Exercises — Gauss's law — integral form, choosing Gaussian surfaces

Shape matter nahi karta — Gauss's law sirf andar trapped charge poochta hai. $\Phi \equiv {\oint }_S\vec{E}\cdot d\vec{A}$ (upar define kiya gaya leakage symbol) use karte hue, $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{3.0\times 10^{-9}}{8.85\times 10^{-12}}\approx 339{\text{N⋅m}}^2/\text{C}.$ Ek nanocoulomb $10^{-9}\text{C}$ hota hai. Humne trapped charge ko ${\epsilon }_0$ se divide kiya; cube ka size aur orientation irrelevant hain.

Zero. Jab $d\vec{A}$ har jagah baaharon point kar raha hai (sign-convention box dekho), baaharon charge se aane wali ek field line jo sphere mein enter karti hai woh $\vec{E}\cdot d\vec{A}<0$ ($-$) deti hai, aur jahan wahi line bahar jaati hai wahan $\vec{E}\cdot d\vec{A}>0$ ($+$) milta hai. Har entry ek exit se match hoti hai (parent derivation ka Step 3), toh cancel ho jaate hain. Koi charge enclosed nahi hai, toh $Q_{\text{enc}}=0$ aur $\Phi =0/{\epsilon }_0=0$. Note: surface par field $\vec{E}$ zero nahi hai — baaharon charge pointwise feel hota hai — lekin net flux cancel ho jaata hai.

Surface ko symmetry se match karo taaki har face par $\vec{E}$ ya to $d\vec{A}$ ke parallel ho (flux $=EA$) ya perpendicular (flux $=0$).

• (a) point charge → concentric sphere (spherical symmetry).
• (b) infinite line → coaxial cylinder (cylindrical symmetry).
• (c) infinite plane → pillbox sheet ko straddling karte hue (planar symmetry).

Yeh ek genuine ambiguity hai — zero-thickness surface par mathematical point na andar hai na baaharon, toh $Q_{\text{enc}}$ well defined nahi hai aur Gauss's law koi clean answer nahi deta.

• Physical resolution: real charges ki finite size hoti hai, toh surface ko ek infinitesimal amount ek taraf khiskaao. Agar surface ek ideal point charge se symmetrically guzarti hai, toh aadhi field lines baaharon jaati hain aur aadhi andar rehti hain, toh tum aadha charge count karte ho: $Q_{\text{enc}}=q/2$, jisse $\Phi =q/2{\epsilon }_0$ milta hai.
• Practical advice: Gaussian surface kisi charge ke through kabhi mat khiincho. Aise choose karo ki har charge unambiguously andar ya baaharon ho; tab saari bookkeeping clean rehti hai.

Surface: radius $r$ ki sphere. Uس par $E$ constant hai aur $d\vec{A}$ ke parallel, toh $\Phi =E(4\pi r^2)$. $E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{4.0\times 10^{-9}}{4\pi (8.85\times 10^{-12})(0.20{)}^2}\approx 899\text{N/C}.$ $\frac{1}{4\pi {\epsilon }_0}\approx 8.99\times 10^9$ use karte hue: $E=8.99\times 10^9\cdot \frac{4.0\times 10^{-9}}{0.04}=899\text{N/C}$.

Figure padho: field arrows curved wall ko perpendicularly cross karte hain ($d\vec{A}$ ke parallel → flux $=EA$) lekin flat end caps ke parallel chalte hain (unhe kabhi pierce nahi karte → flux $=0$).

Surface: coaxial cylinder, radius $r$, length $L$.

• Curved wall: $\vec{E}$ radial $\parallel d\vec{A}$, constant → flux $=E(2\pi rL)$.
• End caps: $\vec{E}\perp d\vec{A}$ → flux $=0$.
• $Q_{\text{enc}}=\lambda L$. $L$ cancel ho jaata hai: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2.5\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.05)}\approx 8990\text{N/C}.$

Surface: sheet ko straddling karta pillbox. Field dono flat faces se bahar nikalti hai, toh $\Phi =2EA$; side wall $0$ deta hai. $Q_{\text{enc}}=\sigma A$ ke saath: $E=\frac{\sigma }{2{\epsilon }_0}=\frac{1.0\times 10^{-6}}{2(8.85\times 10^{-12})}\approx 5.65\times 10^4\text{N/C}.$ Yeh distance se independent hai — field uniform hai.

