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Capacitor Energy Calculator

Find charge Q = CV and stored energy U = ½CV²

Parameters

μFⓘ
Vⓘ
Show Trail

Controls

xⓘ

Calculated Values

Charge:
120.00;μC120.00;μC
Stored Energy:
0.00;J0.00;J

Examples

10 μF at 12 V

Small electrolytic.

  • Stored Energy: 0.000.00

100 μF at 400 V

High-voltage bulk cap.

  • Stored Energy: 8.008.00

Visualization

Energy Stored in a Capacitor

A capacitor stores energy in the electric field between its plates. For parallel-plate capacitors, field E = V/d and energy density u = ½εE² integrate to give U = ½CV².

Charge Q = CV links charge, capacitance, and voltage. Three equivalent energy forms: U = ½CV² = ½QV = Q²/(2C). Use whichever matches known quantities.

Derivation sketch: dW = V dq while charging from 0 to Q; with V = q/C, W = ∫₀^Q (q/C) dq = Q²/(2C) = ½CV². Average voltage during charge is V/2 — hence the factor ½.

RC charging: only half the energy from the battery is stored in C; the other half is dissipated in R during charging. Discharge through R converts stored energy to heat.

Dielectric: inserting dielectric κ increases C = κε₀A/d and energy for fixed V (battery connected) or decreases V for fixed Q (isolated plates).

Applications require rating voltage and energy density. Electrolytic caps store more C per volume but have polarity and ESR.

Key Concepts

  • Q = CV
  • U = ½CV² = ½QV = Q²/(2C)
  • Energy density u = ½εE²
  • RC charge: 50% loss in resistor
  • Dielectric multiplies C by κ
  • Do not exceed rated voltage

Real-World Applications

  • Camera flash and pulsed lasers
  • Defibrillator energy delivery
  • SMPS input bulk capacitors
  • Resonant LC and Tesla coils
  • Touchscreen and MEMS capacitive sensing

Explore Further

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  • Capacitance

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  • RC Circuits

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  • Power Calculator

    Calculate electrical power, energy consumption, and efficiency.

  • Series Circuit

    Analyze series circuits with total resistance, current, and voltage drops.

  • Parallel Circuit

    Analyze parallel circuits with current division and total resistance.

Physics Equations

Charge:
Q=CVQ = CV
Energy:
U=12CV2U = \frac{1}{2}CV^2

Step-by-Step Solution

See how the main results are calculated.

1

Step 1: Convert Capacitance to Farads

Calculator input is in microfarads (μF).

Calculation:

C=10 μF=10×10−6 F=1.0000e−5 FC = 10 \text{ μF} = 10 \times 10^{-6} \text{ F} = 1.0000e-5 \text{ F}

Result:

C=1.0000e−5FC = 1.0000e-5 F

Explanation:

SI unit of capacitance is farad (F). 1 μF = 10⁻⁶ F.

2

Step 2: Stored Charge

Equation:

Q=CVQ = C V

Calculation:

Q=(1.0000e−5)×(12)=1.2000e−4 C=120.0000 μCQ = (1.0000e-5) \times (12) = 1.2000e-4 \text{ C} = 120.0000 \text{ μC}

Result:

Q=120.0000μCQ = 120.0000 μC

Explanation:

Charge on the positive plate equals CV when fully charged to voltage V.

3

Step 3: Select Energy Formula

Choose the form matching your known variables.

Equation:

U=12CV2=12QV=Q22CU = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2C}

Explanation:

All three forms are equivalent. Use U = ½CV² when C and V are known.

4

Step 4: Compute Stored Energy

Substitute C and V into the energy formula.

Calculation:

U=12CV2=12×1.0000e−5×122U = \frac{1}{2} C V^2 = \frac{1}{2} \times 1.0000e-5 \times 12^2
U=7.2000e−4 JU = 7.2000e-4 \text{ J}

Result:

U=7.2000e−4JU = 7.2000e-4 J

Explanation:

Energy is stored in the electric field between plates. The factor ½ arises because voltage builds from 0 to V during charging.

5

Step 5: Verify with U = ½QV

Cross-check using charge and voltage.

Calculation:

U=12QV=12×1.2000e−4×12=7.2000e−4 JU = \frac{1}{2} Q V = \frac{1}{2} \times 1.2000e-4 \times 12 = 7.2000e-4 \text{ J}

Result:

Matches Step 4 ✓

Explanation:

Cross-check confirms consistency between Q, C, and V.

6

Step 6: Practical Note

Relate energy to discharge hazard.

Explanation:

Releasing 7.20e-4 J suddenly can cause spark or component stress. Respect capacitor voltage rating (12 V).

Frequently Asked Questions (FAQ)

Why ½ in the energy formula?

Voltage builds from 0 to V as charge accumulates; average V during the process is V/2.

Can a capacitor shock you?

Large C at high V stores dangerous energy (½CV²). Always discharge high-voltage caps safely.

Energy vs power?

Energy (J) is total stored; power (W) is rate of transfer during charge/discharge.

Two capacitors in series — energy?

Total C is less; with same Q, energy distributes — use U = ½CV² per cap with proper V each.

Real vs ideal capacitor?

Real caps have ESR, leakage, and max ripple current — limits practical energy delivery rate.

Practice MCQs

  1. Doubling voltage (fixed C) multiplies stored energy by:
  2. Doubling capacitance (fixed V) multiplies energy by:
  3. The factor ½ in U = ½CV² arises because:
  4. Energy stored in a capacitor is in:
  5. If Q is fixed and C doubles (isolated plates), energy:
  6. 10 μF at 100 V stores energy approximately:

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