Capacitance — parallel plate derivation, cylindrical, spherical: Exercises

Before we start, three symbols you will meet on every line. If any of these feel unfamiliar, that is exactly what we are about to fix.

The three master formulas we will lean on (all derived in the parent note):

Level 1 — Recognition

Goal: read a formula and plug numbers. No traps of logic yet, just clean substitution.

L1.1 — Plate plug-in

Two parallel plates each of area $A = 0.20 m^{2}$ sit a distance $d = 0.50 mm$ apart with vacuum between them. Find $C$ .

Recall Solution L1.1

Which formula? Flat plates → $C = ε_{0} A / d$ . Convert the gap: $d = 0.50 mm = 0.50 \times 1 0^{- 3} m = 5.0 \times 1 0^{- 4} m$ . $C = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.20 )}{5.0 \times 1 0 ^{- 4}} = \frac{1.77 \times 1 0 ^{- 12}}{5.0 \times 1 0 ^{- 4}} = 3.54 \times 1 0^{- 9} F .$ Answer: $C \approx 3.54 nF$ . A fifth of a square metre still gives only nanofarads — farads are enormous.

L1.2 — Definition drill

A capacitor holds $Q = 6.0 μ C$ when the voltage across it is $V = 3.0 V$ . What is its capacitance?

Recall Solution L1.2

Which formula? The definition itself, $C = Q / V$ — no geometry needed. $C = \frac{6.0 \times 1 0 ^{- 6} C}{3.0 V} = 2.0 \times 1 0^{- 6} F = 2.0 μ F .$ Answer: $C = 2.0 μ F$ . Notice we never asked about the shape — $C = Q / V$ works for any capacitor.

Level 2 — Application

Goal: choose the right shape yourself, do a unit conversion, maybe rearrange.

L2.1 — Solve for the gap

You want a $C = 100 pF$ parallel-plate capacitor from plates of area $A = 1.0 \times 1 0^{- 2} m^{2}$ . What plate separation $d$ do you need?

Recall Solution L2.1

Which formula? $C = ε_{0} A / d$ , but now $d$ is the unknown. Rearrange: $d = ε_{0} A / C$ . $d = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 1.0 \times 1 0 ^{- 2} )}{100 \times 1 0 ^{- 12}} = \frac{8.85 \times 1 0 ^{- 14}}{1.0 \times 1 0 ^{- 10}} = 8.85 \times 1 0^{- 4} m .$ Answer: $d \approx 0.885 mm$ . Sensible — sub-millimetre gaps are exactly how real film capacitors are built.

L2.2 — Coax capacitance per metre

A coaxial cable has inner radius $a = 1.0 mm$ and outer radius $b = 3.0 mm$ , vacuum between. Find its capacitance for a length $L = 1.0 m$ .

Recall Solution L2.2

Which formula? Nested cylinders → $C = \frac{2 π ε _{0} L}{ln ( b / a )}$ . See also Coaxial Cable Transmission Lines. The ratio $b / a = 3.0/1.0 = 3.0$ (the millimetres cancel — only the ratio matters). $ln 3.0 = 1.0986$ . $C = \frac{2 π ( 8.85 \times 1 0 ^{- 12} ) ( 1.0 )}{1.0986} = \frac{5.560 \times 1 0 ^{- 11}}{1.0986} = 5.06 \times 1 0^{- 11} F .$ Answer: $C \approx 50.6 pF$ per metre — a textbook "tens of pF/m" figure.

L2.3 — Spherical shells

Two concentric spheres have radii $a = 4.0 cm$ and $b = 5.0 cm$ . Find $C$ .

Recall Solution L2.3

Which formula? Nested spheres → $C = \frac{4 π ε _{0} ab}{b - a}$ . Convert: $a = 0.040 m$ , $b = 0.050 m$ , $b - a = 0.010 m$ , $ab = 0.040 \times 0.050 = 2.0 \times 1 0^{- 3} m^{2}$ . $C = 4 π (8.85 \times 1 0^{- 12}) \frac{2.0 \times 1 0 ^{- 3}}{0.010} = (1.1127 \times 1 0^{- 10}) (0.20) = 2.23 \times 1 0^{- 11} F .$ Answer: $C \approx 22.3 pF$ .

Level 3 — Analysis

Goal: reason about how $C$ changes, or combine with energy / dielectrics.

L3.1 — Halve the gap, what happens to energy at fixed charge?

A parallel-plate capacitor carries a fixed charge $Q$ . You push the plates from separation $d$ to $d /2$ (charge stays constant — it is disconnected). Does the stored energy rise or fall, and by what factor?

