The short answer

A capacitor's charge is Q = CV and its stored energy is E = ½CV². Those values describe a voltage state; resistance and the circuit path determine how quickly it changes.

On this page
Ideal capacitor energy at five voltsAn ideal 100 microfarad capacitor is charged across a five-volt source. Its upper plate is positive and lower plate negative. The ideal stored energy is one half times capacitance times voltage squared: 0.00125 joule; this is a calculation, not a hardware reading.+−+−5 V100 µFE = ½ CV² = 0.00125 J (ideal)
Ideal capacitor energy at five volts

A voltage across two terminals

A capacitor stores separated charge and energy in an electric field. Capacitance C is measured in farads (F), and voltage V is measured between its two terminals. Q = CV gives the magnitude of charge on either plate for an ideal capacitor: one plate has +Q and the other −Q, not 2Q of net charge created from nothing. Energy is ½CV² in joules (J). These equations describe a state, not a continuous DC current through an ideal charged capacitor.

The micro prefix means one millionth: 100 µF is 0.0001 F. Always convert prefixes before multiplying into SI units.

Q = C × V; E = ½ × C × V²

Worked example: 100 µF at 5 V

Assume an ideal 100 µF capacitor held at 5 V. Its charge magnitude is 0.0001 F × 5 V = 0.0005 C = 500 µC. Its stored energy is ½ × 0.0001 F × (5 V)² = 0.00125 J = 1.25 mJ. Neither figure predicts charging time or peak current.

If the voltage doubles to 10 V while capacitance stays the same, Q doubles to 1,000 µC but E quadruples to 5 mJ.

Q = 500 µC; E = 1.25 mJ (100 µF, 5 V)

RC time constant is a different question

Add a 1 kΩ resistor in series with an ideal 5 V charging source and an initially discharged 100 µF capacitor. The time constant is τ = RC = 1,000 Ω × 0.0001 F = 0.1 s. After one time constant, ideal capacitor voltage reaches about 63% of its eventual 5 V, or about 3.16 V. Initial ideal charging current is 5 mA and then falls.

The 0.1 s is a timing scale, not an energy value.

τ = R × C = 0.1 s; Vc(τ) ≈ 0.632 × 5 V ≈ 3.16 V

Opening a switch can leave a voltage behind

If you open the charging path after the capacitor reaches 5 V, an ideal isolated capacitor remains at 5 V indefinitely; real capacitors leak and may discharge gradually. An open switch therefore does not establish that the capacitor is safe to touch. Providing a properly selected discharge path changes the time behavior; for an ideal 1 kΩ discharge path across 100 µF, the voltage falls to about 37% of its starting value after 0.1 s.

An open switch may isolate it from any load.

Vdischarge(τ) ≈ 0.368 × Vinitial (ideal RC path)

Check rating, polarity, and non-ideal behavior

A physical capacitor has a voltage rating, tolerance, leakage, and often an equivalent series resistance. Polarized electrolytic capacitors have marked terminals and must be used with suitable polarity and within the manufacturer's voltage limits. A ceramic capacitor is often non-polarized, but still has a voltage rating and capacitance that may change with applied voltage. Switch-on surge can exceed component limits.

Do not use an ideal 100 µF example as a universal safety threshold. Unknown or high-voltage capacitors can hold dangerous energy long after power is removed.

Debug a capacitor that appears not to charge

Power off first, check the capacitor's orientation and rating, and trace the actual charging and return paths. Inspect the series resistor and any switch; a split rail can leave one terminal floating. With an appropriate meter in voltage mode and a safe low-voltage setup, measure directly across the capacitor, respecting the meter and component polarity. A slow rise may reflect a larger RC time constant than expected; a steady zero may indicate a missing feed or a short. Turn power off and verify residual voltage before handling the circuit.

  1. Confirm capacitance units and the intended series resistance.
  2. Trace both capacitor terminals to the source and return with power removed.
  3. Measure capacitor voltage safely; compare its trend with the predicted time scale, not with energy units.

Change one variable on paper

Try replacing 1 kΩ with 2 kΩ in the 5 V charging example. The final ideal charge remains 500 µC and final energy remains 1.25 mJ because C and final V have not changed. The time constant becomes 0.2 s, and the initial ideal charging current becomes 2.5 mA. This is a useful counterexample to the idea that a resistor determines how much energy an ideal capacitor ultimately holds. Check source limits and leakage separately.

Sources and further reading

Diagrams are simplified teaching illustrations, not measured results or construction plans. See how these guides are prepared.