Visual walkthrough — Energy stored in capacitor U = ½CV²

1.8.13 · D2 · Physics › Electromagnetism › Energy stored in capacitor U = ½CV²

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Step 0 — Page par jo words hain, woh kya hain?

Koi bhi symbol formula mein aaye usse pehle, chalo pictures ko naam dete hain.

Ab — aur sirf ab jab $q$, $v$ aur $C$ ka naam ho gaya hai — hum woh ek relation state karte hain jo charge aur voltage ko link karta hai, jo har instant par hold karta hai (yeh $Q=CV$ rule hai, running values ke liye likha gaya): $q=Cv⟺v=\frac{q}{C}$

• $q$ — charge jo abhi plate par baitha hai.
• $C$ — capacitor ki fixed capacitance (ek constant).
• $v$ — abhi voltage, yaani current hill height.

Bilkul end mein yahi rule padhta hai $Q=CV$, yaani $V=Q/C$. Figure dekho: jaise charge $q$ left→right badhta hai, hill $v$ seedha-upar proportion mein badhti hai.

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Step 1 — Plate khaali hai. Pehla marble free hai.

KYA. $q=0$ se shuru karo. $v=q/C$ mein plug karo: $v=\frac{0}{C}=0.$

KYUN. Plate par koi charge baithne se pehle, koi field push-back nahi kar raha. Toh charge ka pehla sliver zero height ki hill chadh-ta hai — cost (almost) kuch nahi.

PICTURE. Hill zameen par flat shuru hoti hai. Yahi woh crucial fact hai jo log bhool jaate hain jab woh guess karte hain "energy $=QV$": tum shuru se poora final voltage $V$ nahi pay karte.

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Step 2 — Charge $dq$ ka ek patla sliver add karo.

KYA. Maano plate abhi charge $q$ hold kar rahi hai, toh hill $v=q/C$ hai. Ab ek patla extra sliver push karo, jise $dq$ kaha jaata hai ("charge $q$ ka ek chota sa bit"). Ek charge ko $v$ height ki hill se lift karne ka kaam hai: $dW=vdq=\frac{q}{C}dq.$

• $dW$ — is ek sliver ke liye kiya gaya thoda sa work.
• $v=q/C$ — is waqt hill ki height (jo charge pehle se wahan hai, final charge nahi).
• $dq$ — charge ka patla sliver jo hum add kar rahe hain.

YEH TOOL KYUN — $dW=vdq$ kyun, $W=vq$ nahi? Kyunki hill ki height $v$ tab bhi badal rahi hai jab hum charge pile karte hain. Hum "voltage × charge" ek shot mein multiply nahi kar sakte, kyunki koi ek voltage hai hi nahi — yeh $0$ se final $V$ tak slide karti hai. Calculus ki trick yeh hai: kaam ko itne patले slivers mein kaato ki har ek par $v$ effectively constant ho. Ek sliver ke liye, $dW=vdq$ exactly sahi hai.

PICTURE. Har sliver hill line ke neeche ek patla vertical rectangle hai: uski width $dq$ hai, height $v$ hai, toh area $vdq$ hai — jo hai hi work. Kaam $v$–$q$ line ke neeche ke area ke barabar hai.

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Step 3 — Khaali se bhar-ne tak har sliver stack karo (integrate karo).

KYA. Total work = saare sliver-rectangles ka sum, pehle se ($q=0$) lekar aakhri tak ($q=Q$, final charge). Yeh grand total hai stored energy $U$ — woh character jise humne Step 0 mein naam diya tha, finally apni entry maar raha hai. Infinitely many infinitely thin slivers ko sum karna yehi matlab hai integral sign $\int$ ka: $U={\int }_0^Q\frac{q}{C}dq.$

• ${\int }_0^Q$ — "saare slivers add karo jaise $q$, $0$ se final charge $Q$ tak jaata hai."
• $\frac{q}{C}dq$ — Step 2 se ek sliver ka work.
• $U$ — grand total: stored energy (joules).

KYUN. Stored energy kuch bhi nahi hai siwaaye us saari mehnat ke jo tumne plate fill karne mein ki. Ek ideal capacitor ke andar koi work nahi khota, toh total work = stored energy.

PICTURE. Saare patले rectangles milke straight hill line ke neeche ek triangle tile karte hain. Aur tumhe triangle ka area pehle se pata hai: $\frac{1}{2}\times \text{base}\times \text{height}$.

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Step 4 — Sum karo: triangle ka area.

KYA. Integral karo (yeh sirf "triangle ka area" hai, algebra se kiya gaya): $U=\frac{1}{C}{\int }_0^{Q}qdq=\frac{1}{C}\cdot \frac{q^2}{2}{|}_0^Q=\frac{Q^2}{2C}.$

• $\frac{1}{C}$ — aage nikalke rakha kyunki $C$ constant hai (jab hum fill karte hain tab nahi badalta).
• $\frac{q^2}{2}{|}_0^Q=\frac{Q^2}{2}$ — $\int qdq$ ka running total, $0$ se final $Q$ tak evaluate kiya gaya.
• $\frac{Q^2}{2C}$ — finished stored energy.

