Capacitance — parallel plate derivation, cylindrical, spherical: Exercises

1.8.11 · D4 · Physics › Electromagnetism › Capacitance — parallel plate derivation, cylindrical, spheri

Shuru karne se pehle, teen symbols jo tumhe har line pe milenge. Agar inme se koi unfamiliar lagta hai, to exactly wahi hai jo hum abhi fix karne wale hain.

Teen master formulas jinpar hum rely karenge (sab parent note mein derive hain):

Level 1 — Recognition

Goal: formula padho aur numbers plug karo. Abhi logic ke koi traps nahi, bas clean substitution.

L1.1 — Plate plug-in

Do parallel plates jiनका har ek ka area $A = 0.20 m^{2}$ hai, $d = 0.50 mm$ ki doori par rakhi hain aur unke beech vacuum hai. $C$ nikalo.

Recall Solution L1.1

Kaunsa formula? Flat plates → $C = ε_{0} A / d$ . Gap convert karo: $d = 0.50 mm = 0.50 \times 1 0^{- 3} m = 5.0 \times 1 0^{- 4} m$ . $C = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.20 )}{5.0 \times 1 0 ^{- 4}} = \frac{1.77 \times 1 0 ^{- 12}}{5.0 \times 1 0 ^{- 4}} = 3.54 \times 1 0^{- 9} F .$ Answer: $C \approx 3.54 nF$ . Ek fifth square metre bhi sirf nanofarads deta hai — farads bahut bade hote hain.

L1.2 — Definition drill

Ek capacitor $Q = 6.0 μ C$ hold karta hai jab uske across voltage $V = 3.0 V$ hai. Uski capacitance kya hai?

Recall Solution L1.2

Kaunsa formula? Definition khud, $C = Q / V$ — koi geometry nahi chahiye. $C = \frac{6.0 \times 1 0 ^{- 6} C}{3.0 V} = 2.0 \times 1 0^{- 6} F = 2.0 μ F .$ Answer: $C = 2.0 μ F$ . Notice karo ki humne shape ke baare mein kabhi nahi pucha — $C = Q / V$ kisi bhi capacitor ke liye kaam karta hai.

Level 2 — Application

Goal: khud sahi shape choose karo, unit conversion karo, shayad rearrange bhi karo.

L2.1 — Gap ke liye solve karo

Tumhe $A = 1.0 \times 1 0^{- 2} m^{2}$ area ki plates se $C = 100 pF$ ka parallel-plate capacitor chahiye. Tumhe kitna plate separation $d$ chahiye?

Recall Solution L2.1

Kaunsa formula? $C = ε_{0} A / d$ , lekin ab $d$ unknown hai. Rearrange karo: $d = ε_{0} A / C$ . $d = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 1.0 \times 1 0 ^{- 2} )}{100 \times 1 0 ^{- 12}} = \frac{8.85 \times 1 0 ^{- 14}}{1.0 \times 1 0 ^{- 10}} = 8.85 \times 1 0^{- 4} m .$ Answer: $d \approx 0.885 mm$ . Sensible — sub-millimetre gaps exactly waise hi hain jaise real film capacitors bante hain.

L2.2 — Coax capacitance per metre

Ek coaxial cable ka inner radius $a = 1.0 mm$ aur outer radius $b = 3.0 mm$ hai, beech mein vacuum hai. Length $L = 1.0 m$ ke liye uski capacitance nikalo.

Recall Solution L2.2

Kaunsa formula? Nested cylinders → $C = \frac{2 π ε _{0} L}{ln ( b / a )}$ . Dekho bhi Coaxial Cable Transmission Lines. Ratio $b / a = 3.0/1.0 = 3.0$ (millimetres cancel ho jaate hain — sirf ratio matter karta hai). $ln 3.0 = 1.0986$ . $C = \frac{2 π ( 8.85 \times 1 0 ^{- 12} ) ( 1.0 )}{1.0986} = \frac{5.560 \times 1 0 ^{- 11}}{1.0986} = 5.06 \times 1 0^{- 11} F .$ Answer: $C \approx 50.6 pF$ per metre — ek textbook "tens of pF/m" figure.

