Energy stored in capacitor U = ½CV²: Worked examples

1.8.13 · D3 · Physics › Electromagnetism › Energy stored in capacitor U = ½CV²

Yeh ek problem gym hai parent topic ke liye. Parent ne formula build kiya tha $U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C} .$ Yahan hum har tarah ke questions hit karenge jo exam pooch sakta hai — aur, sabse zaroori baat, hum pehle decide karte hain kaunsa form use karna hai, kyunki galat form choose karna #1 mistake hai.

Scenario matrix

Har capacitor-energy question inhi cells mein se ek hoti hai (ya blend). Neeche ke worked examples labeled hain us cell se jo woh hit karte hain.

Cell	Kya vary karta hai / kya special hai	Kaunsa form	Example
A Plug-and-chug	$C, V$ directly diye hain	$\frac{1}{2} C V^{2}$	Ex 1
B $Q$ fixed, geometry changed	disconnected, $d$ ya $A$ changes	$Q^{2} /2 C$	Ex 2
C $V$ fixed, geometry changed	connected, $d$ ya $A$ changes	$\frac{1}{2} C V^{2}$	Ex 3
D Dielectric insert kiya	$C \to κ C$	B/C pe depend karta hai	Ex 4
E Charge sharing / redistribution	do caps, energy lost	conservation of $Q$	Ex 5
F Zero / degenerate input	$V = 0$ , ya $d \to \infty$ , ya $Q = 0$	limiting behaviour	Ex 6
G Real-world word problem	camera flash / defibrillator	$\frac{1}{2} C V^{2}$ + power	Ex 7
H Exam twist: energy density	field $E$ , volume, $u = \frac{1}{2} ε_{0} E^{2}$	energy per volume	Ex 8
I Series/parallel combo	pehle total $C$ nikalo	$\frac{1}{2} C_{e q} V^{2}$	Ex 9

Ab hum har cell clear karenge.

Ex 1 — Cell A: plug-and-chug

Worked example Sabse basic case

Ek $C = 47 μ F$ capacitor ko $V = 12 V$ tak charge kiya gaya hai. Stored energy $U$ aur charge $Q$ nikalo.

Forecast: microfarads aur ek dozen volts — expect karo ek tiny energy, kuch millijoules. Padhne se pehle guess karo.

Step 1 — Form choose karo. Dono $C$ aur $V$ diye hain, kuch bhi change nahi hota → use karo $U = \frac{1}{2} C V^{2}$ . Yeh step kyun? Koi geometry nahi move karti, koi battery drama nahi, toh $V$ -form sabse seedha rasta hai. Step 2 — Substitute karo. $U = \frac{1}{2} (47 \times 1 0^{- 6}) (12)^{2} = \frac{1}{2} (47 \times 1 0^{- 6}) (144) = 3.384 \times 1 0^{- 3} J .$ Step 3 — Charge $Q = C V$ se. $Q = (47 \times 1 0^{- 6}) (12) = 5.64 \times 1 0^{- 4} C = 564 μ C .$ Yeh step kyun? Problem ne $Q$ maanga tha; isse $U = \frac{1}{2} Q V$ se cross-check bhi ho jaata hai.

Verify: $\frac{1}{2} Q V = \frac{1}{2} (5.64 \times 1 0^{- 4}) (12) = 3.384 \times 1 0^{- 3} J$ ✓ — Step 2 se match karta hai. Units: $F \cdot V^{2} = J$ ✓. Millijoules jaise forecast kiya tha ✓.

Ex 2 — Cell B: $Q$ fixed, plates pulled apart

Worked example Disconnected, separation tripled

Ek parallel-plate capacitor $C_{0} = 20 μ F$ ko $V_{0} = 50 V$ tak charge karke disconnect kar diya. Phir plate gap $d$ ko triple kar diya. Naya energy $U^{'}$ aur tumhara kiya hua work nikalo.

Forecast: tum plates ko unki attraction ke against pull karte ho, toh energy add hoti hai. Upar ya neeche? Guess karo.

Step 1 — $Q$ freeze karo. Disconnected ⇒ charge ja nahi sakta. $Q = C_{0} V_{0} = (20 \times 1 0^{- 6}) (50) = 1.0 \times 1 0^{- 3} C .$ Yeh step kyun? Poori problem isi pe hinge karti hai ki kya constant hai. Woh hai $Q$ . Step 2 — Naya capacitance. Parallel plate capacitor C = ε₀A/d se, $C \propto 1/ d$ . $d$ triple karna ⇒ $C^{'} = C_{0} /3 = 6. \overline{6} μ F .$ Step 3 — $Q$ -form use karo (kyunki $Q$ fixed hai):

U' = \frac{Q^2}{2C'} = 3\,U_0 = 0.075\,\text{J}.$$ **Step 4 — Tumhara kiya hua work.** Energy conservation: tumhara muscle work energy increase ke barabar hai. $$W_{you} = U' - U_0 = 0.075 - 0.025 = 0.050\,\text{J}.$$ *Yeh step kyun?* Koi battery connected nahi hai, toh *tum* hi akele energy source ho. **Verify:** $U'/U_0 = 3$ bilkul match karta hai "$U\propto d$ at fixed $Q$" se (kyunki $U=Q^2/2C$ aur $C\propto1/d$). Energy **upar** gayi, forecast ke saath match karta hai ki attraction ke against pull karne mein work lagta hai ✓.

