Worked examples: geometry, Q=CV, combination networks, RC transient

Three problems. First we compute a parallel-plate capacitance from its geometry; second we use $Q = CV$ to find the charge a cap stores; third we collapse a mixed series-parallel capacitor network down to one equivalent value.

Example 1 — parallel-plate with air dielectric

Two 10 cm × 10 cm square plates ($A = 100\,\text{cm}^2$) face each other 1 mm apart in air ($\varepsilon_r = 1$). Find the capacitance.

1. Convert to SI. $A = 100\,\text{cm}^2 = 0.01\,\text{m}^2$, $d = 1\,\text{mm} = 0.001\,\text{m}$.
2. Apply the formula.
   $$C = \varepsilon_0 \varepsilon_r \frac{A}{d} = (8.854 \times 10^{-12}) \cdot 1 \cdot \frac{0.01}{0.001}$$
   $$C = 8.854 \times 10^{-11}\,\text{F} = 88.5\,\text{pF}$$
3. Sanity check. A hand-sized air-gap capacitor delivers only tens of picofarads. To get into the microfarads you either need a much larger plate area, a much smaller gap, or — the industrial answer — a high-$\varepsilon_r$ dielectric. Swap air for an X7R ceramic at $\varepsilon_r = 3000$ and the same 1 mm × 100 cm² plates become ~0.27 µF.

The widget shows the same geometry live. Slide the dielectric dropdown from “Vacuum / air” to “Ceramic (X7R)” and watch C climb three orders of magnitude without touching the plates.

Example 2 — charge on a loaded cap

A 470 µF electrolytic capacitor is connected across a 12 V supply. How much charge does it hold at steady state, and how many electrons is that?

1. Apply Q = CV directly.
   $$Q = CV = 470 \times 10^{-6}\,\text{F} \cdot 12\,\text{V} = 5.64 \times 10^{-3}\,\text{C}$$
   So 5.64 mC of charge per plate (+Q on one, −Q on the other).
2. Count the electrons. The charge on one electron is $e = 1.602 \times 10^{-19}\,\text{C}$. So the plate holds:
   $$N = \frac{Q}{e} = \frac{5.64 \times 10^{-3}}{1.602 \times 10^{-19}} \approx 3.5 \times 10^{16}\,\text{electrons}$$
   Thirty-five quadrillion electrons sitting on a plate smaller than a fingernail — and a second plate an equal distance short of them. The charge imbalance is tiny in percentage terms but macroscopic in absolute numbers.

Example 3 — Q and V across a combination network

A 12 V supply drives the following network: $C_1 = 100\,\mu\text{F}$ in series with a parallel pair $C_2 = C_3 = 200\,\mu\text{F}$. Find the equivalent capacitance, the charge on each capacitor, and the voltage across each capacitor.

1. Reduce the parallel pair.
   $$C_{23} = C_2 + C_3 = 200 + 200 = 400\,\mu\text{F}$$
2. Combine C₁ and C₂₃ in series.
   $$\frac{1}{C_{eq}} = \frac{1}{100} + \frac{1}{400} = \frac{4+1}{400} = \frac{5}{400}$$
   $$C_{eq} = \frac{400}{5} = 80\,\mu\text{F}$$
3. Total charge. Series elements carry the same charge — it sits on C₁ and on the parallel pair treated as a block:
   $$Q = C_{eq} \cdot V_s = 80\,\mu\text{F} \times 12\,\text{V} = 960\,\mu\text{C}$$
4. Voltage across C₁.
   $$V_1 = \frac{Q}{C_1} = \frac{960\,\mu\text{C}}{100\,\mu\text{F}} = 9.6\,\text{V}$$
5. Voltage across the parallel pair.
   $$V_{23} = \frac{Q}{C_{23}} = \frac{960\,\mu\text{C}}{400\,\mu\text{F}} = 2.4\,\text{V}$$
   Check: $V_1 + V_{23} = 9.6 + 2.4 = 12\,\text{V}$ ✓
6. Charge on each parallel cap. C₂ and C₃ both sit across 2.4 V:
   $$Q_2 = Q_3 = 200\,\mu\text{F} \times 2.4\,\text{V} = 480\,\mu\text{C}$$
   (Check: Q₂ + Q₃ = 480 + 480 = 960 µC = Q ✓)

Example 4 — combining a small capacitor network

Three capacitors: $C_1 = 10\,\mu\text{F}$, $C_2 = 22\,\mu\text{F}$, $C_3 = 100\,\text{nF}$. $C_1$ and $C_2$ are wired in parallel; that combination is in series with $C_3$. Find the equivalent capacitance.

1. Collapse the parallel pair first.
   $$C_{12} = C_1 + C_2 = 10 + 22 = 32\,\mu\text{F}$$
2. Combine with C₃ in series. Use microfarads throughout: $C_3 = 0.1\,\mu\text{F}$.
   $$\frac{1}{C_{eq}} = \frac{1}{32} + \frac{1}{0.1} = 0.03125 + 10 = 10.031$$
   $$C_{eq} = \frac{1}{10.031} \approx 0.0997\,\mu\text{F} \approx 99.7\,\text{nF}$$
3. Sanity check. In series the smallest cap dominates. Here C₃ = 100 nF is two orders of magnitude smaller than $C_{12} = 32\,\mu\text{F}$, so the series equivalent should sit just a hair below 100 nF — and it does (99.7 nF). The parallel pair barely matters because its massive capacitance contributes almost nothing to a reciprocal sum dominated by the tiny C₃.

Example 5 — RC charging time

A 100 µF capacitor charges through a 10 kΩ resistor from a 12 V supply. How long until v_C reaches ~63 % of V_s, and how long until it's effectively fully charged? Sketch the shape of the charging and discharging curves.

1. Compute the time constant.
   $$\tau = R \cdot C = 10\,\text{k}\Omega \cdot 100\,\mu\text{F} = 10^{4} \cdot 10^{-4} = 1\,\text{s}$$
   So one time constant is one second exactly — a useful sanity-check combination.
2. The 63 % rule. At $t = \tau = 1\,\text{s}$ the capacitor has reached $V_s(1 - 1/e) \approx 0.632 \cdot 12 \approx 7.58\,\text{V}$. That number doesn't depend on R, C, or V_s — it's $1 - 1/e$, baked into the exponential.
3. The 5τ rule. After $5\tau = 5\,\text{s}$ the capacitor is effectively fully charged — about 99.3 % of V_s.
4. Curve shape. The charging waveform climbs steeply at first (high initial current pumping charge in fast) and asymptotes to V_s. The discharging waveform is the mirror — it falls quickly at first, then trails off toward zero. Same τ governs both halves.

For an interactive version with sliders, animation, and a scrub cursor head to the Charging & discharging widget on the Simulate page.