Energy stored in capacitor U = ½CV² — Hinglish notes

1.8.13 · Physics › Electromagnetism

KYA store ho raha hai?

Key relation jo hum pehle se jaante hain: $Q = C V$ jahan $C$ capacitance hai (farads mein), $V$ potential difference hai (volts mein), $Q$ charge hai (coulombs mein).

½ ka factor KYU aata hai — scratch se derive karo

Trap yeh hai: log guess karte hain $U = Q V$ kyunki "energy = charge × voltage." Chaliye dekhte hain yeh galat kyun hai aur $\frac{1}{2}$ kahan se aata hai.

Step 1 — Thoda sa charge move karo. Maano capacitor mein pehle se charge $q$ hai. Abhi uska voltage hai $v = \frac{q}{C} .$ Yeh step kyun? Kyunki voltage depend karta hai kitna charge already wahan hai — final charge par nahi. Jaise hum fill karte hain, hill ki height badhalti rehti hai.

Step 2 — Ek aur sliver $d q$ add karne ki work. Potential difference $v$ ke across charge $d q$ move karne mein work lagti hai $d W = v d q = \frac{q}{C} d q .$ Yeh step kyun? Ek potential difference ke through charge move karne ki work hai $d W = (voltage) \times (charge moved)$ . Instantaneous voltage use karo.

Step 3 — Saare slivers add karo (integrate karo). $U = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \cdot \frac{q ^{2}}{2}_{0}^{Q} = \frac{Q ^{2}}{2 C} .$ Yeh step kyun? Total work = saari choti-choti works ka sum, empty ( $q = 0$ ) se full ( $q = Q$ ) tak.

Step 4 — $Q = C V$ use karke rewrite karo. $U = \frac{Q ^{2}}{2 C} = \frac{( C V ) ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V .$

Figure — Energy stored in capacitor U = ½CV²

Galat answer ko steel-man karna

Energy density (energy FIELD mein rehti hai)

Parallel-plate capacitor ke liye, $C = \frac{ε _{0} A}{d}$ aur $V = E d$ hai. Substitute karo: $U = \frac{1}{2} C V^{2} = \frac{1}{2} \cdot \frac{ε _{0} A}{d} \cdot (E d)^{2} = \frac{1}{2} ε_{0} E^{2} (A d) .$ Kyunki $A d$ = plates ke beech volume hai, energy per unit volume hai $u = \frac{U}{volume} = \frac{1}{2} ε_{0} E^{2} .$ Yeh kyun matter karta hai: energy "plates par" nahi hoti — yeh electric field mein khud store hoti hai.

Worked examples

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tum ek shelf par kitaabein stack kar rahe ho, lekin har baar ek kitaab add karne par shelf upar uth jaati hai. Pehli kitaab aasaan hai (shelf neecha hai). Aakhri kitaab tumhe bahut upar uthani padti hai. Toh total effort waisi hai jaise saari kitaabein average shelf height tak uthana — maximum ki aadhi. Ek capacitor bhi aise hi hai: pehla charge push karna aasaan hai, aakhri mushkil hai, toh tum sirf average voltage pay karte ho. Isliye yeh $\frac{1}{2} C V^{2}$ hai, $C V^{2}$ nahi.

Forecast-then-Verify

Flashcards

Energy stored in a capacitor in terms of C and V

U = \frac{1}{2} C V^{2}

Capacitor energy mein ½ ka factor kyun hota hai?

Voltage 0 se V tak badhta hai charge hote waqt, isliye tum sirf average voltage V/2 pay karte ho.

Capacitor energy ke teen equivalent forms

\frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C}

U derive karo: voltage v=q/C par charge dq add karne ki work

d W = \frac{q}{C} d q

; 0→Q integrate karo, milega

Q^{2} /2 C

Battery QV work karti hai lekin capacitor ½QV store karta hai — baaki kahan gaya?

Wire resistance mein heat ke roop mein dissipate ho jaata hai.

Disconnected capacitor, separation double — energy change?

Q fixed, C half,

U = Q^{2} /2 C

double → energy double ho jaati hai.

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

C half,

U = \frac{1}{2} C V^{2}

half ho jaati hai.

Electric field ki energy density

u = \frac{1}{2} ε_{0} E^{2}

(J/m³).

Q constant ho toh kaun sa form use karein?

U = Q^{2} /2 C

Do equal capacitors charge share karte hain — kitni energy lost hoti hai?

Aadhi lost hoti hai (wires mein heat/radiation).

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