Ex 3 — Cell C: $V$ fixed, plates pulled apart

Figure — Energy stored in capacitor U = ½CV²

Worked example Same move, lekin battery connected rehti hai

Same $C_{0} = 20 μ F$ , $V_{0} = 50 V$ , gap tripled, lekin battery connected rehti hai. $U^{'}$ aur jo charge flow karta hai woh nikalo.

Forecast: Ex 2 mein energy upar gayi thi. Kya battery connect karna yeh flip kar dega? Padhne se pehle guess karo.

Step 1 — $V$ freeze karo. Connected battery $V = 50 V$ hamesha pin karti hai. Yeh step kyun? Alag constant ⇒ alag form. Yahi toh matrix ka poora point hai. Step 2 — Naya capacitance (same geometry): $C^{'} = C_{0} /3 = 6. \overline{6} μ F$ . Step 3 — $V$ -form use karo (kyunki $V$ fixed hai): $U_{0} = \frac{1}{2} C_{0} V_{0}^{2} = \frac{1}{2} (20 \times 1 0^{- 6}) (50)^{2} = 0.025 J,$ $U^{'} = \frac{1}{2} C^{'} V_{0}^{2} = \frac{1}{3} U_{0} = 8.33 \times 1 0^{- 3} J .$ Step 4 — Charge jo nikalta hai. $Q_{0} = C_{0} V_{0} = 1 0^{- 3} C, Q^{'} = C^{'} V_{0} = 3.33 \times 1 0^{- 4} C,$ $Δ Q = Q_{0} - Q^{'} = 6.67 \times 1 0^{- 4} C battery mein wapas flow karta hai.$

Verify: Same physical move, Ex 2 se opposite energy result — yahi toh upar wale diagram ka punchline hai: kya constant rakha hai woh sab decide karta hai. Energy neeche $\frac{1}{3}$ ho gayi; charge battery mein wapas drain ho gaya ✓.

Ex 4 — Cell D: dielectric insert kiya

Worked example Ek mein do sub-cases

Ek $C_{0} = 4 μ F$ capacitor ko $V_{0} = 100 V$ tak charge kiya gaya. $κ = 5$ ka ek dielectric use fully fill karta hai (dekho Dielectrics and capacitance). $U^{'}$ nikalo (a) agar pehle disconnect kar diya, (b) agar connected chhodh diya.

Forecast: dielectric hamesha $C$ badhata hai (to $κ C$ ). Kya stored energy dono cases mein rise ya fall karega? Dono guess karo.

(a) Disconnected — $Q$ fixed. $Q = C_{0} V_{0} = (4 \times 1 0^{- 6}) (100) = 4 \times 1 0^{- 4} C, U_{0} = \frac{Q ^{2}}{2 C _{0}} = 0.02 J .$ $C^{'} = κ C_{0} = 20 μ F, U^{'} = \frac{Q ^{2}}{2 C ^{'}} = \frac{U _{0}}{κ} = \frac{0.02}{5} = 4 \times 1 0^{- 3} J .$ Yeh step kyun? $Q$ fixed ⇒ $Q^{2} /2 C$ ; bada $C$ ⇒ chhota $U$ . Slab khicha jaata hai, field kaam karta hai, energy giriti hai.

(b) Connected — $V$ fixed. $U_{0} = \frac{1}{2} C_{0} V_{0}^{2} = 0.02 J, U^{'} = \frac{1}{2} C^{'} V_{0}^{2} = κ U_{0} = 5 (0.02) = 0.1 J .$ Yeh step kyun? $V$ fixed ⇒ $\frac{1}{2} C V^{2}$ ; bada $C$ ⇒ zyada $U$ (battery extra charge pump karti hai).

Verify: disconnected → $U$ girta hai $κ$ se; connected → $U$ badhta hai $κ$ se. Dono "check the constant" rule se agree karte hain. Dono forms se same $U_{0} = 0.02 J$ cross-check karta hai $\frac{1}{2} C_{0} V_{0}^{2} = Q^{2} /2 C_{0}$ ✓.