L2.3 — Spherical shells

Do concentric spheres ke radii $a = 4.0 cm$ aur $b = 5.0 cm$ hain. $C$ nikalo.

Recall Solution L2.3

Kaunsa formula? Nested spheres → $C = \frac{4 π ε _{0} ab}{b - a}$ . Convert karo: $a = 0.040 m$ , $b = 0.050 m$ , $b - a = 0.010 m$ , $ab = 0.040 \times 0.050 = 2.0 \times 1 0^{- 3} m^{2}$ . $C = 4 π (8.85 \times 1 0^{- 12}) \frac{2.0 \times 1 0 ^{- 3}}{0.010} = (1.1127 \times 1 0^{- 10}) (0.20) = 2.23 \times 1 0^{- 11} F .$ Answer: $C \approx 22.3 pF$ .

Level 3 — Analysis

Goal: yeh reason karo ki $C$ kaise change hota hai, ya energy / dielectrics ke saath combine karo.

L3.1 — Gap aadha karo, fixed charge par energy ka kya hoga?

Ek parallel-plate capacitor fixed charge $Q$ carry karta hai. Tum plates ko separation $d$ se $d /2$ tak push karte ho (charge constant rehta hai — yeh disconnected hai). Kya stored energy badhti hai ya ghatti hai, aur kitne factor se?

Recall Solution L3.1

Tools: $C = ε_{0} A / d$ aur fixed-charge energy $U = \frac{Q ^{2}}{2 C}$ (from Energy Stored in a Capacitor). $d$ ko aadha karne se $C$ double ho jaata hai: $C^{'} = ε_{0} A / (d /2) = 2 C$ . $\frac{U ^{'}}{U} = \frac{Q ^{2} / ( 2 C ^{'} )}{Q ^{2} / ( 2 C )} = \frac{C}{C ^{'}} = \frac{C}{2 C} = \frac{1}{2} .$ Answer: energy ghat kar aadhi ho jaati hai. Plates attract karte hain, isliye unhe paas laane se energy release hoti hai — tumhe unhe wapas alag karne ke liye kaam karna padega.

L3.2 — Ek dielectric slide karo

L1.1 ka $3.54 nF$ vacuum capacitor constant $κ = 4.0$ ke dielectric se fill kiya jaata hai. (a) Nayi capacitance? (b) Agar ise fixed voltage $V = 12 V$ par rakha jaaye, to stored charge kaise change hoga?

Recall Solution L3.2

(a) Dielectric $C$ ko $κ$ se multiply karta hai: $C^{'} = κ C = 4.0 \times 3.54 nF = 14.16 nF$ . (b) Fixed $V$ par, $Q = C V$ , isliye charge $C$ ke saath scale karta hai: yeh bhi $\times 4$ badhta hai. Original: $Q = C V = (3.54 \times 1 0^{- 9}) (12) = 4.25 \times 1 0^{- 8} C$ . New: $Q^{'} = C^{'} V = (1.416 \times 1 0^{- 8}) (12) = 1.70 \times 1 0^{- 7} C$ . Answer: $C^{'} \approx 14.2 nF$ ; charge $\approx 42.5 nC$ se $\approx 170 nC$ tak quadruple ho jaata hai.

L3.3 — Coax ka zyaadatar voltage kahaan rehta hai?

$a = 1.0 mm$ , $b = 3.0 mm$ wale coaxial capacitor mein, total voltage ka kitna fraction gap ke inner half mein drop hota hai, yaani $r = a$ se geometric midpoint $r = ab$ tak?

