Capacitance — parallel plate derivation, cylindrical, spherical: Worked examples

1.8.11 · D3 · Physics › Electromagnetism › Capacitance — parallel plate derivation, cylindrical, spheri

Yeh page parent capacitance note ke liye drill hall hai. Wahan humne ek hi chaar-step recipe se teen formulas banaye the. Yahan hum ensure karte hain ki tumhe koi bhi aisa case kabhi na mile jo tumne pehle beat na kiya ho: har sign, har degenerate limit, har real-world twist.

Sab kuch teen boxed results par tika hai (sab Gauss's Law + Electric Potential se derive kiye gaye hain):

Scenario matrix

Kuch bhi work karne se pehle, chalte hain har us tarah ke question ki list banate hain jo yeh topic throw kar sakta hai. Neeche har worked example us cell(s) ke saath tagged hai jise woh cover karta hai.

Cell	Kya cheez ise alag banati hai	Example
A. Plain plug-in	Geometry dono, $C$ chahiye	Ex 1 (plate), Ex 3 (cyl), Ex 5 (sph)
B. Degenerate / zero input	$d \to 0$ , $b \to a$ , $b \to \infty$	Ex 2, Ex 6
C. Limiting behaviour	thin-gap sphere ≈ plate	Ex 6
D. Dielectric ( $κ$ ) inserted	material $C$ ko multiply karta hai	Ex 4
E. Solve backwards	$C$ diya hai, ek dimension nikalni hai	Ex 7
F. Real-world word problem	ek device ko geometry mein translate karo	Ex 8 (coax cable), Ex 5
G. Exam twist / combination	half-filled ya series idea	Ex 9

Humne jaan-bujhkar saatoon letters hit kiye hain. Har baar "Forecast" line follow karo — compute karne se pehle guess karo; isi tarah number yaad rehta hai.

Group A — plain plug-ins (degenerate & limit cases ke saath)

Figure — Capacitance — parallel plate derivation, cylindrical, spherical

Worked example Ex 2 — Degenerate gap

$d \to 0$ (Cell B) Ex 1 wali plates rakho lekin imagine karo ki gap $d \to 0$ ki taraf shrink ho rahi hai. $C$ ka kya hoga, aur physically iska kya matlab hai?

Forecast: $C = ε_{0} A / d$ mein $d$ neeche hai, toh jaise $d \to 0$ hoga fraction blow up hoga — $C \to \infty$ . Kya yeh sense banata hai? Haan: ek given field ke liye potential mein zero climb matlab hai huge charge per volt.

Limit lo. $d \to 0 lim \frac{ε _{0} A}{d} = + \infty.$ Yeh step kyun? Yeh ek formula ki edge test karne ka honest tarika hai — input ko uski boundary tak push karo aur behaviour padho.
Physical meaning. Infinitesimal gap → dono plates almost touch karti hain → charge ko cross move karne mein almost koi kaam nahi → fixed $Q$ par $V \to 0$ → $C = Q / V \to \infty$ . Yeh step kyun? Ek formula limit physics se agree karni chahiye, warna dono mein se ek galat hai.
Real-world catch. Isse pehle ki $d$ truly $0$ tak pahunche, field $E = V / d$ air ki dielectric strength (~ $3 \times 1 0^{6} V/m$ ) se exceed kar jaati hai aur spark ho jaata hai. Toh $C \to \infty$ ek mathematical limit hai, achievable nahi.

Verify: $d = 1 0^{- 6} m$ plug karo: $C = (8.85 \times 1 0^{- 12}) (0.04) /1 0^{- 6} = 3.54 \times 1 0^{- 7} F$ — Ex 1 se already $500 \times$ bada, $1/ d$ blow-up confirm karta hai. ✅

Group D — ek dielectric enter karta hai

Worked example Ex 4 — Plate with a dielectric slab

(Cell D) Ex 1 ki plates ko puri tarah mica se bharo, $κ = 5.4$ . Naya $C$ kya hoga?

Forecast: Ek dielectric field ko polarize karke weaken karta hai, toh same charge kam voltage produce karta hai → bada $C$ . Ex 1 ke $708 pF$ ko $5.4$ se multiply karo → expect karo ~ $3.8 nF$ .

Dielectric rule yaad karo. $C = \frac{κ ε _{0} A}{d} = κ C_{vacuum} .$ Yeh step kyun? $κ$ uss factor ke roop mein define hota hai jisse fully-filled capacitance badhti hai; yeh poore vacuum result ke aage slot ho jaata hai.
Multiply karo. $C = 5.4 \times 708 pF .$ Yeh step kyun? Re-derivation ki zarurat nahi — $κ$ us answer ko scale karta hai jis par hum Ex 1 se bharosa karte hain.
Evaluate. $C = 3.82 \times 1 0^{- 9} F = 3.82 nF .$

Verify: $κ > 1$ toh $C$ zaroor badhni chahiye — badhi ( $708 \to 3820 pF$ ). ✅ Ratio $3820/708 = 5.4 = κ$ . ✅

Group A/B/C — spheres, plate limit samet

Worked example Ex 5 — Spherical, plain plug-in

(Cell A / F) Ek spherical capacitor (metal-shell sensor ke model ke roop mein use kiya gaya) ka inner radius $a = 5.0 cm$ , outer shell $b = 8.0 cm$ , beech mein vacuum hai. $C$ nikalo.

