# Energy Stored in a Capacitor > A charged capacitor holds energy in the electric field between its two plates, and doubling the voltage across it makes the stored energy four times bigger. Page: https://formulabro.yunusemrevurgun.com/electricity/energy-stored-in-capacitor/ Category: Electricity Published: 2026-10-22 Updated: 2026-10-22 ## Formula $$ E = \frac{1}{2} C V^2 $$ Read it out loud: "E equals one half times C times V squared. E is the energy stored in the capacitor, measured in joules. C is the capacitance, measured in farads. V is the voltage across the capacitor, measured in volts. V squared means V times V." ## Before you start - **Capacitor, charge and electric field**: A capacitor is two metal plates separated by an insulator. When it is connected to a voltage, one plate collects positive charge and the other collects an equal amount of negative charge. **Charge** is measured in **coulombs** (C), and one coulomb is a very large number of electrons, about six billion billion. Between the charged plates is an **electric field**, an invisible region that pushes on other charges, and the energy is held in that field. The letter C also stands for coulomb, but in this formula C always means capacitance. - **Voltage and the volt (V)**: **Voltage** is the push that moves charge from one place to another, measured in **volts** (V). One volt means one joule of energy for each coulomb of charge. So a 12 V capacitor transfers 12 joules of energy for every coulomb that moves across its voltage. A 12 V car battery is a familiar example of a voltage that you can measure. - **Energy and the joule (J)**: **Energy** is the ability to make something happen: warm it, move it or light it up. Its unit is the **joule**, written J. One joule is roughly the energy needed to lift a small apple, about 100 grams, one meter straight up. A **watt** (W) is one joule per second, so a 10 W LED bulb uses 10 joules every second. - **Capacitance, farads and microfarads ($\mu$F)**: **Capacitance**, written $C$, is how much charge a capacitor stores for each volt across it, measured in **farads** (F). One farad is one coulomb per volt, which is a very large amount. Smaller capacitors are measured in microfarads, written $\mu$F, where micro means one millionth, $10^{-6}$. So 1000 microfarads is $1000 \times 10^{-6}$ F, which is 0.001 F, one thousandth of a farad. - **Powers of ten and squares, like $10^{-6}$ and $V^2$**: A **power of ten** is 10 with a small raised number, the exponent, that says how many times to multiply by ten. $10^{-6}$ means one divided by ten six times, which is 0.000001. When two powers of ten are multiplied, their exponents add: $10^{3} \times 10^{-6} = 10^{-3}$. A **square**, written $V^{2}$, means the number multiplied by itself, so $12^{2} = 12 \times 12 = 144$. Squaring makes numbers grow quickly, so doubling a voltage makes its square four times bigger. More notation help: https://formulabro.yunusemrevurgun.com/math-basics/ ## What each symbol means - $E$: The **energy** stored in the capacitor, measured in joules (J). It is the answer the formula gives. A joule is a small amount of energy, roughly the energy needed to lift a small apple one meter. - $C$: The **capacitance** of the capacitor, measured in farads (F). Small capacitors are often given in microfarads or picofarads. The worked example uses 1000 microfarads, which is 0.001 farads. - $V$: The **voltage** across the capacitor, measured in volts (V). The formula uses V squared, so the unit is volts squared during the calculation, and the final answer comes out in joules. A 1.5 V AA battery and a 230 V mains outlet are two common voltages you may have heard of. ## The formula piece by piece - $V$: The voltage across the capacitor, in volts. Voltage is the push that moves charge, and one volt means one joule for each coulomb. This is the starting point, because the energy depends on how hard the capacitor is charged. - $V^2$: The voltage multiplied by itself. Squaring makes the energy grow faster than the voltage: at 2 volts the square is 4, and at 3 volts the square is 9. This is the reason doubling the voltage quadruples the energy. - $C$: The capacitance, in farads, which says how much charge the capacitor holds for each volt. A