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Electric Charge (Q)
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Sources & Methodology
The Q = CV formula is a fundamental relationship in electrostatics, verified from standard electrical engineering textbooks and the IEEE standards for capacitor specifications.
Fundamentals of Electric Circuits — Alexander & Sadiku (McGraw-Hill)
Standard university textbook defining Q = CV, capacitor energy storage E = 0.5CV², and the relationship between charge, capacitance, and voltage in circuit analysis
NIST — SI Units: Ampere and Coulomb Definitions
Official SI unit definitions for electric charge (coulomb) and current (ampere) used in all calculations
Methodology: Q (coulombs) = C (farads) × V (volts). Unit conversions: pF = 10−12 F, nF = 10−9 F, µF = 10−6 F, mF = 10−3 F. Charge output displayed in most readable unit. Energy calculated as E = 0.5 × C × V2. All computations use IEEE double-precision floating point.
⏱ Last reviewed: April 2026
How to Calculate Charge from Capacitance and Voltage
The relationship between electric charge, capacitance, and voltage is one of the most fundamental equations in electrical engineering. The formula Q = CV describes how much electric charge a capacitor accumulates when a voltage is applied across it — a relationship that underpins everything from power supply filters to DRAM memory cells.
The Three Forms of the Q = CV Formula
Q = C × V
Find charge: Q (coulombs) = C (farads) × V (volts)
Example: 100 µF at 12V → Q = 100×10−6 × 12 = 1,200 µC (1.2 mC)
Example: 100 µF at 12V → Q = 100×10−6 × 12 = 1,200 µC (1.2 mC)
C = Q / V
Find capacitance: C (farads) = Q (coulombs) ÷ V (volts)
Example: 500 µC at 5V → C = 500×10−6 ÷ 5 = 100 µF
Example: 500 µC at 5V → C = 500×10−6 ÷ 5 = 100 µF
V = Q / C
Find voltage: V (volts) = Q (coulombs) ÷ C (farads)
Example: 2.2 mC stored in 220 µF → V = 0.0022 ÷ 0.00022 = 10 V
Example: 2.2 mC stored in 220 µF → V = 0.0022 ÷ 0.00022 = 10 V
Capacitance Unit Conversion Reference
| Unit | Symbol | Value in Farads | Typical Application |
|---|---|---|---|
| Picofarad | pF | 10−12 F | RF, radio, antenna circuits |
| Nanofarad | nF | 10−9 F | Signal coupling, timing |
| Microfarad | µF | 10−6 F | Power supply filters, audio |
| Millifarad | mF | 10−3 F | Large electrolytic, supercaps |
| Farad | F | 1 F | Supercapacitors, UPS systems |
Charge Stored at Common Capacitor Values and Voltages
| Capacitance | 5V | 12V | 24V | 48V |
|---|---|---|---|---|
| 10 µF | 50 µC | 120 µC | 240 µC | 480 µC |
| 100 µF | 500 µC | 1.2 mC | 2.4 mC | 4.8 mC |
| 470 µF | 2.35 mC | 5.64 mC | 11.28 mC | 22.56 mC |
| 1000 µF | 5 mC | 12 mC | 24 mC | 48 mC |
| 1 F (supercap) | 5 C | 12 C | 24 C | 48 C |
Energy Stored in a Capacitor
Along with charge, the energy stored in a capacitor is calculated using E = 0.5 × C × V². This is equivalent to E = 0.5 × Q × V = Q² / (2C). The quadratic relationship with voltage means doubling the voltage quadruples the stored energy. This is why high-voltage capacitors in camera flashes or defibrillators can store significant energy despite small physical size.
💡 Practical Tip: Never exceed a capacitor's rated voltage. The maximum charge is Qmax = C × Vrated. Applying excess voltage causes dielectric breakdown, which permanently damages the capacitor and can cause it to fail explosively in electrolytic types. Always derate capacitors to 80% of their rated voltage in real designs for reliability.
