Energy stored in a capacitor

Plug a capacitor into a circuit and, for a fraction of a second, current flows — charge piles onto one plate, drains off the other, and then everything goes still. Nothing seems to be happening. But that quiet capacitor is not empty bookkeeping: it is holding real, recoverable energy, squeezed into the electric field between its plates, ready to be released the instant you give it a path. That stored energy is what fires the flash on an old camera in a single bright pulse, what a defibrillator dumps through a patient's chest in a few milliseconds, and what backs up a computer's memory for the split second between a power cut and the backup battery kicking in. Where does that energy come from, and how much is there? A battery does not fill a capacitor for free — every bit of charge it pushes onto a plate has to fight the repulsion of the charge already sitting there, and that fight costs work. CBSE Class 12 asks you to know exactly how much work, using three tidy formulas built from charge $Q$ , potential difference $V$ , and capacitance $C$ — no calculus derivation required, only confident, fast application. This section gives you those formulas, shows the physical reasoning behind the famous factor of one-half (as background insight, not exam-required derivation), reuses and extends the worked numbers from your first pass through this topic, and drills the traps — factor-of-two slips, unit conversions, and the sneaky "energy is not conserved when capacitors share charge" result — that quietly cost marks in board and competitive papers alike.

Key takeaways

Charging a capacitor stores electrostatic potential energy $U$ in the field between its plates; this energy can later be recovered as work, heat, or light.
The syllabus gives you the formula only: $U=\tfrac12 QV=\tfrac12 CV^2=\dfrac{Q^2}{2C}$ — three equivalent forms of the same energy.
Pick the form that matches what the question gives you: $C$ and $V$ known, use $\tfrac12CV^2$ ; $Q$ and $C$ known, use $Q^2/2C$ ; $Q$ and $V$ known, use $\tfrac12QV$ .
The factor of $\tfrac12$ exists because the voltage across the capacitor climbs from $0$ to $V$ during charging, so the average voltage seen by the charge being moved is $V/2$ , not $V$ — never write $U=QV$ .
When two charged capacitors are connected together, total charge is conserved but total energy generally is not — some is dissipated as heat, unless both started at the same potential.
With units in coulombs, volts and farads throughout, $U$ comes out directly in joules; capacitance is very often quoted in $\mu\text{F}$ or $\text{pF}$ , so convert to farads before you square anything.
Key idea: a charged capacitor stores energy $U=\tfrac12QV=\tfrac12CV^2=Q^2/2C$ because the average voltage during charging is half the final voltage, and this stored energy is generally not conserved when charge later redistributes between capacitors.

The core insight

A capacitor charged to potential difference $V$ with charge $Q$ stores electrostatic energy $U=\tfrac12QV=\tfrac12CV^2=Q^2/2C$ , the half arising because the voltage across it rises steadily from zero to $V$ as it charges.

Where it fits in the syllabus

This is the closing quantitative idea of the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance," coming right after series and parallel combinations of capacitors and the parallel-plate capacitor formulas. The 2025-26 syllabus is explicit and narrow here: energy stored in a capacitor is a formula-only result — deriving it by integrating work done against a rising voltage is explicitly out of scope for the board exam, even though such a derivation is short and appears in many textbooks and coaching notes. What you are expected to do is recall the three equivalent formulas, know which one is convenient for a given problem, and apply them accurately, including to situations where charge is shared between capacitors. Because this topic sits at the very end of the chapter, board and competitive questions often combine it with everything before it — series/parallel combination values of $C$ , parallel-plate geometry ( $C=\varepsilon_0A/d$ or $C=K\varepsilon_0A/d$ ), and charge conservation when capacitors are connected. Treat this section as the capstone that pulls those earlier ideas together into one number: the energy in joules.

