Question bank — Applications — sphere, cylinder, infinite plane

False. Gauss's law $\oint \vec{E}\cdot d\vec{A}=Q_{enc}/{\epsilon }_0$ hamesha sach hai; symmetry sirf woh cheez hai jo humein $E$ ko integral se bahar nikalkar solve karne deti hai. Symmetry ke bina bhi law hold karta hai — bas $E$ ko isolate nahi kar sakte.

False. Zero net flux ka matlab hai utne hi arrows andar jaate hain jitne bahar (isliye $Q_{enc}=0$), lekin field ab bhi strong aur non-zero ho sakta hai har point par — socho ek surface ke baare mein jo capacitor plates ke beech uniform field mein baith rahi ho.

Saamanya roop mein False, sach sirf tab jab symmetry $E$ ko surface par uniform banaye. $Q_{enc}$ double karne se total flux double hota hai, lekin agar surface par field uneven hai to individual point values proportionally double ho yeh zaroori nahi — sirf integral pin down hai.

True. $r>R$ ke liye, dono $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$ dete hain. Bahar se, ek spherically symmetric ball aur center par rakha sara charge bilkul alag nahi lagte.

False. Insulator ke andar sirf radius $r$ ke andar ka charge count hota hai, jo $E=\frac{\rho r}{3{\epsilon }_0}$ deta hai, jo center se linearly badhta hai. Yeh sirf exact center par zero hai, poore andar nahi.

False. Yeh $1/s$ se fall off karta hai: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}$. Enclosed charge box length $L$ ke saath badhti hai jabki area bhi $L$ ke saath badhta hai, isliye distance-dependence ka ek power "alag tarike se kharach ho jaata hai" point charge se.

False. Yeh constant hai, $E=\frac{\sigma }{2{\epsilon }_0}$, distance se independent. Jab tum door jaate ho to sheet ka zyada hissa wide angles par dikhta hai, jo har patch ke inverse-square weakening ko exactly cancel karta hai.

False — ek classic trap. Bahari charges net flux mein zero contribute karte hain (unke arrows ek taraf se enter aur doosri taraf se exit karte hain), lekin woh har point par $\vec{E}$ mein bilkul contribute karte hain. Sirf integral unhe nahi dekh sakta, field khud nahi.

False. Yeh $\sigma /{\epsilon }_0$ hai. Conductor apna interior field zero rakhta hai, isliye surface charge ka sara flux ek taraf se bahar nikalna padta hai, jo isolated-sheet value ko double kar deta hai.

False. $\sigma /(2{\epsilon }_0)$ isolated sheet result hai (dono sides par field). Conducting surface, ya do capacitor plates ke beech ka region, $\sigma /{\epsilon }_0$ deta hai. Material aur geometry decide karte hain kaunsa apply hoga.

Error: conducting shell ke andar $Q_{enc}=0$ hai, isliye $E=0$. Point-charge formula sirf bahar ($r>R$) hold karta hai, jahan enclosed charge poora $Q$ hai.

Error: end caps mein zero flux hoti hai kyunki $\vec{E}$ wahan caps ke parallel hoti hai (unke outward normal ke $\perp$). Sirf curved side, area $2\pi sL$, contribute karta hai.

Error: $L$ cancel ho jaata hai — yeh $Q_{enc}=\lambda L$ aur curved area $2\pi sL$ dono ko multiply karta hai. Iska cancel hona proof hai ki field sirf $s$ par depend karta hai, tumhare arbitrary box length par nahi.

Error: cube par $E$ faces par constant nahi hota (corners face centers se door hote hain) aur $\vec{E}$ har jagah $d\vec{A}$ ke parallel nahi hota. Tum $E$ ko integral se bahar nahi nikal sakte — symmetry exploit karne ke liye sphere zaroori hai.

Error: pillbox ke do faces hain, har ek $EA$ flux pass karta hai, isliye left side $2EA$ hai, jo $E=\sigma /(2{\epsilon }_0)$ deta hai. Doosra face bhoolna sabse common plane mistake hai.

Error: $Q_{enc}=(+q)+(-q)=0$ hai, isliye net flux zero hai. Field bilkul bhi zero se door hai — dipole ek rich pattern banata hai — lekin har arrow jo nikalta hai woh wapas bhi aata hai.

Error: andar, sirf radius $r$ ke andar ka fraction enclosed hota hai: $Q_{enc}=Q(r/R{)}^3=\rho \cdot \frac{4}{3}\pi r^3$. Andar poora $Q$ use karna field ko overestimate karta hai.

Error: woh formula ek infinite (ya effectively infinite) line assume karta hai, sirf tab valid jab $s$ wire ki length se bahut chhota ho. $s=2$ m par 10 cm rod se symmetry break ho jaati hai — end caps ab zero flux nahi dete aur Gauss's law $E$ ko isolate nahi kar sakta.

