Energy stored in capacitor U = ½CV²: Question bank

1.8.13 · D5 · Physics › Electromagnetism › Energy stored in capacitor U = ½CV²

Shuru karne se pehle, toolbox ki ek quick reminder, taaki koi bhi symbol bina explain ke use na ho.

Recall Teen forms aur kab kaun sa "safe" hai

$U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C}$ , jahan $U$ stored energy (joules) hai, $C$ capacitance (farads), $V$ plates ke across voltage (volts), $Q$ ek plate par charge (coulombs). Yeh equal tabhi hain jab $Q = C V$ hold kare. Kisi trap mein tumhe woh form chunna hai jinki quantities fixed rehti hain: battery attached → $V$ fixed → $\frac{1}{2} C V^{2}$ use karo; battery removed → $Q$ fixed → $Q^{2} /2 C$ use karo.

True or false — justify karo

Ek capacitor mein double energy store hoti hai agar tum uska voltage double kar do.

False — energy

V^{2}

ke saath jaati hai, isliye

V

double karne se

U

chaar guna ho jaata hai (from

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

). Charge double karna fixed

C

par bhi

U

chaar guna kar deta hai.

Kyunki

U = \frac{1}{2} Q V

hai, energy bas charge times final voltage hai.

False —

\frac{1}{2}

sirf decoration nahi hai. Yeh isliye exist karta hai kyunki charging ke dauran voltage

0

V

tak chada, isliye tumne average voltage

V /2

pay ki, jo

\frac{1}{2} Q V

deta hai,

Q V

nahi.

Agar do capacitors mein same charge

Q

ho, toh jo bada

C

wala hai woh zyaada energy store karta hai.

False — fixed

Q

par

U = Q^{2} /2 C

use karo, isliye bada

C

matlab chhoti

U

. Badi capacitance matlab gentler voltage hill, isliye same charge kam voltage par aur kam energy ke saath baith jaata hai.

Ek battery jo charge

Q

capacitor mein push karti hai, exactly

\frac{1}{2} Q V

kaam karti hai.

False — battery ek constant voltage

V

hold karti hai, isliye woh poora

W_{ba tt} = Q V

kaam karti hai. Sirf aadha (

\frac{1}{2} Q V

) store hota hai; baaki wire resistance mein heat ke roop mein dissipate ho jaata hai.

Energy capacitor ki "metal plates par" store hoti hai.

False — yeh gap ko bharne wale electric field mein store hoti hai, density

u = \frac{1}{2} ε_{0} E^{2}

ke saath. Field hatao (plates short karo) aur energy chali jaati hai, chahe metal abhi bhi wahan ho.

Ek disconnected charged capacitor mein dielectric slide karne se uski stored energy badhti hai.

False —

Q

fixed aur

C \to κ C

ke saath,

U = Q^{2} /2 C

factor

κ

se ghatti hai. Wahi energy drop hai jis wajah se slab apne aap andar khiench jaata hai.

Ek charged capacitor ko identical uncharged capacitor se connect karne par stored energy conserve hoti hai.

False — charge conserve hota hai lekin energy nahi; exactly aadha connecting wires mein heat/radiation ke roop mein lost ho jaata hai, chahe resistance kitna bhi chhota kyun na ho.

Fixed voltage source ke liye, bada capacitor zyaada energy store karta hai.

True — fixed

V

par,

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

linearly

C

ke saath badhta hai. Zyaada plates-worth ki capacitance matlab same voltage par zyaada charge parked.

Error dhundho

"Charge ka aakhri sliver

d q

utna hi kaam leta hai jitna pehla, isliye

U = Q V

Error yeh assume karna hai ki voltage constant hai. Instantaneous voltage

v = q / C

0

se shuru hoke badhta hai, isliye pehle slivers almost free hain aur sirf aakhri wala poora

V

cost karta hai; integrate karne par

\frac{1}{2} Q V

milta hai.

"Jab hum connected capacitor par

d

double karte hain,

Q

same rehta hai, isliye

U = Q^{2} /2 C

use karo."

Error

Q

ko fixed hold karna hai jabki battery attached hai. Battery

V

fix karti hai,

Q

nahi; charge actually battery mein wapas flow karta hai, isliye tumhe

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

use karna chahiye aur

U

aadha ho jaata hai.

"Energy density

u = \frac{1}{2} ε_{0} E

hai, field ka ek power."

Galat power. Energy hamesha field ke square ke saath scale karti hai,

u = \frac{1}{2} ε_{0} E^{2}

— bilkul waise jaise

U \propto V^{2}

aur

V = E d

"Kyunki

U = Q^{2} /2 C

aur

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

equal hain, main inhe kisi bhi situation mein freely swap kar sakta hoon."