Recall Solution L3.3

Tool aur kyun: field $E (r) = \frac{λ}{2 π ε _{0} r}$ hai, $1/ r$ ki tarah girta hai, isliye voltage $ln r$ ki tarah build hota hai. Do radii $r_{1} < r_{2}$ ke beech voltage $ln (r_{2} / r_{1})$ ke proportional hai. Total: $V_{tot} \propto ln (b / a) = ln 3 = 1.0986$ . Inner half ( $r = ab$ tak): $V_{in} \propto ln \frac{ab}{a} = ln b / a = \frac{1}{2} ln (b / a)$ . $\frac{V _{in}}{V _{tot}} = \frac{\frac{1}{2} l n ( b / a )}{l n ( b / a )} = \frac{1}{2} .$ Answer: exactly aadha voltage. Geometric mean $ab$ use karna (arithmetic mean nahi) ek logarithmic drop ko evenly split karta hai — isliye log geometry linear se alag hai.

Level 4 — Synthesis

Goal: do ideas ko saath joodo — series/parallel, limits, ya ek poori derivation.

L4.1 — Layered dielectric = capacitors in series

Ek parallel-plate capacitor (area $A = 0.20 m^{2}$ , total gap $d = 0.50 mm$ ) do stacked slabs se fill hai, har ek ki thickness $d /2$ : lower slab ka $κ_{1} = 2.0$ hai, upper ka $κ_{2} = 6.0$ . Total capacitance nikalo.

Recall Solution L4.1

Idea aur kyun: do slabs ke beech ki boundary har jagah same potential par hai, isliye hum wahan ek thin conductor imagine kar sakte hain. Yeh device ko do capacitors in series mein split karta hai, har ek ka gap $d /2$ hai (dekho Capacitors in Series and Parallel). Har slab: $C_{i} = \frac{κ _{i} ε _{0} A}{d /2} = \frac{2 κ _{i} ε _{0} A}{d}$ . Base value $\frac{ε _{0} A}{d} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.20 )}{5.0 \times 1 0 ^{- 4}} = 3.54 \times 1 0^{- 9} F$ (hamara purana L1.1 number). $C_{1} = 2 (2.0) (3.54 nF) = 14.16 nF$ , $C_{2} = 2 (6.0) (3.54 nF) = 42.48 nF$ . Series combine: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}}$ . $\frac{1}{C} = \frac{1}{14.16} + \frac{1}{42.48} = 0.07062 + 0.02354 = 0.09416 (nF^{- 1}),$ $C = \frac{1}{0.09416} = 10.62 nF .$ Answer: $C \approx 10.6 nF$ . Sanity check: total dono mein se chhhote se bhi neeche hai — yeh series ka hallmark hai.

L4.2 — Isolated-sphere limit

Spherical-shell formula se shuru karo aur dikhao kya hota hai jab outer shell infinity tak chali jaati hai ( $b \to \infty$ ). Phir $a = 0.10 m$ radius ke isolated conducting sphere ki capacitance compute karo.

Recall Solution L4.2

Limit lo. $C = \frac{4 π ε _{0} ab}{b - a}$ . Upar aur neeche $b$ se divide karo: $C = \frac{4 π ε _{0} a}{1 - a / b} .$ $b \to \infty$ ke saath, $a / b \to 0$ , isliye denominator $\to 1$ : $C \to 4 π ε_{0} a .$ Interpretation: akela sphere phir bhi capacitance rakhta hai — uski "doosri plate" infinity hai. Number: $C = 4 π (8.85 \times 1 0^{- 12}) (0.10) = (1.1127 \times 1 0^{- 10}) (0.10) = 1.11 \times 1 0^{- 11} F \approx 11.1 pF .$ Answer: $C \approx 11.1 pF$ .

Level 5 — Mastery

Goal: multi-step, derive-and-optimize, ya ek subtle limit. Poori reasoning required.

