Applications — sphere, cylinder, infinite plane

1.8.7 · Physics › Electromagnetism

Intuition Badi idea (WHY ye kaam karta hai)

Gauss's law $\oint E \cdot d A = \frac{Q _{e n c}}{ε _{0}}$ hamesha true hai, lekin ye hamein $E$ solve karne deta hai tabhi jab symmetry $E$ ko magnitude mein constant bana de aur ek acchi tarah se choose ki gayi surface ke parallel ya perpendicular rakh de. Teen classic geometries mein se har ek mein ek aisi symmetry hoti hai jo ye kaam karti hai:

Sphere → spherical symmetry → ek concentric spherical Gaussian surface use karo.
Cylinder / line → axial (cylindrical) symmetry → ek coaxial cylindrical surface use karo.
Infinite plane → planar symmetry → plane ko straddle karta hua ek pillbox use karo.

80/20 rule: jab ek baar wo surface choose kar lo jahan $E$ constant hai, $\oint E \cdot d A$ collapse hokar $E \times (area)$ ban jaata hai, aur poora problem algebra ban jaata hai.

0. Master tool

Figure — Applications — sphere, cylinder, infinite plane

1. Sphere

Maano total charge $Q$ hai radius $R$ wale sphere par (ya toh surface par charge wala conducting shell, YA uniformly charged insulator jisme volume density $ρ$ ho).

Derivation (bahar, $r > R$ ): $\oint E \cdot d A = E \oint d A = E (4 π r^{2})$ Ye step kyun? $E$ surface par constant hai isliye factor out ho jaata hai; $\oint d A$ bas sphere ki area hai.

$Q_{e n c} / ε_{0} = Q / ε_{0}$ ke barabar rakho: $E (4 π r^{2}) = \frac{Q}{ε _{0}} \Rightarrow E_{o u t} = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$

Conducting shell ke andar ( $r < R$ ): $Q_{e n c} = 0$ , isliye $E_{in} = 0$ .

Uniformly charged insulator ke andar ( $r < R$ ): sirf radius $r$ ke andar wala charge count hota hai. $Q_{e n c} = ρ \cdot \frac{4}{3} π r^{3}, ρ = \frac{Q}{\frac{4}{3} π R ^{3}}$ $E (4 π r^{2}) = \frac{ρ \frac{4}{3} π r ^{3}}{ε _{0}} \Rightarrow E_{in} = \frac{ρ r}{3 ε _{0}} = \frac{Q r}{4 π ε _{0} R ^{3}}$ Ye step kyun? Solid ball ke andar, enclosed charge $r^{3}$ ki tarah badhta hai jabki area $r^{2}$ ki tarah, isliye $E \propto r$ bachta hai — ye center se linearly badhta hai.

2. Cylinder / Infinite Line

Ek line charge linear density $λ$ (C/m) ke saath, ya ek lamba charged cylinder.

Derivation (line charge): Gaussian cylinder radius $s$ , length $L$ . $\oint E \cdot d A = curved side E (2 π s L) + two caps 0$ Ye step kyun? Curved area $= 2 π s L$ ; end caps kuch nahi dete kyunki field unke parallel hai.

Enclosed charge $Q_{e n c} = λ L$ : $E (2 π s L) = \frac{λ L}{ε _{0}} \Rightarrow E = \frac{λ}{2 π ε _{0} s}$ Ye step kyun? $L$ cancel ho jaata hai — field ko koi fark nahi padta tumhara box kitna lamba hai, jo prove karta hai ki choice legitimate thi.

3. Infinite Plane

Ek flat sheet surface charge density $σ$ (C/m²) ke saath.

Derivation: pillbox face area $A$ ke saath. $\oint E \cdot d A = E A + E A = 2 E A$ Ye step kyun? Field dono faces se nikalta hai (do contributions $E A$ ); side walls mein $E ⊥ d A$ ⇒ zero.

Enclosed charge $= σ A$ : $2 E A = \frac{σ A}{ε _{0}} \Rightarrow E = \frac{σ}{2 ε _{0}}$ Ye step kyun? $A$ cancel ho jaata hai — confirm karta hai ki field uniform hai, sheet se distance se independent!

Common mistakes

Recall Feynman: 12-saal ke bachche ko explain karo

Ek bag imagine karo (Gauss's surface). Electric "arrows" ki total number jo bag se bahar nikal rahi hain sirf is par depend karti hain ki bag ke andar kitna charge hai — bahar ka kuch matter nahi. Ab ek bag pick karo jiska shape charge se match kare: ball of charge ke liye ball-shaped bag, wire ke liye tube bag, sheet ke liye flat box. Matching shape ki wajah se, arrows evenly bahar nikalte hain, isliye unhe count karna easy hai. Door se ball of charge ek tiny dot jaisi lagti hai ( $1/ r^{2}$ ). Wire ka field slowly fade hota hai ( $1/ s$ ). Aur ek giant flat sheet equally hard push karta hai chahe tum kitni bhi dur khade ho — kyunki hamesha tumhare around aur sheet hai!

Flashcards

Gauss's law integral form mein

\oint E \cdot d A = Q_{e n c} / ε_{0}

Uniformly charged sphere ke bahar field (r>R)

E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}

— point charge jaisa

Conducting sphere ke andar field

E = 0

(saara charge surface par,

Q_{e n c} = 0

)

Uniformly charged insulating ball ke andar field (r<R)

E = \frac{ρ r}{3 ε _{0}}

, r ke saath linearly badhta hai

Infinite line charge ka field

E = \frac{λ}{2 π ε _{0} s}

, radial,

1/ s

ki tarah fall hota hai

Line charge ke liye end caps zero flux kyun dete hain

E

caps ke perpendicular hai (

E ⊥ d A

)

Infinite charged sheet ka field

E = \frac{σ}{2 ε _{0}}

, uniform, distance se independent

Conductor surface ke just bahar field

E = \frac{σ}{ε _{0}}

(isolated-sheet value se double)

Equal & opposite parallel plates ke beech field

E = \frac{σ}{ε _{0}}

andar,

0

bahar

Sheet field distance ke saath kyun kam nahi hota

pillbox area cancel ho jaata hai; sheet ka wider angular view inverse-square ko offset karta hai

Sphere, line, plane ke liye kaunsa power law

1/ r^{2}

1/ s

, constant

Cylinder problem ke liye kaunsi symmetry use hoti hai

axial/cylindrical → coaxial Gaussian cylinder

Connections

Gauss's Law — master equation jo yahan apply hui
Electric Flux — $\oint E \cdot d A$ ki definition
Coulomb's Law — sphere result point-charge field recover karta hai
Conductors in Electrostatics — kyun andar $E = 0$ , bahar $σ / ε_{0}$
Parallel Plate Capacitor — two-sheet superposition use karta hai
Electric Potential — $V$ paane ke liye in $E$ fields ko integrate karo
Symmetry in Physics — surface choice ke liye deciding principle