bigger capacitance stores more energy at the same voltage, in a straight line: twice the capacitance stores twice the energy. - $C V^2$: Capacitance times voltage squared. This product is twice the stored energy, because the one half in the full formula comes from the way the voltage rises from zero while the capacitor charges. - $\frac{1}{2}$: One half. It comes from charging: the voltage starts at zero and climbs to its final value, so the average voltage during charging is only half of the final voltage. - $E = \frac{1}{2} C V^2$: The whole formula in one sentence: the energy stored in a capacitor equals one half times its capacitance times the square of the voltage across it. - $E = \frac{Q^2}{2C}$: A standard second form, using the charge $Q$ in coulombs instead of the voltage. It gives the same energy, because $Q = CV$ links the charge to the capacitance and the voltage. ## What the formula is doing, and why A charged capacitor stores energy in the electric field between its plates. Charging it takes work: the source pushes charge onto one plate and pulls an equal charge off the other. The inputs to the formula are the capacitance $C$ and the voltage $V$ across the capacitor. The output is the energy $E$, in joules, that stays in the field. That energy can be released later, for example in a short flash of light. **Why square the voltage?** The charge on a capacitor is $Q = CV$, so more voltage means more charge. The energy is the voltage multiplied by the charge that moves through it. Both the voltage and the charge grow with $V$, and multiplying them gives $V \times V$, which is $V^2$. Doubling the voltage on the same capacitor therefore makes the energy four times bigger, and tripling the voltage makes it nine times bigger. **Why one half?** The voltage starts at zero and climbs in step with the charge, so during charging the average voltage is half of the final voltage, $\frac{V}{2}$. Multiplying that average by the total charge, $CV$, gives $\frac{V}{2} \times CV = \frac{1}{2} C V^2$. If you used the full voltage for the whole charge, you would get $CV^2$, which is twice the true energy. **Two other forms.** The same energy can be written as $\frac{Q^2}{2C}$ or as $\frac{1}{2} Q V$. They agree because $Q = CV$: put $CV$ in place of $Q$ in either one and you get $\frac{1}{2} C V^2$ back. Use whichever pair of quantities you happen to know. **When it applies.** The formula assumes a fixed capacitance, which is true for most ordinary capacitors. It gives the energy stored at one moment, when the voltage is $V$. High-voltage capacitors can keep a dangerous amount of energy even after the power is off, so they must be discharged safely before anyone works on them. ## An everyday picture Think of a stretched spring. The energy stored in a spring is $\frac{1}{2} k x^2$, where $x$ is how far the spring is stretched and $k$ is its stiffness. The voltage plays the part of the stretch $x$, and the capacitance plays the part of the stiffness $k$. A stiffer spring stores more energy for the same stretch, and a bigger capacitance stores more energy for the same voltage. Stretching the spring twice as far stores four times the energy, in the same way that doubling the voltage quadruples the energy in a capacitor. ## Worked example: Energy in a 1000 microfarad capacitor charged to 12 volts Imagine a made-up example with easy numbers: a capacitor with $C = 1000\ \mu\text{F}$ charged to $V = 12\ \text{V}$, which is about the voltage of a car battery. How much energy is stored? We find the charge first, then check the answer with the other form of the formula, $\frac{Q^2}{2C}$. 1. **Write the formula and list the inputs.** The formula is $E = \frac{1}{2} C V^2$. We have $C = 1000\ \mu\text{F}$ and $V = 12\ \text{V}$. The capacitance is in microfarads, but the formula needs farads, so the first job is to convert it before any multiplying. 2. **Convert the capacitance to farads.** Micro means one millionth, $10^{-6}$. So $1000\ \mu\text{F} = 1000 \times 10^{-6}$ F. Writing 1000 as $10^{3}$ and multiplying the powers of ten adds the exponents: $10^{3} \times 10^{-6} = 10^{3 + (-6)} = 10^{-3}$. So $C = 10^{-3}$ F, which is 0.001 farads, one thousandth of a farad. 3. **Square the voltage.