$\frac{1}{2}$ kyun? Triangle dekho. Uski base final charge $Q$ hai, aur uski height final voltage $V$ hai — Step 0 se yaad karo ki $V\equiv v_{\text{final}}=Q/C$. Uska area hai: $\frac{1}{2}\times \underset{\text{base }=q_{\text{final}}}{\underbrace{Q}}\times \underset{\text{height }=V=v_{\text{final}}}{\underbrace{\frac{Q}{C}}}=\frac{Q^2}{2C}.$ $\frac{1}{2}$ "triangle ke area" ka $\frac{1}{2}$ hai. Yahi poora mystery hai.

PICTURE. Do boxes compare karo: GALAT guess "energy $=QV$" poora rectangle hai (base $Q$, height final voltage $V$). SAHI answer triangle hai — exactly uska aadha. Missing half woh upper-left triangle hai jo tumhe kabhi pay nahi karna pada, kyunki early charges ne low hill chadhi.

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Step 5 — Teeno famous forms mein likhte hain.

KYA. Final relation $Q=CV$ use karo (yaani $q_{\text{final}}=C{v}_{\text{final}}$) symbols swap karne ke liye aur ek hi coin ke teeno faces pao: $U=\frac{Q^2}{2C}=\frac{(CV{)}^2}{2C}=\frac{1}{2}C{V}^2=\frac{1}{2}QV.$

• $\frac{Q^2}{2C}$ — tab use karo jab final charge $Q$ fixed ho (capacitor disconnect ho).
• $\frac{1}{2}C{V}^2$ — tab use karo jab final voltage $V$ fixed ho (abhi bhi battery par ho).
• $\frac{1}{2}QV$ — "triangle" form: $Q$ times average voltage $V/2$.

TEEN FORMS KYUN? Woh identical numbers hain, lekin har ek tab convenient hai jab ek alag quantity constant rakhi ja rahi ho. Sahi choose karna tumhari algebra bachata hai (parent ke disconnect/reconnect examples dekho).

PICTURE. Wohi triangle, teen tarah relabel kiya — area kabhi nahi badalta, sirf axes par names badte hain.

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Step 6 — Degenerate cases (koi gap kabhi mat chhodna)

Hum un corners check karte hain jahan cheezein toot sakti hain.

Case A: $q=0$ (khaali plate). Toh $v=0/C=0$ aur $U=0$. Koi charge nahi, koi hill nahi, koi energy nahi. ✓ Triangle ka zero base → zero area.

Case B: $C\to \infty$ (ek "gentle-hill" giant capacitor). Toh $v=q/C\to 0$ kisi bhi finite $q$ ke liye: hill kabhi nahi uthti, toh $U=Q^2/2C\to 0$. Ek capacitor itna bada ki wire jaisa lagta hai, essentially koi energy per coulomb store nahi karta. ✓

Case C: $C\to 0$ (ek "steep-hill" tiny capacitor). Ab $v=q/C$ upar shoot karta hai: hill ek cliff hai. Fixed final $Q$ ke liye, $U=Q^2/2C\to \infty$. Near-zero capacitor par charge cramming karna enormous work leta hai. ✓

Case D: negative charge se charging. Maano hum instead negative charge deliver karte hain, toh final charge $Q=-|Q|$ hai aur har running $q$ negative hai. Do sign-flips hoti hain aur woh cancel ho jaate hain, jo miss karna aasaan hai:

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Ek-picture summary

Upar ki saari cheezein ek figure mein compress ho jaati hain: straight line $v=q/C$, tiny sliver $vdq$, woh triangle jo yeh fill karta hai, aur us triangle ke area ke liye teeno naam ($Q,V$ ke saath jo final charge aur voltage hain).

Recall Feynman retelling — poora walk plain words mein

Tum marbles ek shelf par load kar rahe ho, lekin shelf upar float karti hai jitne zyada marbles tum pile karte ho (woh hai voltage rising: $v=q/C$). Pehla marble zameen par ek shelf par girta hai — free. Aakhri marble ko poori tarah top height $V$ tak uthana padta hai (final voltage). Toh agar tum "lift height" ko "marble number" ke against plot karo, tumhe $0$ se $V$ tak ek straight ramp milega. Kul mehnat — jo exactly stored energy $U$ hai — us ramp ke neeche area hai, aur ek straight ramp ke neeche area ek triangle hai, jo full rectangle ka aadha hai. $Q\times V$ ka aadha $\frac{1}{2}QV$ hai. Ise $Q=CV$ ke saath rename karo aur tumhe $\frac{1}{2}C{V}^2$ ya $Q^2/2C$ milta hai. Famous $\frac{1}{2}$ kabhi magic nahi tha — yeh "triangle ke area" se aaya $\frac{1}{2}$ hai, jo yeh kehne jaisa hai ki "tumne sirf kabhi average hill height, $V/2$, pay ki."

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Connections

• Parent: Energy stored in a capacitor
• 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