Yeh equal sirf us waqt hain jab

Q = C V

hold kare. Kisi change-of-state problem mein tumhe track karna hai ki kaun si quantity fixed hai; tab do forms opposite trends predict karte hain, isliye blindly swap karne se galat jawab milte hain.

"Battery koi energy lose nahi karti kyunki capacitor woh sab store kar leta hai jo battery supply karti hai."

Battery

Q V

supply karti hai lekin capacitor sirf

\frac{1}{2} Q V

store karta hai; energy conservation demand karta hai ki baaki

\frac{1}{2} Q V

kaheen jaaye — wire heating (Joule loss) mein, jo kabhi khatam nahi hoti.

"Ek disconnected capacitor ki plates khinchne mein koi kaam nahi lagta kyunki charge move nahi karta."

Opposite charged plates ek doosre ko attract karti hain, isliye unhe alag karne ka matlab hai attraction ke against push karna — tum positive kaam karte ho, jo exactly wajah hai ki

U

kyun badhta hai (

U = Q^{2} /2 C

mein

C

shrink ho raha hai).

Why questions

Exactly

\frac{1}{2}

ka factor kyun aata hai aur, maan lo,

\frac{1}{3}

kyun nahi?

Kyunki voltage hill charge ke saath linearly badhti hai (

v = q / C

0

V

tak ek straight line ka average exactly

V /2

hota hai; ek linear ramp hamesha one-half deta hai.

Energy kyun lost hoti hai jab connecting wires mein almost zero resistance ho?

Total heat dissipated sirf redistributed charge aur voltage difference par depend karta hai,

R

par nahi; chhota

R

sirf same energy ko faster dump karta hai (bada current), isliye loss resistance se independent hai.

Dielectric ek disconnected capacitor mein bina kisi ke push kiye kyun khiench jaata hai?

Systems lower potential energy ki taraf move karte hain; slab insert karne se

U

ghatta hai (fixed

Q

par bade

C

ke zariye), isliye field slab par kaam karta hai, use andar khienchta hai.

Tum sirf final charge ko final voltage se multiply karke stored energy kyun nahi nikal sakte?

Kyunki charging ke dauran voltage apni final value par nahi tha — woh

0

V

tak sweep kiya, isliye pehle add hue charge ne bahut chhota "price" pay kiya compared to last mein add hue charge se.

Stored energy charges mein ki bajaye field mein kyun hoti hai?

U = \frac{1}{2} ε_{0} E^{2} \times (volume)

rewrite karne par pata chalta hai ki energy field-squared times us region ke proportional hai jise woh fill karta hai — jahan bhi field exist karta hai, energy bhi hoti hai, chahe usne kya create kiya ho.

Voltage double karne mein chaar guna energy kyun lagti hai lekin sirf double charge?

Charge voltage mein linear hai (

Q = C V

), lekin energy mein charge aur badhta hua voltage dono involved hain, jo

V^{2}

deta hai — ek quadratic, isliye four ka factor.

Edge cases

Ek uncharged capacitor (

Q = 0

) ki stored energy kya hai?

Zero, har form se (

\frac{1}{2} C \cdot 0^{2} = 0

); plates ke beech koi field nahi toh energy reside karne ki koi jagah nahi.

Agar

V \to 0

lekin

C \to \infty

aise ki

Q = C V

finite rahe, toh

U

ka kya hoga?

U = \frac{1}{2} Q V \to 0

kyunki

V \to 0

; ek enormous capacitor negligible voltage par real charge hold karta hai lekin almost koi energy nahi.

Fixed amount of stored energy

U

ke liye,

C

aur

V

ke beech kya trade-off hai?

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

se, fixed

U

par

C

aur

V^{2}

inversely tied hain; tum same energy store kar sakte ho chhote

C

ko high

V

par ya bade

C

ko low

V

par.

Plate separation

d \to 0

hone par fixed charge

Q

ke saath

U

ka kya hota hai?

C = ε_{0} A / d \to \infty

, isliye

U = Q^{2} /2 C \to 0

; touching plates voltage hill collapse kar deti hain aur field khatam ho jaata hai, koi stored energy nahi bachti.

Do capacitors ko connect karne wali perfect (zero-resistance) wire ki limit mein, kya "half lost" result abhi bhi valid hai?

Haan — ideal-wire limit mein bhi exactly aadha lost hota hai; ek full treatment dikhata hai ki missing energy electromagnetic radiation ke roop mein escape hoti hai resistive heat ki jagah, isliye woh lost hoti hai waise bhi.

Agar tum ek capacitor ko resistor ke zariye infinitely slowly charge karo, toh heat ke roop mein kitna waste hoga?

Abhi bhi

\frac{1}{2} Q V

— fixed-voltage battery se charging ke dauran lost fraction speed se independent hai; dheema hona current change karta hai, total heat nahi.

Connections

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