sticks.

Group A — plain plug-ins (with the degenerate & limit cases woven in)

Figure — Capacitance — parallel plate derivation, cylindrical, spherical

Worked example Ex 2 — The degenerate gap

$d \to 0$ (Cell B) Keep Ex 1's plates but imagine shrinking the gap toward $d \to 0$ . What happens to $C$ , and physically what does that mean?

Forecast: $C = ε_{0} A / d$ has $d$ downstairs, so as $d \to 0$ the fraction blows up — $C \to \infty$ . Does that make sense? Yes: zero climb in potential for a given field means huge charge per volt.

Take the limit. $d \to 0 lim \frac{ε _{0} A}{d} = + \infty.$ Why this step? This is the honest way to test a formula's edge — push the input to its boundary and read off the behaviour.
Physical meaning. An infinitesimal gap → the two plates almost touch → almost no work to move charge across → $V \to 0$ at fixed $Q$ → $C = Q / V \to \infty$ . Why this step? A formula limit must agree with the physics, or one of them is wrong.
The real-world catch. Before $d$ truly reaches $0$ , the field $E = V / d$ exceeds the dielectric strength of the air (~ $3 \times 1 0^{6} V/m$ ) and it sparks. So $C \to \infty$ is a mathematical, not achievable, limit.

Verify: Plug $d = 1 0^{- 6} m$ : $C = (8.85 \times 1 0^{- 12}) (0.04) /1 0^{- 6} = 3.54 \times 1 0^{- 7} F$ — already $500 \times$ bigger than Ex 1, confirming the $1/ d$ blow-up. ✅

Group D — a dielectric enters

Group A/B/C — spheres, including the plate limit

Worked example Ex 5 — Spherical, plain plug-in

(Cell A / F) A spherical capacitor (used as a model for a metal-shell sensor) has inner radius $a = 5.0 cm$ , outer shell $b = 8.0 cm$ , vacuum between. Find $C$ .

Forecast: Radii are centimetres, and spheres are geometrically "generous." Guess a few picofarads.

Assemble the combination. $ab = (0.05) (0.08) = 4.0 \times 1 0^{- 3} m^{2}$ , and $b - a = 0.03 m$ . Why this step? The spherical formula packages geometry as $ab / (b - a)$ — "product over difference" (from the $1/ a - 1/ b$ voltage integral).
Plug in. $C = 4 π ε_{0} \frac{ab}{b - a} = \frac{1}{8.99 \times 1 0 ^{9}} \cdot \frac{4.0 \times 1 0 ^{- 3}}{0.03} .$ Why this step? $4 π ε_{0} = 1/ (8.99 \times 1 0^{9})$ — using the Coulomb constant keeps arithmetic clean.
Evaluate. $C = \frac{0.1333}{8.99 \times 1 0 ^{9}} = 1.48 \times 1 0^{- 11} F = 14.8 pF .$

Verify: $ab / (b - a)$ has units $m^{2} / m = m$ , times $F/m$ = F. ✅ A few pF, as forecast. ✅

Worked example Ex 6 — Two spherical limits:

$b \to a$ and $b \to \infty$ (Cells B & C) (i) Show a thin-gap sphere ( $b = a + t$ , small $t$ ) reduces to the parallel-plate formula. (ii) Take $b \to \infty$ and recover the isolated sphere.

Forecast: A tiny gap between two big spheres should look locally flat — like plates. And a lone sphere's "other plate" is infinitely far, so we expect a finite $C = 4 π ε_{0} a$ .

Set $b = a + t$ with $t ≪ a$ . Then $b - a = t$ and $ab = a (a + t) \approx a^{2}$ . Why this step? "Thin gap" means $t$ is negligible next to $a$ , so $ab \approx a^{2}$ .
Substitute. $C = \frac{4 π ε _{0} ab}{b - a} \approx \frac{4 π ε _{0} a ^{2}}{t} = \frac{ε _{0} ( 4 π a ^{2} )}{t} = \frac{ε _{0} A}{t},$ where $A = 4 π a^{2}$ is the sphere's surface area and $t$ the gap. Why this step? This is exactly $C_{plate} = ε_{0} A / d$ — geometry we already trust. Curved plates look flat up close.
Now the far limit $b \to \infty$ . $C = \frac{4 π ε _{0} ab}{b - a} = \frac{4 π ε _{0} a}{1 - a / b} b \to \infty 4 π ε_{0} a .$ Why this step? Dividing top and bottom by $b$ isolates the $b \to \infty$ behaviour cleanly; the $a / b$ term dies.