Q = CV in Real-World Circuit Design
- Power supply filtering: Large µF capacitors store charge to smooth rectified AC voltage, delivering current during low-voltage periods
- RC timing circuits: Charge time T = R × C determines the time constant; the capacitor charges to 63.2% of supply voltage in one time constant
- Flash photography: Large capacitors (hundreds of µF at 300-400V) store energy for strobe discharge in microseconds
- DRAM memory: Each bit stored as charge on a capacitor of a few femtofarads (fF), refreshed thousands of times per second
- Supercapacitors (ultracaps): 1-3000 F capacitors with low voltage (2-5V) used for energy recovery in EVs and UPS systems
Frequently Asked Questions
The formula is Q = C × V, where Q is electric charge in coulombs (C), C is capacitance in farads (F), and V is voltage in volts (V). For example, a 100 µF capacitor at 12V stores Q = 100×10−6 × 12 = 0.0012 coulombs (1.2 mC).
Capacitance C = Q ÷ V. If a capacitor stores 0.005 coulombs at 10V, its capacitance = 0.005 ÷ 10 = 0.0005 F = 500 µF. This rearrangement of Q = CV lets you determine an unknown capacitance if you know the stored charge and applied voltage.
Voltage V = Q ÷ C. A 220 µF capacitor with 0.0033 coulombs of stored charge has V = 0.0033 ÷ 0.00022 = 15V. This is useful in circuit analysis to find the voltage across a capacitor when you know its charge state.
Charge is measured in coulombs (C). In practical electronics it is often expressed in millicoulombs (mC = 10−3 C), microcoulombs (µC = 10−6 C), or nanocoulombs (nC = 10−9 C). One coulomb equals one ampere of current flowing for one second.
Capacitance (C) is the inherent ability of a capacitor to store charge per volt — a fixed physical property measured in farads. Charge (Q) is the actual amount stored, which depends on both the capacitance and the applied voltage: Q = C × V. The same capacitor stores different charges at different voltages.
Maximum charge Qmax = C × Vrated. A 1000 µF capacitor rated at 50V can store at most Q = 0.001 × 50 = 0.05 coulombs (50 mC). Exceeding the rated voltage causes dielectric breakdown and can destroy the capacitor.
Energy E = 0.5 × C × V² = 0.5 × Q × V = Q² ÷ (2C). For a 100 µF capacitor at 12V: E = 0.5 × 0.0001 × 144 = 0.0072 joules (7.2 mJ). Energy is proportional to the square of voltage, so doubling voltage quadruples stored energy.
Q = CV is used to size capacitors for energy storage, calculate RC circuit time constants, design power supply filters, determine charge in timing circuits, analyze DRAM memory cells, and compute energy stored in supercapacitors for EV and UPS applications.
In parallel: capacitances add (Ctotal = C1 + C2 + ...), so more total charge can be stored at the same voltage. In series: capacitances decrease (1/Ctotal = 1/C1 + 1/C2 + ...) and each capacitor stores the same charge Q = Ctotal × Vtotal.
A picofarad (pF) = 10−12 F stores picocoulombs at typical voltages. A microfarad (µF) = 10−6 F stores microcoulombs — one million times more charge. A 1 pF capacitor at 5V stores 5 pC; a 1 µF capacitor at 5V stores 5 µC.
An ideal capacitor holds charge indefinitely. In practice, leakage resistance causes gradual discharge. Time constant τ = R × C, where after time τ the voltage drops to 37% of initial. A 1000 µF cap with 1 MΩ leakage has τ = 1000 seconds, losing significant charge after 15-20 minutes.
Current I = C × dV/dt — capacitor current equals capacitance times the rate of voltage change. When charging at constant current, Q = I × t. A 1A current for 1 second deposits 1 coulomb of charge. This is why capacitors block DC but pass AC signals.
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