L5.1 — Coax ka optimal inner radius

Ek coaxial cable ka fixed outer radius $b$ hai. Peak field (jo breakdown/failure set karta hai) inner conductor par hota hai, $E_{m a x} = E (a) = \frac{λ}{2 π ε _{0} a}$ . Cable ke across fixed voltage $V$ ke liye, woh inner radius $a$ choose karo jo peak field minimise kare. Dikhao ki $a = b / e$ .

Recall Solution L5.1

$E_{m a x}$ ko $V$ ke terms mein set up karo. Parent derivation se, $V = \frac{λ}{2 π ε _{0}} ln (b / a)$ , isliye $\frac{λ}{2 π ε _{0}} = \frac{V}{ln ( b / a )}$ . Substitute karo: $E_{m a x} = \frac{1}{a} \cdot \frac{V}{l n ( b / a )} = \frac{V}{a l n ( b / a )} .$ Derivative kyun? Hum chahte hain ki $a$ ke vary karne par $E_{m a x}$ sabse chhhota ho. Minimise karne ka matlab hai jahan slope $d E_{m a x} / d a = 0$ ho. Equivalently, kyunki $V$ ek constant multiplier hai, hum denominator $f (a) = a ln (b / a)$ ko maximise karte hain. Differentiate $f (a) = a ln b - a ln a$ : $f^{'} (a) = ln b - (ln a + a \cdot \frac{1}{a}) = ln b - ln a - 1 = ln (b / a) - 1.$ $f^{'} (a) = 0$ set karo: $ln (b / a) = 1 \Rightarrow b / a = e \Rightarrow a = b / e .$ Confirm karo ki yeh $f$ ka maximum hai: $f^{''} (a) = - 1/ a < 0$ , isliye $f$ concave hai — yeh denominator ka max hai, hence $E_{m a x}$ ka minimum. ✅ Answer: $a = b / e \approx 0.368 b$ . Yeh high-voltage coax design ke liye ek genuine engineering rule hai.

L5.2 — Ek number daalo

$b = 3.0 mm$ lo aur $V = 10 kV$ rakho. (a) Optimal $a$ nikalo. (b) Us $a$ par minimum peak field $E_{m a x}$ nikalo.

Recall Solution L5.2

(a) $a = b / e = 3.0/2.71828 = 1.1036 mm \approx 1.10 mm$ . (b) Optimum par $ln (b / a) = 1$ , isliye $E_{m a x} = \frac{V}{a l n ( b / a )} = \frac{V}{a \cdot 1} = \frac{V}{a} .$ $E_{m a x} = \frac{10 \times 1 0 ^{3} V}{1.1036 \times 1 0 ^{- 3} m} = 9.06 \times 1 0^{6} V/m .$ Answer: $a \approx 1.10 mm$ , $E_{m a x} \approx 9.06 \times 1 0^{6} V/m \approx 9.1 MV/m$ . (Air ke ~3 MV/m breakdown se comfortably neeche sirf tab hoga jab gap mein solid dielectric ho — isliye real coax PTFE use karta hai, air nahi.)

Recall Quick self-quiz (answers cover karo)

Series capacitors kaise combine hote hain? ::: Reciprocals add hote hain: $1/ C = \sum 1/ C_{i}$ — total sabse chhhote se bhi neeche hota hai. Fixed $Q$ par, plate gap ko aadha karne se energy ka kya hota hai? ::: Aadhi ho jaati hai ( $U = Q^{2} /2 C$ , aur $C$ double hota hai). Isolated-sphere capacitance? ::: $C = 4 π ε_{0} a$ ( $b \to \infty$ se). Minimum peak field ke liye optimal coax inner radius? ::: $a = b / e$ . Coax formula mein $b / a$ unit-free kyun hai? ::: Yeh same quantities ka ratio hai — dimensionless, isliye mm/mm = m/m.

Parent topic par wapas jao · related: Gauss's Law, Electric Potential, Energy Stored in a Capacitor, Capacitors in Series and Parallel, Dielectrics and Polarization, Coaxial Cable Transmission Lines.