Capacitance — parallel plate derivation, cylindrical, spherical: Visual walkthrough

1.8.11 · D2 · Physics › Electromagnetism › Capacitance — parallel plate derivation, cylindrical, spheri

Hum ek hi sawaal ka jawab denge: do flat metal plates ek volt par kitna charge soakh sakti hain?

Step 1 — "Plate par charge" actually dikhta kaise hai

KYA: Hum do identical flat metal plates lete hain, unhe $d$ distance par rakhte hain, aur ek battery se ek plate se doosri par charge move karte hain.

KYU: Metal charge ko freely aur evenly phailne deta hai, isliye surface uniformly coated ho jaati hai — yahi uniformity ki wajah se math simple rehti hai. Agar coating uneven hoti, toh field ek mess hoti.

PICTURE: Figure dekho. Top plate par pale-yellow $+$ marks bichhe hain, bottom par blue $-$ marks. Woh evenly phaile hain — har jagah same bheed.

Figure — Capacitance — parallel plate derivation, cylindrical, spherical

Kyunki charge evenly phaila hua hai, hum ise total $Q$ se nahi balki surface ki bheed se describe karte hain:

Step 2 — Har charge ek arrow banata hai: electric field

KYA: $+ Q$ plate test charges ko door dhakelta hai; $- Q$ plate unhe apni taraf kheenchti hai. Gap mein, dono effects ek hi direction mein point karti hain — $+$ plate se seedha neeche $-$ plate ki taraf.

KYU: Hume field chahiye kyunki field "charge" aur "voltage" ke beech ka bridge hai. Field nahi → push nahi → voltage difference nahi. Field poori recipe mein middle-man hai.

PICTURE: Figure mein gap ke andar ke arrows sab ek hi length ke hain aur seedhe neeche point karte hain — ek bilkul uniform stack of parallel arrows. Plates ke bahar arrows cancel ho jaate hain aur essentially kuch bhi nahi hota.

Dhyan do: humne claim kiya ki field uniform hai. Abhi hum ise prove nahi kar sake. Yahi agla step hai, aur jo tool ise prove karta hai woh hai Gauss's law.

Step 3 — Gauss's law kyun, aur pillbox trick

KYA: Hum ek tiny imaginary sealed box — ek pillbox — positive plate mein aadha ghusate hain, taaki ek flat face metal ke andar ho aur doosra gap mein float kare.

KYU pillbox: Solid metal ke andar field zero hoti hai (charges tab tak chalte rehte jab tak push cancel na ho jaaye), isliye koi field andar wali face se nahi nikal sakta. Field char patli side walls ke saath slide karti hai, isliye unse bhi kuch nahi nikal raha. Sirf gap-face hi field ko pass karne deti hai. Ek surface, ek term.

PICTURE: Figure mein pillbox dikhaya gaya hai: ek chota blue cylinder. Top cap (pink) plate mein daba hua hai — wahaan zero field. Side walls par arrows unke saath slide karte hain — zero flux. Sirf bottom cap (yellow) jo gap mein hai, uspar arrows seedhe ghus rahe hain.

Toh Gauss's law ka poora left-hand side sirf ek term mein collapse ho jaata hai: $field \times area of the one gap-face E \cdot A_{box} = \frac{σ A _{box} charge trapped = crowdedness \times that same area}{ε _{0}}$

$A_{box}$ = pillbox face ka area (ek number jo humne choose kiya; ise cancel hona chahiye).
$σ A_{box}$ = andar sealed charge = bheed × face ka plate ka hissa.

Step 4 — Area cancel hota hai: ek uniform field paida hoti hai

KYA: Banaya hua area $A_{box}$ dono sides par aata hai. Ise divide kar do.

KYU yeh matter karta hai: Agar hamaara jawab ek imaginary box ke size par depend karta jo humne khud banaya, toh physics bakwaas hoti. Cancellation nature ka yeh kehna hai ki field nahi care karti ki tum kahan dekh rahe ho — yeh gap mein har jagah same hai. Yahi exactly "uniform" ka matlab hai.

$E = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}$

Term by term:

$E$ — gap mein field strength, aur jawab mein koi $r$ nahi, koi $x$ nahi → yeh har jagah constant hai.
$σ = Q / A$ — humne Step 1 se bheed substitute ki.
$ε_{0}$ — conversion constant.

PICTURE: Figure mein do alag pillboxes dikhaye hain, mota aur patla. Dono same $E$ dete hain — arrow ki length identical hai. Alag box, same field: uniformity ka proof.

Step 5 — Field se voltage tak: pahaad chadna

KYA: Bottom plate se top plate tak march karo, $d$ ki distance, poore raaste field feel karte hue.

KYU yahan aasaan hai: Field constant hai (Step 4), isliye "distance par $E$ add karna" sirf $E$ times $d$ hai — koi calculus gymnastics nahi chahiye. Constant integrand integral ko ek seedhi multiplication bana deta hai.

$V = \int_{0}^{d} E d x = E d = \frac{Q d}{ε _{0} A}$

$\int_{0}^{d} \dots d x$ — gap ke har sliver par sum, $0$ se $d$ tak.
$E d$ — constant field × gap width = total voltage climbed.
$d$ — plate separation (m). Bada gap → lambi climb → bada $V$ .

