Energy stored in capacitor U = ½CV²

WHAT is being stored?

Key relation we already know: $Q = CV$ where $C$ is capacitance (farads), $V$ the potential difference (volts), $Q$ the charge (coulombs).

WHY there's a factor of ½ — derive from scratch

The trap: people guess $U = QV$ because "energy = charge × voltage." Let's see why that's wrong and where $\tfrac12$ appears.

Step 1 — Move a tiny bit of charge. Suppose the capacitor already holds charge $q$ . Its voltage right now is $v = \frac{q}{C}.$ Why this step? Because voltage depends on how much charge is already there — not the final charge. The hill height changes as we fill it.

Step 2 — Work to add one more sliver $dq$ . Moving charge $dq$ across potential difference $v$ costs work $dW = v\,dq = \frac{q}{C}\,dq.$ Why this step? Work to move charge through a potential difference is $dW = (\text{voltage})\times(\text{charge moved})$ . Use the instantaneous voltage.

Step 3 — Add up all the slivers (integrate). $U = \int_0^Q \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{q^2}{2}\Big|_0^Q = \frac{Q^2}{2C}.$ Why this step? Total work = sum of all little works from empty ( $q=0$ ) to full ( $q=Q$ ).

Step 4 — Rewrite using $Q = CV$ . $U = \frac{Q^2}{2C} = \frac{(CV)^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV.$

Figure — Energy stored in capacitor U = ½CV²

Steel-manning the wrong answer

Energy density (energy lives in the FIELD)

For a parallel-plate capacitor, $C = \dfrac{\varepsilon_0 A}{d}$ and $V = Ed$ . Substitute: $U = \frac12 C V^2 = \frac12\cdot\frac{\varepsilon_0 A}{d}\cdot(Ed)^2 = \frac12 \varepsilon_0 E^2 (Ad).$ Since $Ad$ = volume between plates, energy per unit volume is $u = \frac{U}{\text{volume}} = \frac{1}{2}\varepsilon_0 E^2.$ Why this matters: energy isn't "on the plates" — it's stored in the electric field itself.

Worked examples

Recall Feynman: explain to a 12-year-old

Imagine stacking books on a shelf, but the shelf rises higher every time you add a book. The first book is easy (shelf is low). The last book you must lift really high. So the total effort is like lifting all the books to the average shelf height — half the maximum. A capacitor is the same: the first charge is easy to push on, the last is hard, so you only pay the average voltage. That's why it's $\tfrac12 CV^2$ and not $CV^2$ .

Forecast-then-Verify

Flashcards

Energy stored in a capacitor in terms of C and V

U = \tfrac12 CV^2

Why is there a factor of ½ in the capacitor energy?

Voltage rises from 0 to V as it charges, so you only pay the average voltage V/2.

Three equivalent forms of capacitor energy

\tfrac12 CV^2 = \tfrac12 QV = \dfrac{Q^2}{2C}

Derive U: work to add charge dq at voltage v=q/C

dW=\frac{q}{C}dq

; integrate 0→Q gives

Q^2/2C

A battery does work QV but capacitor stores ½QV — where's the rest?

Dissipated as heat in wire resistance.

Disconnected capacitor, separation doubled — energy change?

Q fixed, C halves,

U=Q^2/2C

doubles → energy doubles.

Connected capacitor (V fixed), separation doubled — energy?

C halves,

U=\tfrac12 CV^2

halves.

Energy density of the electric field

u = \tfrac12 \varepsilon_0 E^2

(J/m³).

Which form to use when Q is held constant?

U = Q^2/2C

Two equal capacitors sharing charge — fraction of energy lost?

Half is lost (heat/radiation in wires).

Connections

Capacitance and Q = CV
Parallel plate capacitor C = ε₀A/d
Dielectrics and capacitance
Energy density of electric field
Work done by a battery and Joule heating
Capacitors in series and parallel

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, capacitor ko charge karna ek spring ko dabane jaisa hai. Jab plate khaali hoti hai to pehla charge daalna easy hai kyunki voltage zero hai. Lekin jaise-jaise charge bharta jaata hai, voltage badhta jaata hai, aur har naye charge ko zyada "hill" chadhni padti hai. Isliye total work nikaalne ke liye hum instantaneous voltage $v = q/C$ ko use karke integrate karte hain, $0$ se $Q$ tak — aur answer aata hai $U = \tfrac12 CV^2$ .

Yeh $\tfrac12$ kahaan se aaya? Bahut simple — voltage $0$ se $V$ tak linearly badhta hai, to average voltage sirf $V/2$ hota hai. Isliye energy $= Q \times \frac{V}{2} = \tfrac12 QV$ . Log galti se $U = QV$ likh dete hain, jo battery ke fixed voltage ke liye theek hai par capacitor ke liye galat, kyunki capacitor ka voltage constant nahi rehta.

Ek mast point: jab battery se charge karte ho, battery $QV$ ka kaam karti hai par capacitor sirf $\tfrac12 QV$ store karta hai. Baaki aadha energy wire ke resistance mein heat ban kar nikal jaata hai — chahe resistance kitna bhi chhota ho. Aur ek aur cheez: energy actually plates ke beech ke electric field mein store hoti hai, jiska density $u = \tfrac12 \varepsilon_0 E^2$ hota hai.

Exam tip: teeno forms yaad rakho — $\tfrac12 CV^2$ jab $V$ pata ho, $\tfrac12 QV$ jab dono pata ho, aur $\frac{Q^2}{2C}$ jab capacitor disconnected ho (charge fixed). Galat form choose karna sabse common mistake hai!

WHAT is being stored?

WHY there's a factor of ½ — derive from scratch

Steel-manning the wrong answer

Energy density (energy lives in the FIELD)

Worked examples

Forecast-then-Verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Electromagnetism

Connections