⚡ Full Lesson · Electricity & Magnetism
C = Q/V · Energy = ½CV²

Capacitors

A capacitor stores energy in an electric field — and its series/parallel combination rules are the OPPOSITE of resistors.

The Memory Trick

💡 C = Q/V, and Combination Rules Flip

Capacitance measures how much charge a capacitor stores per volt applied: C = Q/V, measured in Farads. The energy stored in a charged capacitor is ½CV² (equivalently ½QV). A key detail worth memorizing carefully: capacitor combination rules are the OPPOSITE of resistor combination rules — in series, capacitor reciprocals add; in parallel, capacitor values add directly.

Why It Works
Series capacitors effectively increase the total plate separation (since charge must travel through the whole series chain), which decreases overall capacitance — that's why reciprocals add for series, just like resistors in parallel. Parallel capacitors effectively increase total plate area, which increases capacitance directly — that's why values add for parallel, just like resistors in series.
Step by Step

Working With Capacitors

1
Parallel plate capacitance formula
C = ε₀A/d, where A is plate area and d is plate separation — larger plates or closer spacing both increase capacitance.
Doubling the plate area of a parallel-plate capacitor doubles its capacitance, all else equal.
2
Series capacitors — reciprocals add
1/C_total = 1/C₁ + 1/C₂ + ... — total capacitance is always LESS than the smallest individual capacitor in a series combination.
Two identical 10 μF capacitors in series produce a total capacitance of just 5 μF — half of either individual value.
3
Parallel capacitors — values add directly
C_total = C₁ + C₂ + ... — total capacitance is simply the sum, always greater than any individual capacitor.
Two identical 10 μF capacitors in parallel produce a total capacitance of 20 μF — double either individual value.
🏥 Worked Example
A 4 μF capacitor is charged to 50 V. How much charge does it store, and how much energy?
1
Find charge using C = Q/V: Q = CV = 4×10⁻⁶ × 50 = 2×10⁻⁴ C = 200 μC.
2
Find energy using ½CV²: U = ½ × 4×10⁻⁶ × 50².
3
Solve: U = ½ × 4×10⁻⁶ × 2500 = 5×10⁻³ J = 5 mJ — the total energy stored in the capacitor's electric field at this charge and voltage.
📌 Exam Application
Exams test correctly applying C = Q/V and energy = ½CV², and specifically remembering that capacitor combination rules are the OPPOSITE of the analogous resistor rules (reciprocals for series, direct addition for parallel).
⚠️ Most Common Capacitors Mistakes
The most common trap is applying resistor combination rules directly to capacitors without flipping them — remember capacitors are the mirror image: series capacitors combine like PARALLEL resistors (reciprocals), and parallel capacitors combine like SERIES resistors (direct addition).
✓ Quick Self-Test
1) Write the definition of capacitance. C = Q/V. 2) Write the formula for energy stored in a charged capacitor. U = ½CV² (equivalently ½QV). 3) How do capacitors combine in series — do values add directly, or do reciprocals add? Reciprocals add: 1/C_total = 1/C₁ + 1/C₂. 4) How do capacitors combine in parallel? Values add directly: C_total = C₁ + C₂. 5) How does capacitor combination behavior compare to resistor combination behavior? It's the opposite — capacitors in series behave like resistors in parallel (reciprocals add), and capacitors in parallel behave like resistors in series (values add directly).
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