Worked example Famous "missing half"

$C_{1} = 6 μ F$ jo $V_{1} = 200 V$ tak charged hai, use ek uncharged $C_{2} = 3 μ F$ se connect kiya (from Capacitors in series and parallel yeh ek parallel join hai). Final voltage, final energy, aur lost energy nikalo.

Forecast: charge conserve hota hai lekin energy nahi — kuch heat/sparks ke roop mein wires mein jaata hai. Kitna? Fraction guess karo.

Step 1 — Charge conserve karo (jab woh touch karte hain tab aur kuch fixed nahi rehta). $Q_{t o t} = C_{1} V_{1} = (6 \times 1 0^{- 6}) (200) = 1.2 \times 1 0^{- 3} C .$ Step 2 — Common final voltage (parallel ⇒ dono pe same $V_{f}$ ): $V_{f} = \frac{Q _{t o t}}{C _{1} + C _{2}} = \frac{1.2 \times 1 0 ^{- 3}}{9 \times 1 0 ^{- 6}} = 133. \overline{3} V .$ Yeh step kyun? Jab join hote hain, unhe ek voltage share karni padti hai; total charge distribute hota hai taaki voltages equal ho jayein. Step 3 — Energies. $U_{i} = \frac{1}{2} C_{1} V_{1}^{2} = \frac{1}{2} (6 \times 1 0^{- 6}) (200)^{2} = 0.12 J,$ $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{f}^{2} = \frac{1}{2} (9 \times 1 0^{- 6}) (133. \overline{3})^{2} = 0.08 J .$ Step 4 — Loss. $Δ U = U_{i} - U_{f} = 0.12 - 0.08 = 0.04 J heat/radiation ke roop mein lost.$

Verify: fraction kept $= U_{f} / U_{i} = 0.08/0.12 = 2/3$ . General formula: $U_{f} / U_{i} = C_{1} / (C_{1} + C_{2}) = 6/9 = 2/3$ ✓. Agar caps equal hote toh exactly aadha lost hota — parent ka Example 4 ✓.

Ex 6 — Cell F: zero aur degenerate inputs

Worked example Edge cases jo naive formulas tod dete hain

Teen limits evaluate karo: (a) $V = 0$ , (b) $Q$ fixed jabki $d \to \infty$ (plates infinitely door khicho), (c) $C \to 0$ fixed $Q$ pe.

Forecast: inme se kaunsa infinity tak blow up karta hai, aur kaunsa quietly zero deta hai? Guess karo.

(a) $V = 0$ (uncharged). $U = \frac{1}{2} C (0)^{2} = 0$ . Kyun? Abhi tak hill pe koi charge nahi → koi work nahi kiya. Sanity: $Q = C V = 0$ bhi, toh $Q^{2} /2 C = 0$ ✓ (consistent despite $0/0$ -jaisi form — $Q$ zyada tezi se zero hoti hai).

(b) $Q$ fixed, $d \to \infty$ . $C = ε_{0} A / d \to 0$ , toh $U = \frac{Q ^{2}}{2 C} \to \infty$ . Kyun? Fixed charges ko hamesha ke liye door kheenchna unbounded work maangta hai — attraction kabhi itni tezi se khatam nahi hoti. Physically tum infinite $d$ reach nahi kar sakte; formula honestly warn karta hai ki energy lagatar badhti jaati hai.

(c) $C \to 0$ fixed $Q$ pe. Same as (b): $U = Q^{2} /2 C \to \infty$ . Ek tiny capacitance jo real charge hold karti hai wo enormous energy enormous voltage pe store karti hai ( $V = Q / C \to \infty$ ).

Verify (b ke liye numeric spot-check): $Q = 1 0^{- 6} C$ ke saath, $C = 1 nF$ pe: $U = \frac{( 1 0 ^{- 6} ) ^{2}}{2 ( 1 0 ^{- 9} )} = 5 \times 1 0^{- 4} J$ ; $C = 1 pF$ pe: $U = 0.5 J$ — $1000 \times$ jump jab $C$ $1000 \times$ shrink hua, confirming $U \propto 1/ C$ ✓.

Ex 7 — Cell G: real-world word problem

Worked example Camera flash

Ek camera flash $C = 150 μ F$ capacitor use karta hai jo $V = 300 V$ tak charged hai. Woh apni saari energy flash tube mein $t = 2 ms$ mein dump karta hai. Stored energy aur deliver ki gayi average power nikalo.

Forecast: chhota capacitor, lekin $300 V$ aur bahut chhota dump time — modest energy expect karo lekin badi peak power. Power guess karo (watts ya kilowatts?).