Recall Solution L3.1

Tools: $C = ε_{0} A / d$ and the fixed-charge energy $U = \frac{Q ^{2}}{2 C}$ (from Energy Stored in a Capacitor). Halving $d$ doubles $C$ : $C^{'} = ε_{0} A / (d /2) = 2 C$ . $\frac{U ^{'}}{U} = \frac{Q ^{2} / ( 2 C ^{'} )}{Q ^{2} / ( 2 C )} = \frac{C}{C ^{'}} = \frac{C}{2 C} = \frac{1}{2} .$ Answer: the energy drops to half. The plates attract, so moving them closer releases energy — you would have to do work to pull them back apart.

L3.2 — Slide in a dielectric

The $3.54 nF$ vacuum capacitor from L1.1 is filled with a dielectric of constant $κ = 4.0$ . (a) New capacitance? (b) If it is held at a fixed voltage $V = 12 V$ , how does the stored charge change?

Recall Solution L3.2

(a) Dielectric multiplies $C$ by $κ$ : $C^{'} = κ C = 4.0 \times 3.54 nF = 14.16 nF$ . (b) At fixed $V$ , $Q = C V$ , so charge scales with $C$ : it also rises by $\times 4$ . Original: $Q = C V = (3.54 \times 1 0^{- 9}) (12) = 4.25 \times 1 0^{- 8} C$ . New: $Q^{'} = C^{'} V = (1.416 \times 1 0^{- 8}) (12) = 1.70 \times 1 0^{- 7} C$ . Answer: $C^{'} \approx 14.2 nF$ ; charge quadruples from $\approx 42.5 nC$ to $\approx 170 nC$ .

L3.3 — Where does most of the coax voltage live?

For a coaxial capacitor with $a = 1.0 mm$ , $b = 3.0 mm$ , what fraction of the total voltage is dropped across the inner half of the gap, i.e. from $r = a$ out to the geometric midpoint $r = ab$ ?

Recall Solution L3.3

Tool & why: the field is $E (r) = \frac{λ}{2 π ε _{0} r}$ , falling as $1/ r$ , so voltage builds as $ln r$ . Voltage between two radii $r_{1} < r_{2}$ is proportional to $ln (r_{2} / r_{1})$ . Total: $V_{tot} \propto ln (b / a) = ln 3 = 1.0986$ . Inner half (up to $r = ab$ ): $V_{in} \propto ln \frac{ab}{a} = ln b / a = \frac{1}{2} ln (b / a)$ . $\frac{V _{in}}{V _{tot}} = \frac{\frac{1}{2} l n ( b / a )}{l n ( b / a )} = \frac{1}{2} .$ Answer: exactly half the voltage. Using the geometric mean $ab$ (not the arithmetic mean) splits a logarithmic drop evenly — this is why log geometry is different from linear.

Level 4 — Synthesis

Goal: stitch two ideas together — series/parallel, limits, or a full derivation.

L4.1 — Layered dielectric = capacitors in series

A parallel-plate capacitor (area $A = 0.20 m^{2}$ , total gap $d = 0.50 mm$ ) is filled by two stacked slabs, each of thickness $d /2$ : the lower slab has $κ_{1} = 2.0$ , the upper $κ_{2} = 6.0$ . Find the total capacitance.

Recall Solution L4.1

Idea & why: the boundary between the two slabs sits at the same potential everywhere across it, so we may imagine a thin conductor there. That splits the device into two capacitors in series, each of gap $d /2$ (see Capacitors in Series and Parallel). Each slab: $C_{i} = \frac{κ _{i} ε _{0} A}{d /2} = \frac{2 κ _{i} ε _{0} A}{d}$ . Base value $\frac{ε _{0} A}{d} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.20 )}{5.0 \times 1 0 ^{- 4}} = 3.54 \times 1 0^{- 9} F$ (our old L1.1 number). $C_{1} = 2 (2.0) (3.54 nF) = 14.16 nF$ , $C_{2} = 2 (6.0) (3.54 nF) = 42.48 nF$ . Series combine: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$ . $\frac{1}{C} = \frac{1}{14.16} + \frac{1}{42.48} = 0.07062 + 0.02354 = 0.09416 (nF^{- 1}),$ $C = \frac{1}{0.09416} = 10.62 nF .$ Answer: $C \approx 10.6 nF$ . Check the sanity: the total sits below the smaller of the two — the hallmark of series.

L4.2 — Isolated-sphere limit

Start from the spherical-shell formula and show what happens as the outer shell recedes to infinity ( $b \to \infty$ ). Then compute the capacitance of an isolated conducting sphere of radius $a = 0.10 m$ .