Verify (numeric, part i): Take $a = 0.05 m$ , $t = 0.5 mm$ . Exact spherical: $ab = 0.05 \times 0.0505 = 2.525 \times 1 0^{- 3}$ , $b - a = 5 \times 1 0^{- 4}$ , $C = (2.525 \times 1 0^{- 3} /5 \times 1 0^{- 4}) / (8.99 \times 1 0^{9}) = 5.62 \times 1 0^{- 10} F$ . Plate approx: $ε_{0} (4 π a^{2}) / t = (8.85 \times 1 0^{- 12}) (0.0314) /5 \times 1 0^{- 4} = 5.56 \times 1 0^{- 10} F$ . Agree to ~ $1%$ . ✅

Group E — solve backwards

Group F — real-world word problem

Worked example Ex 8 — Capacitance of a real coax cable per metre

(Cell F) A coaxial cable has inner conductor diameter $1.0 mm$ , insulation (polyethylene, $κ = 2.25$ ) out to a shield of inner diameter $4.0 mm$ . What is the capacitance per metre?

Forecast: From Ex 3, vacuum coax with $b / a = 3$ gave $\sim 50 pF/m$ ; here $b / a = 4$ (bigger log, smaller $C$ ) but $κ = 2.25$ boosts it. Expect ~ $100 pF/m$ — the standard textbook cable figure.

Radii from diameters. $a = 0.50 mm$ , $b = 2.0 mm$ , so $b / a = 4$ , $ln 4 = 1.386$ . Why this step? The formula uses radii; a diameter is twice the radius. Both halve, so the ratio $b / a$ is unchanged — but always convert to avoid slips.
Insert dielectric. With filler, $C = \frac{2 π κ ε _{0} L}{ln ( b / a )}$ ; set $L = 1 m$ . Why this step? Same $κ$ -multiplies-everything rule as Ex 4, applied to the cylindrical result.
Evaluate. $C = \frac{2 π ( 2.25 ) ( 8.85 \times 1 0 ^{- 12} ) ( 1 )}{1.386} = 9.03 \times 1 0^{- 11} F = 90.3 pF/m .$

Verify: Without $κ$ this would be $40.1 pF/m$ (matches parent note's coax number); $\times 2.25 = 90.3 pF/m$ . ✅ In the "tens-to-hundred pF per metre" band engineers quote. ✅

Group G — the exam twist

Worked example Ex 9 — Two dielectrics stacked: a series trick

(Cell G) A parallel-plate gap $d = 1.0 mm$ , area $A = 0.02 m^{2}$ , is filled half-and-half by thickness: a $0.5 mm$ layer of $κ_{1} = 2$ then a $0.5 mm$ layer of $κ_{2} = 4$ , stacked so the field crosses both in turn. Find $C$ .

Forecast: Stacking two slabs along the field is like two capacitors in series (same charge flows through both). Series gives less than either alone, so expect a modest few hundred pF.

Recognise the structure. Each layer alone is a capacitor: $C_{i} = κ_{i} ε_{0} A / (d /2)$ . The interface carries the same $\pm Q$ that the plates do → they are in series. Why this step? "Same $Q$ , voltages add" is the definition of series; slabs stacked along the field share the charge exactly.
Compute each. With $d /2 = 0.5 \times 1 0^{- 3} m$ : $C_{1} = \frac{2 ( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{5 \times 1 0 ^{- 4}} = 7.08 \times 1 0^{- 10} F,$ $C_{2} = \frac{4 ( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{5 \times 1 0 ^{- 4}} = 1.416 \times 1 0^{- 9} F .$
Series combine. $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} \Rightarrow C = \frac{C _{1} C _{2}}{C _{1} + C _{2}} .$ Why this step? Series capacitances add as reciprocals (voltages add at fixed $Q$ ).
Evaluate. $C = \frac{( 7.08 \times 1 0 ^{- 10} ) ( 1.416 \times 1 0 ^{- 9} )}{7.08 \times 1 0 ^{- 10} + 1.416 \times 1 0 ^{- 9}} = 4.72 \times 1 0^{- 10} F = 472 pF .$

Verify: Series result must be smaller than the smaller part ( $C_{1} = 708 pF$ ): $472 < 708$ . ✅ Equivalent single- $κ$ check: the effective $κ_{eff} = \frac{2 κ _{1} κ _{2}}{κ _{1} + κ _{2}} = \frac{2 ( 8 )}{6} = 2.667$ ; then $C = κ_{eff} ε_{0} A / d = (2.667) (8.85 \times 1 0^{- 12}) (0.02) /1 0^{- 3} = 4.72 \times 1 0^{- 10} F$ . ✅ Same number two ways.

Recall Scenario-matrix self-test

Which formula's answer blows up as the gap shrinks to zero? ::: All three grow, but plate is cleanest: $C = ε_{0} A / d \to \infty$ as $d \to 0$ . A thin-gap spherical capacitor reduces to which formula, and why? ::: Parallel plate $ε_{0} A / t$ — curved plates look flat up close ( $ab \approx a^{2}$ , $A = 4 π a^{2}$ ). Two dielectric slabs stacked along the field combine as? ::: Series — same charge, reciprocals add; result is smaller than either slab alone. Inserting a dielectric of constant $κ$ does what to any capacitance? ::: Multiplies it by $κ$ (field weakens, $V$ drops at fixed $Q$ ). Isolated sphere of radius $a$ has capacitance? ::: $4 π ε_{0} a$ — the $b \to \infty$ limit of the spherical formula.

For the energy side of these devices, continue to Energy Stored in a Capacitor.