Step 1 — Energy (kuch nahi badalta, $V$ known → $\frac{1}{2} C V^{2}$ ): $U = \frac{1}{2} (150 \times 1 0^{- 6}) (300)^{2} = \frac{1}{2} (150 \times 1 0^{- 6}) (9 \times 1 0^{4}) = 6.75 J .$ Step 2 — Average power $=$ energy $\div$ time: $P_{a v g} = \frac{U}{t} = \frac{6.75}{2 \times 1 0 ^{- 3}} = 3375 W \approx 3.4 kW .$ Yeh step kyun? Power energy per unit time hai; chhota discharge hi flashes ko itna bright banata hai.

Verify: units $J / s = W$ ✓. Battery-sized device se $3.4 kW$ bilkul wahi reason hai kyun capacitors (batteries nahi) flashes drive karte hain — batteries itni tezi se energy release nahi kar sakti. "Big power" ka forecast confirmed ✓.

Ex 8 — Cell H: energy density (field view)

Worked example Energy actually kahan rehti hai

Ek parallel-plate capacitor ka plate area $A = 0.02 m^{2}$ hai, gap $d = 1 mm$ hai, itna charged hai ki field $E = 2 \times 1 0^{5} V/m$ hai. Energy density of electric field use karo: energy density $u$ aur total energy $U$ nikalo. $ε_{0} = 8.85 \times 1 0^{- 12} F/m$ lo.

Forecast: energy gap ke volume mein spread hai. Tiny volume, toh tiny total energy? Guess karo.

Step 1 — Energy per cubic metre: $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (2 \times 1 0^{5})^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (4 \times 1 0^{10}) = 0.177 J/m^{3} .$ Yeh step kyun? $u = \frac{1}{2} ε_{0} E^{2}$ kehta hai energy field mein store hoti hai, plates pe nahi. Step 2 — Plates ke beech ke volume se multiply karo: $Vol = A d = (0.02) (1 0^{- 3}) = 2 \times 1 0^{- 5} m^{3},$ $U = u \cdot Vol = (0.177) (2 \times 1 0^{- 5}) = 3.54 \times 1 0^{- 6} J .$

Verify ( $\frac{1}{2} C V^{2}$ se cross-check): $C = ε_{0} A / d = (8.85 \times 1 0^{- 12}) (0.02) /1 0^{- 3} = 1.77 \times 1 0^{- 10} F$ ; $V = E d = (2 \times 1 0^{5}) (1 0^{- 3}) = 200 V$ ; $\frac{1}{2} C V^{2} = \frac{1}{2} (1.77 \times 1 0^{- 10}) (200)^{2} = 3.54 \times 1 0^{- 6} J$ ✓ — field view aur plate view bilkul agree karte hain.

Ex 9 — Cell I: series/parallel twist

Worked example Pehle

$C_{e q}$ nikalo Teen capacitors: $C_{1} = 2 μ F$ aur $C_{2} = 2 μ F$ series mein, woh combination parallel mein $C_{3} = 3 μ F$ ke saath. Poora $V = 100 V$ ke across hai. Total stored energy nikalo. (Rules: Capacitors in series and parallel.)

Forecast: energy use karne se pehle network ko ek $C_{e q}$ mein collapse karna zaroori hai. $3 μ F$ se bada ya chhota? Guess karo.

Step 1 — Series pair (reciprocals add hote hain): $\frac{1}{C _{12}} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow C_{12} = 1 μ F .$ Yeh step kyun? Series capacitors charge share karte hain, toh unke inverse capacitances add hote hain — pair dono mein se kisi se bhi kamzor hoti hai. Step 2 — $C_{3}$ ke saath parallel (capacitances add hoti hain): $C_{e q} = C_{12} + C_{3} = 1 + 3 = 4 μ F .$ Step 3 — Energy (poore network ke across $V$ fixed hai): $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} (4 \times 1 0^{- 6}) (100)^{2} = 0.02 J .$

Verify: $C_{e q} = 4 μ F > 3 μ F$ , sahi hai kyunki parallel branch sirf capacitance add karti hai ✓. Energy $0.02 J$ ; units $F \cdot V^{2} = J$ ✓.

Recall Quick self-test: cell name karo, phir form

Ek capacitor disconnect hai aur $κ = 2$ ka slab insert kiya. Kaunsa form aur kya $U$ rise ya fall karega? Cell D (disconnected) → $Q$ fixed → $Q^{2} /2 C$ use karo → $C$ double hoti hai → $U$ aadha ho jaata hai. ::: Cell D; $U$ aadha ho jaata hai. Do identical charged-and-uncharged caps join hue — energy ka fraction kitna bachta hai? ::: Cell E; exactly aadha. Camera flash: battery nahi, capacitor kyun? ::: Cell G; sirf capacitor milliseconds mein energy dump kar sakta hai → kilowatt power.

Connections

Energy stored in capacitor U = ½CV² (index 1.8.13)
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

Worked examples — Energy stored in capacitor U = ½CV²