Recall Solution L4.2

Take the limit. $C = \frac{4 π ε _{0} ab}{b - a}$ . Divide top and bottom by $b$ : $C = \frac{4 π ε _{0} a}{1 - a / b} .$ As $b \to \infty$ , $a / b \to 0$ , so the denominator $\to 1$ : $C \to 4 π ε_{0} a .$ Interpretation: a lone sphere still has capacitance — its "other plate" is infinity. Number: $C = 4 π (8.85 \times 1 0^{- 12}) (0.10) = (1.1127 \times 1 0^{- 10}) (0.10) = 1.11 \times 1 0^{- 11} F \approx 11.1 pF .$ Answer: $C \approx 11.1 pF$ .

Level 5 — Mastery

Goal: multi-step, derive-and-optimize, or a subtle limit. Full reasoning required.

L5.1 — Optimal inner radius of a coax

A coaxial cable has a fixed outer radius $b$ . The peak field (which sets breakdown/failure) sits at the inner conductor, $E_{m a x} = E (a) = \frac{λ}{2 π ε _{0} a}$ . For a fixed voltage $V$ across the cable, choose the inner radius $a$ that minimises the peak field. Show $a = b / e$ .

Recall Solution L5.1

Set up $E_{m a x}$ in terms of $V$ . From the parent derivation, $V = \frac{λ}{2 π ε _{0}} ln (b / a)$ , so $\frac{λ}{2 π ε _{0}} = \frac{V}{ln ( b / a )}$ . Substitute: $E_{m a x} = \frac{1}{a} \cdot \frac{V}{l n ( b / a )} = \frac{V}{a l n ( b / a )} .$ Why a derivative? We want the smallest $E_{m a x}$ as $a$ varies. Minimising means finding where the slope $d E_{m a x} / d a = 0$ . Equivalently, since $V$ is a constant multiplier, we maximise the denominator $f (a) = a ln (b / a)$ . Differentiate $f (a) = a ln b - a ln a$ : $f^{'} (a) = ln b - (ln a + a \cdot \frac{1}{a}) = ln b - ln a - 1 = ln (b / a) - 1.$ Set $f^{'} (a) = 0$ : $ln (b / a) = 1 \Rightarrow b / a = e \Rightarrow a = b / e .$ Confirm it is a maximum of $f$ : $f^{''} (a) = - 1/ a < 0$ , so $f$ is concave — this is indeed a max of the denominator, hence a minimum of $E_{m a x}$ . ✅ Answer: $a = b / e \approx 0.368 b$ . This is a genuine engineering rule for high-voltage coax design.

L5.2 — Put a number on it

Take $b = 3.0 mm$ and hold $V = 10 kV$ . (a) Find the optimal $a$ . (b) Find the minimum peak field $E_{m a x}$ at that $a$ .

Recall Solution L5.2

(a) $a = b / e = 3.0/2.71828 = 1.1036 mm \approx 1.10 mm$ . (b) At the optimum $ln (b / a) = 1$ , so $E_{m a x} = \frac{V}{a l n ( b / a )} = \frac{V}{a \cdot 1} = \frac{V}{a} .$ $E_{m a x} = \frac{10 \times 1 0 ^{3} V}{1.1036 \times 1 0 ^{- 3} m} = 9.06 \times 1 0^{6} V/m .$ Answer: $a \approx 1.10 mm$ , $E_{m a x} \approx 9.06 \times 1 0^{6} V/m \approx 9.1 MV/m$ . (Comfortably under the ~3 MV/m breakdown of air only if a solid dielectric fills the gap — which is why real coax uses PTFE, not air.)

Recall Quick self-quiz (cover the answers)

Series capacitors combine how? ::: Reciprocals add: $1/ C = \sum 1/ C_{i}$ — total is below the smallest. At fixed $Q$ , halving the plate gap does what to energy? ::: Halves it ( $U = Q^{2} /2 C$ , and $C$ doubles). Isolated-sphere capacitance? ::: $C = 4 π ε_{0} a$ (from $b \to \infty$ ). Optimal coax inner radius for minimum peak field? ::: $a = b / e$ . Why is $b / a$ in the coax formula unit-free? ::: It is a ratio of like quantities — dimensionless, so mm/mm = m/m.

Return to the parent topic · related: Gauss's Law, Electric Potential, Energy Stored in a Capacitor, Capacitors in Series and Parallel, Dielectrics and Polarization, Coaxial Cable Transmission Lines.