The physical picture

Picture an uncharged capacitor being connected to a battery. At the very first instant, the plates carry no charge, so there is no opposing field and the battery can push the first bit of charge across almost freely. But as charge accumulates, the plates develop a potential difference of their own — given by $V=q/C$ for whatever charge $q$ sits there at that instant — and this opposing voltage grows steadily as charging proceeds. Every subsequent bit of charge has to be forced across against a stiffer and stiffer resistance, right up until the capacitor's voltage matches the battery's own EMF $V$ and charging stops. The total work done by the battery over this whole process becomes the electrostatic potential energy $U$ stored in the capacitor. Because charge is not pushed against a constant voltage but against one that rises linearly from $0$ to $V$ , it turns out (and the syllabus asks you to accept this, not derive it) that the total work equals charge times the average of the initial and final voltages, not the final voltage itself:

U = \frac{1}{2}QV

where: $Q$ is the total charge finally stored on either plate (coulombs), and $V$ is the final, steady potential difference across the capacitor (volts) — the same $V$ that appears in $Q=CV$ . The physical reading is that $U$ is exactly half of what you would get from naively multiplying the final charge by the final voltage, $QV$ , because the charge was not all moved across the full voltage — most of it was moved when the voltage was still small. Since $Q=CV$ , this single result can be rewritten using any two of the three quantities $Q$ , $V$ , $C$ , giving two more completely equivalent forms:

U = \frac{1}{2}CV^2 \qquad \text{and} \qquad U = \frac{Q^2}{2C}

where: $C$ is the capacitance (farads). All three expressions are exactly the same number written three different ways — substituting $Q=CV$ into any one of them reproduces the other two. Which one to reach for depends purely on what the problem hands you: given $C$ and $V$ , use $\tfrac12CV^2$ ; given $Q$ and $C$ , use $Q^2/2C$ ; given $Q$ and $V$ directly, use $\tfrac12QV$ . With $Q$ in coulombs, $V$ in volts and $C$ in farads, $U$ comes out in joules.

Figure

A bar chart comparing three equal-height bars labelled half Q V, half C V squared and Q squared over two C, each producing the same energy value, sitting above a small triangular voltage-versus-charge graph showing the shaded triangular area under the rising line as the stored energy

A related idea, worth knowing but not the working tool for problems, is that this energy can also be pictured as living inside the electric field itself, spread through the volume between the plates rather than sitting "on" the charge. For a parallel-plate capacitor this leads to an energy-per-unit-volume proportional to $E^2$ . The syllabus does not ask you to derive or use this energy-density picture quantitatively — stick to the $Q$ , $V$ , $C$ formulas for every calculation.

Derivation AWhere the factor of one-half comes from

The syllabus explicitly does not require you to reproduce this derivation on a board exam — energy stored in a capacitor is listed as formula-only, no derivation. It is included here purely so the factor of $\tfrac12$ stops looking like something to memorise blindly and becomes something you understand, which also makes it far harder to forget under exam pressure. The goal is to show why the total work done charging a capacitor to charge $Q$ and voltage $V$ comes out to $\tfrac12QV$ rather than the naive $QV$ , using nothing beyond the definition $V=q/C$ and a simple averaging argument — no calculus is strictly needed if you accept that the voltage rises in a straight line from $0$ to $V$ .

Step 1. Picture charging as many tiny steps. At some intermediate moment the plate already carries charge $q$ (somewhere between $0$ and the final $Q$ ), so by the definition of capacitance the voltage across the capacitor at that moment is $v=q/C$ . Moving one more small packet of charge across this voltage costs a small amount of work, and that voltage keeps climbing as $q$ grows.

Step 2. Notice the voltage rises in a straight line. Because $v=q/C$ and $C$ is constant, $v$ is directly proportional to $q$ — a straight line from $(0,0)$ when the plate is empty to $(Q,V)$ when charging is complete. This is the key geometric fact: the voltage does not jump straight to its final value $V$ and stay there; it climbs steadily alongside the charge.

Step 3. Total work is charge times average voltage. When a quantity is dragged against a resistance that rises in a straight line from $0$ to $V$ , the total work done equals the total charge moved times the average of the starting and ending voltage — and for a straight line, that average is simply $\left(0+V\right)/2=V/2$ . So the total work, which becomes the stored energy $U$ , is $U=Q\times\dfrac{V}{2}=\dfrac{1}{2}QV$ .

Step 4. Rewrite using $Q=CV$ . Substituting the charge–voltage relation into $U=\tfrac12QV$ gives $U=\tfrac12(CV)V=\tfrac12CV^2$ ; substituting $V=Q/C$ instead gives $U=\tfrac12Q(Q/C)=Q^2/2C$ . All three forms trace back to the same triangular area under the straight-line voltage-versus-charge graph — exactly the shaded triangle in the figure above, whose area is base times height over two, i.e. $Q$ times $V$ over two.