Forecast: Radii centimetres mein hain, aur spheres geometrically "generous" hote hain. Guess karo kuch picofarads.

Combination assemble karo. $ab = (0.05) (0.08) = 4.0 \times 1 0^{- 3} m^{2}$ , aur $b - a = 0.03 m$ . Yeh step kyun? Spherical formula geometry ko $ab / (b - a)$ ke roop mein package karta hai — "product over difference" ( $1/ a - 1/ b$ voltage integral se).
Plug in. $C = 4 π ε_{0} \frac{ab}{b - a} = \frac{1}{8.99 \times 1 0 ^{9}} \cdot \frac{4.0 \times 1 0 ^{- 3}}{0.03} .$ Yeh step kyun? $4 π ε_{0} = 1/ (8.99 \times 1 0^{9})$ — Coulomb constant use karne se arithmetic clean rehti hai.
Evaluate. $C = \frac{0.1333}{8.99 \times 1 0 ^{9}} = 1.48 \times 1 0^{- 11} F = 14.8 pF .$

Verify: $ab / (b - a)$ ki units $m^{2} / m = m$ hain, times $F/m$ = F. ✅ Kuch pF, forecast jaisa. ✅

Worked example Ex 6 — Do spherical limits:

$b \to a$ aur $b \to \infty$ (Cells B & C) (i) Dikhaao ki ek thin-gap sphere ( $b = a + t$ , chhota $t$ ) parallel-plate formula mein reduce ho jaata hai. (ii) $b \to \infty$ lo aur isolated sphere recover karo.

Forecast: Do bade spheres ke beech tiny gap locally flat dikhna chahiye — plates jaisa. Aur akele sphere ka "dusra plate" infinitely door hai, toh hum expect karte hain finite $C = 4 π ε_{0} a$ .

$b = a + t$ set karo jahan $t ≪ a$ . Tab $b - a = t$ aur $ab = a (a + t) \approx a^{2}$ . Yeh step kyun? "Thin gap" ka matlab hi hai ki $t$ , $a$ ke next neglect kiya ja sakta hai, toh $ab \approx a^{2}$ .
Substitute karo. $C = \frac{4 π ε _{0} ab}{b - a} \approx \frac{4 π ε _{0} a ^{2}}{t} = \frac{ε _{0} ( 4 π a ^{2} )}{t} = \frac{ε _{0} A}{t},$ jahan $A = 4 π a^{2}$ sphere ki surface area hai aur $t$ gap hai. Yeh step kyun? Yeh exactly $C_{plate} = ε_{0} A / d$ hai — geometry jis par hum pehle se bharosa karte hain. Curved plates close up flat dikhti hain.
Ab far limit $b \to \infty$ . $C = \frac{4 π ε _{0} ab}{b - a} = \frac{4 π ε _{0} a}{1 - a / b} b \to \infty 4 π ε_{0} a .$ Yeh step kyun? Top aur bottom ko $b$ se divide karne par $b \to \infty$ behaviour cleanly isolate ho jaata hai; $a / b$ term mar jaata hai.

Verify (numeric, part i): Lo $a = 0.05 m$ , $t = 0.5 mm$ . Exact spherical: $ab = 0.05 \times 0.0505 = 2.525 \times 1 0^{- 3}$ , $b - a = 5 \times 1 0^{- 4}$ , $C = (2.525 \times 1 0^{- 3} /5 \times 1 0^{- 4}) / (8.99 \times 1 0^{9}) = 5.62 \times 1 0^{- 10} F$ . Plate approx: $ε_{0} (4 π a^{2}) / t = (8.85 \times 1 0^{- 12}) (0.0314) /5 \times 1 0^{- 4} = 5.56 \times 1 0^{- 10} F$ . ~ $1%$ tak agree karte hain. ✅

Group E — ulta solve karo

Group F — real-world word problem

Worked example Ex 8 — Ek real coax cable ki capacitance per metre

(Cell F) Ek coaxial cable ka inner conductor diameter $1.0 mm$ hai, insulation (polyethylene, $κ = 2.25$ ) ek shield ke inner diameter $4.0 mm$ tak jaati hai. Per metre capacitance kya hai?

Forecast: Ex 3 se, $b / a = 3$ wale vacuum coax ne $\sim 50 pF/m$ diya tha; yahan $b / a = 4$ hai (bada log, chhota $C$ ) lekin $κ = 2.25$ use boost karta hai. Expect karo ~ $100 pF/m$ — standard textbook cable figure.