** Squaring means multiplying a number by itself, so $V^2 = (12\ \text{V})^2 = 12 \times 12 = 144\ \text{V}^2$. The unit is volts squared, which we carry along until the end. Squaring is the reason a higher voltage stores so much more energy: 12 squared is 144, but 24 squared is 576, which is four times as large. 4. **Multiply the pieces.** Half of 144 is 72, and then we multiply by the capacitance: $E = \frac{1}{2} \times 10^{-3}\ \text{F} \times 144\ \text{V}^2 = 72 \times 10^{-3}\ \text{J} = 0.072\ \text{J}$. Multiplying by $10^{-3}$ moves the decimal point three places to the left, from 72 to 0.072, so the energy is 0.072 joules. 5. **Check the units.** A farad is a coulomb per volt, so $\text{F} \cdot \text{V}^2 = \frac{\text{C}}{\text{V}} \cdot \text{V}^2 = \text{C} \cdot \text{V}$. A coulomb times a volt is a joule, because a volt is one joule per coulomb. So the answer is in joules, which is the correct unit for energy. 6. **Find the charge on the plates.** The charge is $Q = CV = 10^{-3}\ \text{F} \times 12\ \text{V} = 0.012\ \text{C}$. The power of ten moves the decimal point again: $12 \times 10^{-3} = 0.012$. That is 12 millicoulombs, because milli means one thousandth, so 0.012 coulombs is 12 millicoulombs. 7. **Check with the other form of the formula.** Using $\frac{Q^2}{2C}$, first square the charge: $0.012 \times 0.012 = 0.000144 = 1.44 \times 10^{-4}$. Then divide by $2 \times 10^{-3}$: $\frac{1.44 \times 10^{-4}}{2 \times 10^{-3}} = 0.72 \times 10^{-4 - (-3)} = 0.72 \times 10^{-1} = 0.072$ J. When dividing powers of ten, subtract the exponents, so $-4$ minus $-3$ is $-1$. 8. **Convert the answer to millijoules.** Milli means one thousandth, so 1 joule is 1000 millijoules. Multiplying $0.072\ \text{J}$ by 1000 gives $72\ \text{mJ}$. Small energies are often easier to read in millijoules, and this capacitor stores 72 of them. **Answer:** The capacitor stores **0.072 J**, which is **72 millijoules**. Its charge is 0.012 C, or 12 millicoulombs, and both forms of the formula give the same energy. If the voltage were doubled to 24 V, the energy would be four times bigger, which is **288 millijoules**. ## Common mistakes - Using $V$ instead of $V^2$. Doubling the voltage makes the energy four times bigger, so the answer does not merely double. - Forgetting the one half. Without it the answer is twice the true energy. - Forgetting the micro prefix. $1000\ \mu\text{F}$ is $10^{-3}$ F, not 1000 F. Skipping the prefix makes the answer a million times too big. - Treating the capacitor like a battery. A capacitor's voltage falls as it discharges, so it does not supply a steady voltage the way a battery does. ## Where you will meet it - Camera flashes, which store energy in a capacitor and release it in a short burst of light. - Defibrillators, which store energy in a high-voltage capacitor before delivering a shock to the heart. - Power supplies, where stored energy keeps a device running through very short breaks in the input power. - Spot welders and pulsed lasers, which release a large store of energy very quickly. ## Questions people ask ### Why is the voltage squared? The charge on the plates grows in step with the voltage, and the energy is voltage times charge. Two things that both grow with $V$ multiply to give $V$ times $V$, which is $V^2$. ### Is the energy lost when the capacitor is disconnected? No. An ideal capacitor keeps its charge and its energy after it is disconnected. Real capacitors leak a little, so the stored energy falls slowly over time. ### Can a capacitor stay dangerous after the power is off? Yes. A high-voltage capacitor can keep a dangerous charge after the power is switched off. It must be discharged safely, with a resistor or a tool made for the job, before anyone touches its terminals. ## Related formulas - [Capacitance](https://formulabro.yunusemrevurgun.com/electricity/capacitance/) - [Energy Stored in an Inductor](https://formulabro.yunusemrevurgun.com/electricity/energy-stored-in-inductor/) - [Electric Power](https://formulabro.yunusemrevurgun.com/electricity/electric-power/) ## Sources - [Capacitor, Wikipedia](https://en.wikipedia.org/wiki/Capacitor) - [Capacitance, Wikipedia](https://en.wikipedia.org/wiki/Capacitance) From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.