Question bank — Gauss's law — integral form, choosing Gaussian surfaces

False — $\vec{E}$ andar aur bahar ke sab charges ka total field hai. Bahar ke charges surface ke har point par $\vec{E}$ ko change karte hain; woh sirf net flux mein zero add karte hain.

False — zero net flux ka matlab sirf yeh hai ki utni hi lines andar jaati hain jitni bahar nikalti hain. Ek dipole ke around (ya uniform external field mein) surface par har jagah strong $\vec{E}$ ho sakta hai phir bhi net flux zero ho sakta hai.

True — law $\oint \vec{E}\cdot d\vec{A}=Q_{\text{enc}}/{\epsilon }_0$ hamesha hold karti hai; yeh Coulomb + superposition se follow hoti hai. Validity unconditional hai — sirf $E$ ke liye solvability ko symmetry chahiye.

False — validity aur solvability alag hain. Bina symmetry ke $E$ ko integral se bahar nahi nikaal sakte, isliye equation, sach hone ke bawajood, $E$ ki koi value nahi deti (jaise ek finite rod).

False — net flux sirf total $Q_{\text{enc}}$ par depend karta hai, naa ki woh andar kahan baitha hai. Trapped charge ko re-arrange karna $\vec{E}$ ko point-wise change karta hai lekin total leakage nahi.

True — ek infinite plane kisi bhi doori se same dikhta hai (constant solid-angle coverage), isliye field lines kabhi spread out nahi hoti; $E$ uniform rehta hai.

True — ek conductor ke andar metal mein field zero hota hai, isliye saara flux sirf ek outer face se escape karta hai ($EA=\sigma A/{\epsilon }_0$). Isolated sheet dono faces se leak karti hai, field ko aadha kar deti hai.

False — $Q_{\text{enc}}=\lambda L$ aur curved area $2\pi rL$ dono $L$ ke saath scale karte hain, isliye $L$ cancel ho jaata hai. Answer $E=\lambda /(2\pi {\epsilon }_{0}r)$ length-independent hai.

Dono contributions same magnitude ke $+$ phir $-$ hain, isliye woh cancel hokar zero hote hain, add nahi. External charge exactly zero net flux contribute karta hai.

End caps axis ke along face karte hain, jabki $\vec{E}$ radially outward point karta hai — caps ke normal ke perpendicular. Unka flux $\vec{E}\cdot d\vec{A}=0$ hai, naa ki $E\pi r^2$.

Sheet ko straddle karne wala symmetric pillbox dono faces se leak karta hai: total flux $2EA=\sigma A/{\epsilon }_0$, jisse $E=\sigma /2{\epsilon }_0$ milta hai. One-face pillbox sirf conductor ke liye valid hai, jahan inner face field-free metal mein hota hai.

Off-centre charge ke saath $E$ sphere par constant nahi hai, isliye aap $E$ ko integral se factor out nahi kar sakte. Flux phir bhi $q/{\epsilon }_0$ hai (Gauss hold karta hai), lekin yeh surface aapko $E$ nahi de sakti.

Ek finite rod mein translational symmetry nahi hoti — field purely radial nahi hai aur length ke saath vary karti hai, isliye $E$ integral se bahar nahi aa sakta. Gauss sach hai lekin yahan useless hai; aapko Coulomb's law ka direct integration karna padega.

$Q_{\text{enc}}=Q{r}^3/R^3$, $r^2$ se zyaada tezi se badhta hai: $E\propto r^3/r^2=r$, isliye $E$ centre par linearly zero tak shrink karta hai, blow up nahi karta.

Zero net enclosed charge ka matlab hai andar equal positive aur negative charge hai (jaise ek dipole), zaruri nahi ki koi bhi charge nahi. Sirf algebraic sum zero hai.

Coulomb's $E\propto 1/r^2$ field ko exactly usi tarah shrink karta hai jaise sphere ka area $4\pi r^2$ badhta hai, toh unka product $r$-independent hai — yeh cancellation Gauss's law ka core hai.

Bahar ke charge ki har field line jo enter karti hai, use exit bhi karna padta hai (yeh sirf charge par terminate hoti hai, aur andar koi nahi hai usse pakadne ke liye), isliye uska inward aur outward flux cancel ho jaata hai — jabki point-wise woh $\vec{E}$ ko bend karta rehta hai.

Sirf woh surface jo field ki symmetry ke saath aligned ho, $E$ ko ya toh constant-and-parallel ($\Phi =EA$) ya perpendicular ($\Phi =0$) rakh sakti hai har face par, jo ki $E$ ko integral se bahar nikaalane ka akela tarika hai. Dekhen Symmetry in physics.

Ek patch ko tilt karna uski area ko $1/\cos \theta$ se multiply karta hai lekin flux-catching component ko $\cos \theta$ se, aur zyaada doori field ko exactly compensating $r^2$ se dilute karti hai — saare shape effects cancel ho jaate hain, sirf total solid angle $4\pi$ bachta hai.

Plane ka field dono pillbox faces se escape karta hai (do leaking surfaces → $2EA$); cylinder ke caps radial field ke perpendicular hain aur kuch bhi leak nahi karte, isliye sirf curved wall count karti hai.

Divergence theorem $\oint \vec{E}\cdot d\vec{A}$ ko $\nabla \cdot \vec{E}$ ke volume integral mein convert karta hai; ise $Q_{\text{enc}}/{\epsilon }_0=\int (\rho /{\epsilon }_0)dV$ ke barabar karna har volume ke liye $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$ force karta hai — differential form aur Maxwell's equations mein se ek.

Cavity ke andar ek Gaussian sphere koi charge enclose nahi karti ($Q_{\text{enc}}=0$), aur spherical symmetry constant radial $E$ ko $E\cdot 4\pi r^2=0$ satisfy karne ke liye force karti hai, toh $E=0$. Dekhen Electric field of conductors.

Exactly zero — $Q_{\text{enc}}=0$ toh $\oint \vec{E}\cdot d\vec{A}=0$, chahe external charge ki wajah se surface par har jagah $\vec{E}\ne 0$ ho.

$E=q/(4\pi {\epsilon }_0{r}^2)\to \infty$ — idealized point charge ek genuine singularity deta hai; real charges finite size ke hote hain, isliye yeh divergence ek modelling artefact hai, physics nahi.

Zero — $r\to 0$ radius ki ek Gaussian sphere vanishing charge enclose karti hai, aur symmetry se field kisi bhi direction mein point nahi kar sakta, toh $E=Qr/(4\pi {\epsilon }_0{R}^3)\to 0$.

Ill-defined / degenerate — "enclosed" ambiguous hai jab charge boundary par ho; idealization break ho jaata hai. Practice mein charge ko split karein ya surface ko perturb karein taaki charge clearly andar ya bahar ho.

Symmetry se ek point charge flat boundary par apni half lines dono taraf bhejta hai, toh har surface $q/2{\epsilon }_0$ pakadti hai — lekin yeh sirf symmetric placement ke liye kaam karta hai aur surface ko shift karke best avoid kiya jaata hai.

Zero net flux, kyunki $Q_{\text{enc}}=+q+(-q)=0$; field surface par highly non-uniform hai lekin $+q$ se jaane wali har outgoing line andar $-q$ par return karti hai.

Woh exactly $q/{\epsilon }_0$ rehta hai — flux kabhi $r$ par depend nahi karta. Field $1/r^2$ se weaken hota hai lekin sphere ki area $r^2$ se grow karti hai, product fixed rakhti hai.