PICTURE: Figure voltage ko ek staircase / ramp ke roop mein dikhata hai: kyunki $E$ constant hai, potential ek seedhi slanted line ke roop mein upar jaati hai — bottom plate par $0$ se top plate par $V$ tak. Us line ki slope hi $E$ hai.

Step 6 — Divide karo, aur $Q$ gayab hote dekho

KYA: $V$ ko definition $C = Q / V$ mein daalo.

KYU magic hoti hai: $E \propto Q$ , isliye $V \propto Q$ . Jab hum $Q$ ko kisi cheez se divide karte hain jo $Q$ ke proportional hai, toh $Q$ cancel ho jaata hai. Jo bachta hai woh kitna charge load kiya iske par depend nahi kar sakta — sirf shape aur material par. Yahi capacitance ki gehri sach hai.

$C = \frac{Q}{V} = \frac{Q}{\frac{Q d}{ε _{0} A}} = \frac{ε _{0} A}{d}$

PICTURE: Figure cancellation visually dikhata hai — ek $Q$ numerator box mein aur ek $Q$ $V$ box ke andar, dono par chalk stroke cross karta hua, sirf pure geometry glowing chhodta hua.

Step 7 — Edge cases (reader ko kabhi akela mat chodo)

Case $A \to 0$ (plates ko kuch bhi nahi tak sikkodo): $C = ε_{0} A / d \to 0$ . Koi plate nahi → charge store karne ki jagah nahi → zero capacitance. Sanity ✅.

Case $d \to 0$ (plates touch karti hain): $C \to \infty$ . Zero climb ka matlab hai koi bhi charge zero voltage banata hai, isliye $Q / V$ blow up karta hai. Reality mein plates short out ho jaati hain ya dielectric breakdown hoti hai — formula honestly warn karta hai ki design unphysical hai.

Case $d \to \infty$ (plates ko door kheencho): $C \to 0$ . Climb enormous ho jaati hai, isliye same charge ek huge voltage banata hai → tiny capacitance. Plates ek doosre ko "dekhna" band kar deti hain.

Fixed charge, phir squeeze karo ( $d$ chota): $C$ badhta hai, aur stored energy $U = Q^{2} /2 C$ (dekho Energy Stored in a Capacitor) girta hai. Isliye squeeze karna energy release karta hai — yahi reason hai ki dono plates ek doosre ko attract karti hain. Formula secretly force ko bhi contain karta hai.

PICTURE: Figure $C$ versus $d$ plot karta hai — ek $1/ d$ curve. Yeh $d \to 0$ par infinity ki taraf spike karta hai aur $d$ badhne par zero ki taraf flatten hota hai. Teen chalk dots upar ke teen limits mark karte hain.

Ek-picture summary

Ek board par poori derivation: charge → (Gauss + pillbox) → uniform field → (integrate over $d$ ) → voltage → (divide, $Q$ cancels) → $C = ε_{0} A / d$ .

Recall Feynman retelling — simple words mein poora walk

Humne do metal sheets face to face rakhe aur ek se doosre par electrons pump kiye, isliye ek sheet plus se bhari hai aur doosri minus se — aur evenly, kyunki metal charge ko phaila deta hai. Woh bheed ek push (field) banati hai jo gap ko fill karti hai, aur kyunki dono sheets ek hi direction mein push karti hain, push har jagah same hai — arrows ka ek even stack. "Har jagah same" wala part humne ek clever imaginary box (pillbox) se prove kiya: field sirf uske gap-face se bahar nikal sakti hai, aur box ka banaya hua area cancel ho jaata hai, jo nature ka yeh kehna hai ki field depend nahi karti ki tum kahan dekh rahe ho. Phir humne ek test charge ko gap ke paar march karaya; kyunki push constant hai, voltage sirf push times gap-width hai. Aakhir mein humne charge ko voltage se divide kiya — aur numerator ka charge voltage ke andar chuppe charge se cancel ho gaya, sirf shape bachti hai: area over gap, times nature ka constant. Badi plates aur patla gap ek zyada "generous" capacitor banate hain. Aur jab gap zero ho jaata hai toh number blow up karta hai, jab plates gayab ho jaati hain toh zero ho jaata hai — formula poori sach batata hai, apne tod-phod ke points samait.

Recall Quick self-check

Plates ke beech field uniform kyun hoti hai? ::: Ek pillbox Gauss surface $E = σ / ε_{0}$ deta hai jismein box area cancel ho jaata hai — koi position dependence nahi, isliye $E$ har jagah same hai. $Q$ final formula se kyun gayab ho jaata hai? ::: $E \propto Q$ isliye $V \propto Q$ ; $Q$ ko kisi cheez se divide karna jo $\propto Q$ hai woh ise cancel kar deta hai, sirf pure geometry bachti hai. $d \to 0$ par $C$ ka kya hota hai? ::: $C = ε_{0} A / d \to \infty$ — lekin physically dielectric pehle breakdown kar deta hai. $A = 2 m^{2}$ , $d = 0.5 mm$ ke liye $C$ kya hai? ::: $C = ε_{0} (2) / (5 \times 1 0^{- 4}) \approx 3.54 \times 1 0^{- 8} F = 35.4 nF$ .

Isi field ke related deep-uses: Coaxial Cable Transmission Lines, Capacitors in Series and Parallel.