Understand capacitance and the farad: Worked examples

Before anything, one reminder of what each symbol is, in plain words:

The scenario matrix

Every question this topic can throw at you falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it different	Example
A. Solve for $C$	given $Q$ and $V$	Ex 1
B. Solve for $Q$	given $C$ and $V$ (rearrange)	Ex 2
C. Solve for $V$	given $Q$ and $C$ (rearrange)	Ex 3
D. Unit-scale trap	tiny pF / huge F — powers of ten	Ex 4
E. Geometry design	use $C = ε A / d$ , pick $A$ , $d$ , $ε_{r}$	Ex 5
F. Degenerate / limiting	$V \to 0$ , $d \to 0$ , $Q = 0$ — what happens?	Ex 6
G. Energy	$E = \frac{1}{2} C V^{2}$ , and $E$ vs $V$ scaling	Ex 7
H. Real-world word problem	camera flash — charge, energy, discharge	Ex 8
I. Exam twist	charge fixed, then change $d$ — what stays, what moves	Ex 9

The one identity that generates cells A–C is the capacitance triangle: cover the quantity you want, read off the rest.

Figure — Understand capacitance and the farad

Cells A, B, C — the three rearrangements

Cell D — the unit-scale trap

Worked example Example 4 — Cell D: how big is a farad, really?

(a) A parallel-plate cap of $C = 2 pF$ at $V = 12 V$ — find $Q$ . (b) A supercap of $C = 3 F$ at the same $12 V$ — find $Q$ . Compare.

Forecast: the two charges differ by a colossal factor. Guess how many powers of ten.

Both use $Q = C V$ . Why this step? Same rearrangement (cell B); the lesson is in the exponents.
Picofarad ( $p = 1 0^{- 12}$ ): $Q_{a} = (2 \times 1 0^{- 12}) (12) = 2.4 \times 1 0^{- 11} C = 24 pC$
Farad (no prefix): $Q_{b} = (3) (12) = 36 C$ Why this step? Shows a whole-farad device holds ordinary-object amounts of charge.
Ratio: $Q_{b} / Q_{a} = 36/ (2.4 \times 1 0^{- 11}) = 1.5 \times 1 0^{12}$ . Why this step? Quantifies the [!mistake] from the parent note — a farad is fat: about a trillion times more charge than a picofarad cap at the same volts.

Verify: $Q_{a} = 2.4 \times 1 0^{- 11}$ C, $Q_{b} = 36$ C, ratio $1.5 \times 1 0^{12}$ . ✓ Units $F \cdot V = C$ . ✓

Cell E — geometry design

Worked example Example 5 — Cell E: build a target capacitance

You need $C = 1 nF$ . Your plates are $A = 5 \times 1 0^{- 3} m^{2}$ and you'll fill the gap with a plastic film of $ε_{r} = 3.0$ (Dielectrics and Permittivity). What gap $d$ do you need? ( $ε_{0} = 8.85 \times 1 0^{- 12} F/m$ .)

Forecast: to hit a nanofarad — recall Ex 3 in the parent needed a $0.1 mm$ gap for less capacitance without a dielectric. So expect a small gap, but eased by the $ε_{r} = 3$ .

Start from $C = \frac{ε _{0} ε _{r} A}{d}$ and solve for $d$ : $d = \frac{ε _{0} ε _{r} A}{C}$ Why this step? $d$ is the only free knob; rearrange to isolate it.
Substitute: $d = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 3.0 ) ( 5 \times 1 0 ^{- 3} )}{1 \times 1 0 ^{- 9}} = 1.33 \times 1 0^{- 4} m = 0.13 mm$ Why this step? Numerator is $ε A$ (a capacitance-times-length), dividing by $C$ leaves metres.

The picture shows why the knobs push $C$ up or down.

Verify: Put it back: $C = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 3.0 ) ( 5 \times 1 0 ^{- 3} )}{1.33 \times 1 0 ^{- 4}} = 1.0 \times 1 0^{- 9} F = 1 nF$ . ✓ Units of $d$ : $\frac{( F/m ) ( m ^{2} )}{F} = m$ . ✓

Cell F — degenerate and limiting cases

Worked example Example 6 — Cell F: the edge cases you must not fear

Reason through each without plugging blindly.

(a) $Q = 0$ (uncharged). What is $C$ ?

If $Q = 0$ then $V = 0$ too (no charge → no push). So $C = Q / V = 0/0$ looks undefined. Why this step? Spotting the $0/0$ is the point — the formula breaks but the physics doesn't.
$C$ is not $0/0$ ; it's a fixed geometric property $ε A / d$ that exists even with zero charge. The bucket's size doesn't vanish when it's empty. Why this step? Reinforces the parent's core claim: $C$ is geometry, independent of charge.

(b) $V \to 0$ at fixed $C$ . Then $Q = C V \to 0$ . The charge fades smoothly to zero, no blow-up. Why: $Q$ scales linearly with $V$ .

(c) $d \to 0$ (plates touching). In $C = ε A / d$ , as $d \to 0$ , $C \to \infty$ .

Mathematically the capacitance diverges. Why this step? Limiting behaviour of the geometry formula.
Physically it fails: real plates short out (field $E = V / d$ blows up and the insulator breaks down) before $d$ reaches zero. So $C \to \infty$ is a limit, not a buildable device. Why this step? Cover the case and its real-world ceiling — see Gauss's Law and Electric Fields for the field that does the breaking.