Kyunki Gaussian surface ko source ki symmetry se match karna chahiye taaki $E$ constant rahe aur ya to parallel ya perpendicular ho $d\vec{A}$ se har piece par. Sphere yeh sirf radial (point/ball) fields ke liye achieve karta hai; line's field coaxial cylinder par uniform hai, aur plane's pillbox par.

Curved side par $\vec{E}$ radially point karta hai, seedha $d\vec{A}$ ke saath, isliye $\vec{E}\cdot d\vec{A}=EdA$ (full contribution). Flat caps par $\vec{E}$, $d\vec{A}$ ke perpendicular hai, isliye $\vec{E}\cdot d\vec{A}=0$.

Source geometry power law set karta hai. Ek point apna flux outward spheres se bhejta hai jinki area $r^2$ se badhti hai, $E$ ko $1/r^2$ se dilute karta hai. Infinite plane ka flux sirf do fixed pillbox faces of area $A$ cross karta hai, jo distance ke saath nahi badhte — isliye kuch bhi dilute nahi karta.

Ek isolated sheet dono sides par equally radiate karta hai, apna flux split karta hai ($\frac{\sigma }{2{\epsilon }_0}$ har taraf). Conductor apna interior $E=0$ par rakhta hai, isliye surface charge ka sara flux ek taraf se bahar nikalta hai — wahi total flux half area se double karta hai $E$ ko.

Kyunki $Q_{enc}=\lambda L$ aur curved area $=2\pi sL$ dono $L$ ke saath scale karte hain. Iska cancel hona confirm karta hai ki answer ek arbitrary choice se independent hai jo humne ki thi — ek self-consistency check ki hamaari surface legitimate thi.

Symmetry se (Symmetry in Physics): spherically symmetric charge par koi special sideways direction nahi hota, isliye koi bhi non-radial component ke paas ek taraf point karne ka koi reason nahi hoga — woh zero hona hi chahiye. Yeh argument, algebra nahi, woh cheez hai jo problem ko solvable banati hai.

Plates ke beech dono single-sheet fields ($\frac{\sigma }{2{\epsilon }_0}$ each) ek hi direction mein point karte hain aur $\frac{\sigma }{{\epsilon }_0}$ mein add ho jaate hain. Bahar, $+\sigma$ sheet ka field aur $-\sigma$ sheet ka field opposite directions mein point karte hain aur exactly cancel ho jaate hain.

Zero. $r\to 0$ radius ki Gaussian sphere negligible charge enclose karti hai ($\propto r^3$) jabki uski area $\propto r^2$ hai, isliye $E=\frac{\rho r}{3{\epsilon }_0}\to 0$. Har direction equivalent hai, isliye field symmetry se bhi cancel ho jaata hai.

Insulating ball ke liye inside formula $\frac{\rho R}{3{\epsilon }_0}$ aur outside formula $\frac{Q}{4\pi {\epsilon }_0{R}^2}$ same value par milte hain — field continuous hai. Conducting shell ke liye yeh jump karta hai $0$ (andar) se $\frac{\sigma }{{\epsilon }_0}$ (bilkul bahar) tak charged surface ke across.

Koi bhi real line infinitely thin nahi hoti, isliye asli wire ka finite radius hota hai jahan formula apply karna band ho jaata hai (ab tum conductor ke andar ho, $E=0$). Divergence zero-thickness idealization ka ek artifact hai.

$Q_{enc}=0$ hai, isliye net flux zero hai, chahe atom ek real (dipole-like) field produce kare. Flux sirf net charge count karta hai.

Bilkul andar, $Q_{enc}=0\Rightarrow E=0$; bilkul bahar, $Q_{enc}=Q\Rightarrow E=\frac{Q}{4\pi {\epsilon }_0{R}^2}$. Shell apne andar ek sharp discontinuity create karti hai $\frac{\sigma }{{\epsilon }_0}$ ke size ki.

Haan — net flux phir bhi $Q/{\epsilon }_0$ hai kyunki $Q_{enc}=Q$. Lekin ab tum $E$ ke liye solve nahi kar sakte: off-center sphere par field constant nahi hai, isliye integral $E\times \text{area}$ mein collapse nahi hoga.

Haan — flux sirf total $Q_{enc}=2q$ par depend karta hai, isliye double hota hai. Lekin field pattern har jagah "ek charge ka double" nahi hota; sirf integrated flux ka clean doubling law follow karta hai.

Nahi. Ek external charge zero net flux contribute karta hai (uske arrows enter aur leave karte hain). Yeh box par points par $\vec{E}$ ko zaroor perturb karta hai, lekin $\oint \vec{E}\cdot d\vec{A}$ phir bhi sirf enclosed sheet charge se $\sigma A/{\epsilon }_0$ ke barabar hoga.