Worked examples

Board level - single charged capacitor

A capacitor of capacitance $C = 4\ \mu\text{F}$ is charged to a potential difference of $V = 50\ \text{V}$ . Find the energy stored in the capacitor, and the charge on it.

Step 1. Identify what is given and pick the matching formula. Both $C$ and $V$ are given, so use $U=\tfrac12CV^2$ , converting $C$ to farads first: $C = 4\times10^{-6}\ \text{F}$ .

Step 2. Substitute and compute. $U = \dfrac12\times(4\times10^{-6})\times(50)^2 = \dfrac12\times4\times10^{-6}\times2500 = \dfrac12\times10^{-2}\ \text{J}$ .

Step 3. State the energy answer. $U = 5\times10^{-3}\ \text{J} = \textbf{5 mJ}$ .

Step 4. Find the charge using $Q=CV$ . $Q = 4\times10^{-6}\times50 = 2\times10^{-4}\ \text{C} = \textbf{200 }\mu\textbf{C}$ .

Step 5. Cross-check with a second formula, as an examiner would expect for full marks on a "show your working" question. Using $U=\tfrac12QV$ : $U=\tfrac12\times(2\times10^{-4})\times50=\tfrac12\times10^{-2}=5\times10^{-3}\ \text{J}$ , matching Step 3 exactly.

Competitive level - charge sharing between capacitors

A parallel-plate capacitor with $C = 10\ \mu\text{F}$ is charged so that it carries $Q = 300\ \mu\text{C}$ of charge. It is then disconnected from the battery and connected in parallel to a second, identical, initially uncharged $10\ \mu\text{F}$ capacitor. Find (a) the energy stored before connection, (b) the common potential and total energy after connection, and (c) how much energy is lost.

Step 1. Find the initial energy using $Q^2/2C$ , since $Q$ and $C$ are both known directly. $U_i = \dfrac{(300\times10^{-6})^2}{2\times10\times10^{-6}} = \dfrac{9\times10^{-8}}{2\times10^{-5}} = 4.5\times10^{-3}\ \text{J} = \textbf{4.5 mJ}$ .

Step 2. Apply charge conservation to find the common potential. When the two capacitors are joined in parallel, the total charge $300\ \mu\text{C}$ redistributes over the combined capacitance $C_{total}=10+10=20\ \mu\text{F}$ , so $V_{common} = \dfrac{Q_{total}}{C_{total}} = \dfrac{300\ \mu\text{C}}{20\ \mu\text{F}} = \textbf{15 V}$ .

Step 3. Compute the new total energy using $\tfrac12CV^2$ with the combined capacitance and the common voltage. $U_f = \dfrac12\times(20\times10^{-6})\times(15)^2 = \dfrac12\times20\times10^{-6}\times225 = \textbf{2.25 mJ}$ .

Step 4. Compute the energy lost and interpret it. $\Delta U = U_i - U_f = 4.5-2.25 = \textbf{2.25 mJ lost}$ — exactly half the original energy vanishes, dissipated as heat and a spark in the connecting wire, even though the total charge of $300\ \mu\text{C}$ is fully conserved throughout. Timing note: a full charge-sharing problem like this should take about 3 minutes in a competitive exam — one line each for $U_i$ , $V_{common}$ , $U_f$ , and $\Delta U$ ; do not waste time re-deriving why energy is lost, just state the redistribution result and move on.

Common traps

Trying to derive $U=\tfrac12CV^2$ from first principles under exam time pressure — the syllabus asks only for the formula, not the derivation; know the three forms cold and apply them, rather than reconstructing the reasoning shown above.
Writing $U=QV$ instead of $U=\tfrac12QV$ — the missing factor of $\tfrac12$ is the single most common mark-losing slip in this topic, and it exists precisely because the voltage climbs from $0$ to $V$ during charging rather than sitting at $V$ throughout.
Assuming energy is conserved whenever charged capacitors are connected together — total charge is always conserved, but total energy is generally not, unless the two capacitors were already at exactly the same potential before connecting.
Losing powers of ten because capacitance is quoted in $\mu\text{F}$ or $\text{pF}$ — always convert to farads before squaring a voltage or dividing charge-squared by capacitance; a missed conversion routinely produces an answer wrong by a factor of $10^6$ or $10^{12}$ .
Mixing up which capacitance to use after a series or parallel combination — always compute the equivalent $C$ for the combination first, then apply the energy formula to that equivalent value, not to one of the original individual capacitors.