Verify: at $V = 0$ : $Q = C \cdot 0 = 0$ for any finite $C$ . ✓ Limit sign: $ε A / d \to + \infty$ as $d \to 0^{+}$ . ✓

Cell G — energy

Cell H — real-world word problem

Worked example Example 8 — Cell H: camera flash

A camera flash uses a $C = 1000 μ F$ capacitor charged to $V = 300 V$ . (a) What charge is stored? (b) What energy is dumped into the flash tube? (c) If the flash lasts $t = 2 ms$ , what is the average power?

Forecast: guess the energy — is it "AA-battery" scale or "single burst" scale?

Charge: $Q = C V = (1000 \times 1 0^{- 6}) (300) = 0.3 C$ . Why this step? Cell B rearrangement; gives the charge that will flow through the tube.
Energy: $E = \frac{1}{2} C V^{2} = \frac{1}{2} (1000 \times 1 0^{- 6}) (300)^{2} = 45 J$ . Why this step? $\frac{1}{2} C V^{2}$ because $C$ and $V$ are known — this is the light+heat delivered.
Average power $= \frac{E}{t} = \frac{45}{2 \times 1 0 ^{- 3}} = 22, 500 W = 22.5 kW$ . Why this step? Power is energy per time; the tiny discharge time turns a modest 45 J into kilowatts of instantaneous brightness.

Verify: $E = \frac{1}{2} Q V = \frac{1}{2} (0.3) (300) = 45 J$ — matches via the alternate energy form. ✓ Power units: $J / s = W$ . ✓

Cell I — the exam twist

Worked example Example 9 — Cell I: charge trapped, then pull the plates apart

A capacitor is charged to $Q = 40 μ C$ , then disconnected from the supply (so $Q$ is now fixed). Its initial gap is $d_{1}$ giving $C_{1} = 2 μ F$ at $V_{1} = 20 V$ . You now pull the plates to double the gap. Find the new $C_{2}$ , $V_{2}$ , and energy $E_{2}$ .

Forecast: which quantity is locked ( $Q$ or $V$ )? Guess whether $V$ goes up or down when you widen the gap.

Consistency check first: $Q = C_{1} V_{1} = (2 \times 1 0^{- 6}) (20) = 40 μ C$ . ✓ Why this step? Confirm the starting numbers agree before changing anything.
Doubling $d$ in $C = ε A / d$ halves capacitance: $C_{2} = \frac{1}{2} C_{1} = 1 μ F$ . Why this step? $C \propto 1/ d$ — geometry, not charge, sets $C$ .
Charge is trapped ( $Q$ fixed), so $V_{2} = \frac{Q}{C _{2}} = \frac{40 \times 1 0 ^{- 6}}{1 \times 1 0 ^{- 6}} = 40 V$ . Why this step? With $Q$ constant and $C$ halved, $V = Q / C$ must double. This is the trap: many expect $V$ to stay at 20 V.
Energy: $E_{2} = \frac{Q ^{2}}{2 C _{2}} = \frac{( 40 \times 1 0 ^{- 6} ) ^{2}}{2 ( 1 \times 1 0 ^{- 6} )} = 8 \times 1 0^{- 4} J = 0.8 mJ$ . Compare $E_{1} = \frac{Q ^{2}}{2 C _{1}} = \frac{( 40 \times 1 0 ^{- 6} ) ^{2}}{2 ( 2 \times 1 0 ^{- 6} )} = 0.4 mJ$ . Why this step? Use $Q^{2} / (2 C)$ because $Q$ is the fixed quantity here. Energy doubled — the extra came from your hand's work pulling the attracting plates apart.

Verify: $C_{2} = 1 μ F$ , $V_{2} = 40 V$ ; check $E_{2}$ via $\frac{1}{2} C_{2} V_{2}^{2} = \frac{1}{2} (1 \times 1 0^{- 6}) (40)^{2} = 8 \times 1 0^{- 4} J$ . ✓ Both energy forms agree.

Active Recall

Recall Which rearrangement for which unknown?

Given $Q$ and $V$ , find capacitance ::: $C = Q / V$ Given $C$ and $V$ , find charge ::: $Q = C V$ Given $Q$ and $C$ , find voltage ::: $V = Q / C$ Disconnected cap, gap doubled — what happens to $V$ ? ::: $C$ halves, $Q$ fixed, so $V$ doubles Doubling voltage multiplies stored energy by what? ::: Four ( $E \propto V^{2}$ ) As $d \to 0$ , what does $C = ε A / d$ do, and why can't we build it? ::: $C \to \infty$ ; the insulator breaks down / plates short first

Connections

Parent: Capacitance & the Farad — the tools these examples exercise.
Electric Charge and the Coulomb — the $Q$ in every example.
Voltage and Potential Difference — the $V$ that floats in Cell I.
Dielectrics and Permittivity — $ε_{r}$ in the geometry design (Ex 5).
Gauss's Law and Electric Fields — the field that breaks down in the $d \to 0$ limit.
Capacitors in Series and Parallel — next when combining these.
Energy Storage in Circuits — the flash problem (Ex 8) in action.