Capacitors and Combinations

A capacitor stores charge — and energy — by separating positive and negative plates. Two questions: how much charge does it hold per volt (its capacitance), and how do capacitors combine in circuits?

Concept

Capacitance of a conductor system is defined as $C = \frac{Q}{V},$ the charge stored per unit potential difference. Unit: farad (F) $= C/V$. A microfarad ($\mu F = 10^{-6}\,F$) is more typical for lab capacitors.

Parallel-plate capacitor. Two plates of area $A$ separated by $d$, with a dielectric of constant $K$ between: $C = \frac{K\varepsilon_0 A}{d}.$ Notice $C$ grows with $A$ and shrinks with $d$. Cars have small flat plates; supercapacitors have huge effective area from porous materials.

Series combination: capacitors $C_1, C_2, \dots, C_n$ in series carry the same charge $Q$ but split the voltage: $\frac{1}{C_\text{series}} = \sum_i \frac{1}{C_i}.$ $C_\text{series} <$ smallest $C_i$.

Parallel combination: capacitors in parallel share the same voltage but split the charge: $C_\text{parallel} = \sum_i C_i.$ $C_\text{parallel} >$ largest $C_i$.

Derivation

Parallel-plate. Each plate has surface charge density $\sigma = Q/A$. Field between plates (with dielectric of constant $K$): $E = \sigma/(K\varepsilon_0)$. Voltage: $V = Ed = Qd/(K\varepsilon_0 A)$. So $C = Q/V = K\varepsilon_0 A/d.$

Series. Charge conservation on the isolated middle plates forces every capacitor to carry the same $Q$. The total voltage is $V = V_1 + V_2 + \dots = \frac{Q}{C_1} + \frac{Q}{C_2} + \dots = Q\sum\frac{1}{C_i}.$ Equating $V = Q/C_\text{series}$ gives the reciprocal sum.

Parallel. Both plates are at common voltage $V$. Total charge $Q = C_1 V + C_2 V + \dots = V\sum C_i$. So $C_\text{parallel} = \sum C_i$.

Worked Example

Three capacitors $2\,\mu F$, $3\,\mu F$, $6\,\mu F$:

All in series: $\frac{1}{C} = \frac{1}{2}+\frac{1}{3}+\frac{1}{6} = 1 \Longrightarrow C = 1\,\mu F.$

All in parallel: $C = 2+3+6 = 11\,\mu F$.

Mixed (series of $2\,\mu F$ with parallel of $3\,\mu F$ and $6\,\mu F$):

$3 + 6 = 9\,\mu F$ parallel. Then $2\,\mu F$ in series with $9\,\mu F$: $C = \frac{2\times 9}{2+9} = \frac{18}{11} \approx 1.64\,\mu F.$

If connected across $11\,V$, the charge on the series equivalent is $Q = 11\times 18/11 = 18\,\mu C$. This is the charge on the $2\,\mu F$ alone (and the total of $3$ and $6\,\mu F$ together).

Common Confusions

• In series, $C$ decreases; in parallel, $C$ increases. This is opposite of resistors!
• In series, all capacitors carry the same Q, not the same V.
• In parallel, all capacitors share the same V, not the same Q.
• Parallel-plate $C$ formula assumes uniform field — good only when $d \ll \sqrt A$.
• Doubling plate area doubles $C$; doubling separation halves $C$.

Key Takeaways

• $C = Q/V$, unit farad.
• Parallel-plate: $C = K\varepsilon_0 A/d$.
• Series: $1/C = \sum 1/C_i$, $C$ smaller than smallest.
• Parallel: $C = \sum C_i$, $C$ larger than largest.
• Dielectric multiplies $C$ by $K$.