Common trapCharge is always conserved when capacitors are connected — stored energy is not

It is tempting to treat energy the way you treat charge: as something that simply carries over unchanged when capacitors are wired together. It does not. Whenever charge redistributes between capacitors that started at different potentials, some energy is unavoidably dissipated as heat in the connecting wires (or radiated as a spark), even in an idealised circuit with no explicit resistor shown. Only when both capacitors already share the same potential before connecting is no energy lost. On an exam, never assume $U_{\text{before}}=U_{\text{after}}$ for a charge-sharing problem — always compute both sides using charge conservation for $Q$ and the energy formula for $U$ separately.

You should be able to…

State all three equivalent energy formulas, $U=\tfrac12QV$ , $U=\tfrac12CV^2$ , $U=Q^2/2C$ , and pick the correct one instantly based on which two of $Q$ , $V$ , $C$ a problem gives you.
Calculate the energy stored in a single capacitor from any two of $Q$ , $V$ , $C$ , converting $\mu\text{F}$ or $\text{pF}$ to farads before computing.
Explain in words, without deriving it from calculus, why the factor of $\tfrac12$ appears — the average voltage during charging is half the final voltage.
Work out the common potential and the new total energy after two capacitors are connected in parallel, using charge conservation followed by the energy formula.
Recognise and quantify that energy is generally lost (as heat) when charged capacitors redistribute charge, even though total charge itself is always conserved.
Avoid the standard numerical traps: dropped factor of $\tfrac12$ , unit conversion slips, and using the wrong capacitance after a series/parallel combination.

Ask the tutor: want to work through the energy-loss calculation for two capacitors that both start out charged to different nonzero potentials, or check your own numbers on a fresh problem?

Why is there a half in the energy formula but not in the potential energy of two point charges?

Good catch — the two situations are physically different. For two fixed point charges, the potential energy $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$ describes bringing one fixed-size charge from infinity against the field of another fixed-size charge — nothing is growing during the process, so there is no averaging to do. In a capacitor, by contrast, the charge itself accumulates gradually and the opposing voltage grows as it accumulates, from $0$ up to $V$ . That growth is exactly what produces the averaging and the resulting factor of $\tfrac12$ . Different physical process, different formula.

If I short-circuit a charged capacitor, where does the stored energy go?

It has to go somewhere, since energy cannot simply vanish. In a real circuit, the connecting wire always has some resistance, however small, and the stored energy is converted almost entirely into heat in that resistance as the discharge current flows — often visible as a small spark if the capacitor was charged to a high enough voltage. In an idealised zero-resistance short circuit the discharge would in principle be instantaneous and the energy would still end up dissipated, though the idealisation breaks down physically; real short circuits always have some resistance, however tiny, that absorbs the energy as heat.

Is the field-energy-density idea, energy per volume proportional to E squared, something I need for board exams?

No — treat it as useful background, not examinable working. The 2025-26 CBSE syllabus lists only the formula-based energy result in terms of $Q$ , $V$ , $C$ ; the idea that energy density in a parallel-plate capacitor is proportional to $E^2$ is a natural extension that shows up in some reference books and in later study of electromagnetism, but it is not part of the prescribed scope here. For every board question, stick to $U=\tfrac12QV=\tfrac12CV^2=Q^2/2C$ .

Why does connecting two capacitors always lose energy unless they started at the same potential — where does the lost energy actually come from physically?

Think of it as an irreversible mixing process, much like pouring hot and cold water together. When the two capacitors are at different potentials and you connect them, charge rushes from the higher-potential plate toward the lower-potential one until they equalise, and that rush of charge through a wire with any resistance whatsoever generates heat, just like current through any resistor does. The specific amount lost, worked out in the competitive-level example above, always comes out as a fixed fraction of the smaller starting energy for two equal capacitors — exactly half is lost when one starts charged and the other starts at zero. If the two capacitors start at the same potential, there is no charge rush at all, hence no heat and no energy loss.