Diameters se radii nikalo. $a = 0.50 mm$ , $b = 2.0 mm$ , toh $b / a = 4$ , $ln 4 = 1.386$ . Yeh step kyun? Formula radii use karta hai; diameter radius ka double hota hai. Dono aadhe ho jaate hain, toh ratio $b / a$ unchanged rehta hai — lekin slips se bachne ke liye hamesha convert karo.
Dielectric insert karo. Filler ke saath, $C = \frac{2 π κ ε _{0} L}{ln ( b / a )}$ ; $L = 1 m$ set karo. Yeh step kyun? Wahi $κ$ -multiplies-everything rule jaisa Ex 4 mein, cylindrical result par apply kiya gaya.
Evaluate. $C = \frac{2 π ( 2.25 ) ( 8.85 \times 1 0 ^{- 12} ) ( 1 )}{1.386} = 9.03 \times 1 0^{- 11} F = 90.3 pF/m .$

Verify: $κ$ ke bina yeh $40.1 pF/m$ hota (parent note ke coax number se match karta hai); $\times 2.25 = 90.3 pF/m$ . ✅ Engineers jo "tens-to-hundred pF per metre" band quote karte hain, us mein hai. ✅

Group G — exam twist

Worked example Ex 9 — Do dielectrics stacked: ek series trick

(Cell G) Ek parallel-plate gap $d = 1.0 mm$ , area $A = 0.02 m^{2}$ , aadha-aadha thickness se bhara hai: $κ_{1} = 2$ ka $0.5 mm$ layer phir $κ_{2} = 4$ ka $0.5 mm$ layer, stacked taaki field dono ko baar baar cross kare. $C$ nikalo.

Forecast: Field ke saath do slabs stack karna do capacitors ko series mein rakhne jaisa hai (same charge dono mein se flow karta hai). Series dono mein se kisi ek se bhi kam deta hai, toh expect karo modest kuch sau pF.

Structure pehchano. Akela akela layer ek capacitor hai: $C_{i} = κ_{i} ε_{0} A / (d /2)$ . Interface wohi $\pm Q$ carry karta hai jo plates karte hain → yeh series mein hain. Yeh step kyun? "Same $Q$ , voltages add" series ki definition hai; field ke saath stacked slabs charge exactly share karte hain.
Har ek compute karo. $d /2 = 0.5 \times 1 0^{- 3} m$ ke saath: $C_{1} = \frac{2 ( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{5 \times 1 0 ^{- 4}} = 7.08 \times 1 0^{- 10} F,$ $C_{2} = \frac{4 ( 8.85 \times 1 0 ^{- 12} ) ( 0.02 )}{5 \times 1 0 ^{- 4}} = 1.416 \times 1 0^{- 9} F .$
Series combine karo. $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} \Rightarrow C = \frac{C _{1} C _{2}}{C _{1} + C _{2}} .$ Yeh step kyun? Series capacitances reciprocals ke roop mein add hoti hain (fixed $Q$ par voltages add hote hain).
Evaluate. $C = \frac{( 7.08 \times 1 0 ^{- 10} ) ( 1.416 \times 1 0 ^{- 9} )}{7.08 \times 1 0 ^{- 10} + 1.416 \times 1 0 ^{- 9}} = 4.72 \times 1 0^{- 10} F = 472 pF .$

Verify: Series result chhote wale part se chhota hona chahiye ( $C_{1} = 708 pF$ ): $472 < 708$ . ✅ Equivalent single- $κ$ check: effective $κ_{eff} = \frac{2 κ _{1} κ _{2}}{κ _{1} + κ _{2}} = \frac{2 ( 8 )}{6} = 2.667$ ; phir $C = κ_{eff} ε_{0} A / d = (2.667) (8.85 \times 1 0^{- 12}) (0.02) /1 0^{- 3} = 4.72 \times 1 0^{- 10} F$ . ✅ Do tareekon se same number.

Recall Scenario-matrix self-test

Kaunse formula ka answer gap zero hone par blow up karta hai? ::: Teeno badhte hain, lekin plate sabse clean hai: $C = ε_{0} A / d \to \infty$ jaise $d \to 0$ . Thin-gap spherical capacitor kaunse formula mein reduce ho jaata hai, aur kyun? ::: Parallel plate $ε_{0} A / t$ — curved plates close up flat dikhti hain ( $ab \approx a^{2}$ , $A = 4 π a^{2}$ ). Field ke saath stacked do dielectric slabs kaise combine hote hain? ::: Series mein — same charge, reciprocals add hote hain; result dono slabs se chhota hota hai. Constant $κ$ wala dielectric insert karna kisi bhi capacitance ka kya karta hai? ::: Usse $κ$ se multiply karta hai (field weaken hoti hai, fixed $Q$ par $V$ girta hai). Radius $a$ wale isolated sphere ki capacitance kya hai? ::: $4 π ε_{0} a$ — spherical formula ka $b \to \infty$ limit.

In devices ke energy side ke liye, Energy Stored in a Capacitor par jaate raho.