(a) Andar ($r<R$): radius $r$ ki Gaussian sphere sirf apne volume ke andar wala charge trap karti hai. $V_{\text{enc}}=\frac{4}{3}\pi r^3$ aur $V_{\text{total}}=\frac{4}{3}\pi R^3$ (upar define kiye) ke saath, $\frac{V_{\text{enc}}}{V_{\text{total}}}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\frac{r^3}{R^3}\Rightarrow Q_{\text{enc}}=Q\frac{r^3}{R^3}.$ $\frac{4}{3}\pi$ cancel ho jaata hai, clean $r^3/R^3$ milta hai. Phir $E(4\pi r^2)=\frac{Q{r}^3/R^3}{{\epsilon }_0}\Rightarrow E=\frac{Qr}{4\pi {\epsilon }_0{R}^3}.$ $E=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}(0.05)}{(0.10{)}^3}=2697\text{N/C}.$ (b) Baaharon ($r>R$): poora $Q$ enclosed hai; yeh point charge ki tarah behave karta hai: $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}}{(0.20{)}^2}=1349\text{N/C}.$ Degenerate check: $r=0$ (centre) par, $E=0$ — symmetry se koi direction preferred nahi hai, aur indeed $E\propto r\to 0$ (figure mein origin dot).

(a) Andar ($r<R$): Gaussian sphere koi bhi charge enclose nahi karti (saara charge outer shell par hai), toh $Q_{\text{enc}}=0$ aur $E=0.$ Uniform shell ke andar field har jagah exactly zero hoti hai — opposite sides se pulls cancel ho jaate hain. (b) Baaharon ($r>R$): saara $Q$ enclosed hai, point-charge field: $E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{6.0\times 10^{-9}}{(0.20{)}^2}=1349\text{N/C}.$ L3.1 se contrast: baaharon, solid aur shell dono agree karte hain (dono point charge ki tarah dikhte hain). Andar, solid sphere mein $E\ne 0$ hai lekin shell mein $E=0$ — enclosed charge alag hota hai.

Enclosed charge signed sum hai: $Q_{\text{enc}}=+3+(-8)=-5\text{nC}$. (a) Flux: $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{-5.0\times 10^{-9}}{8.85\times 10^{-12}}\approx -565{\text{N⋅m}}^2/\text{C}.$ Flux negative hai — $d\vec{A}$ baaharon point karte hue, negative $\Phi$ matlab field net andar ki taraf point karti hai. Field lines andar net negative charge par terminate ho rahi hain. (b) Magnitude aur direction: $|E|=\frac{|Q_{\text{enc}}|}{4\pi {\epsilon }_0{r}^2}=8.99\times 10^9\cdot \frac{5.0\times 10^{-9}}{(0.20{)}^2}=1124\text{N/C},$ radially inward point karte hue (net negative charge ki taraf).

Har sheet akele $\frac{\sigma }{2{\epsilon }_0}$ magnitude ki uniform field banata hai, $+$ sheet se door aur $-$ sheet ki taraf point karte hue. Superpose karo (vectors add karo):

• Beech (red arrows region): dono fields same direction mein point karti hain ($+$ se $-$ ki taraf), toh add hoti hain: $E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{2.0\times 10^{-6}}{8.85\times 10^{-12}}\approx 2.26\times 10^5\text{N/C}.$
• Baaharon: dono fields opposite directions mein point karti hain aur cancel ho jaati hain: $E=0.$ Isliye ek parallel-plate capacitor apni field gap mein confined rakhta hai.

Kyun hum flux contributions add kar sakte hain (superposition): Real field, wire ki field aur point charge ki field ka vector sum hai, $\vec{E}={\vec{E}}_{\text{wire}}+{\vec{E}}_{\text{point}}$ (yeh Coulomb's law plus forces/fields ka superposition hai). Flux integral $\vec{E}$ mein linear hai: $\oint ({\vec{E}}_{\text{wire}}+{\vec{E}}_{\text{point}})\cdot d\vec{A}=\oint {\vec{E}}_{\text{wire}}\cdot d\vec{A}+\oint {\vec{E}}_{\text{point}}\cdot d\vec{A}.$ Integral ke andar sum, sum of integrals mein split ho jaata hai, toh total flux sirf woh flux hai jo har charge akele produce karta — isliye flux superpose karte hain. Har piece apne enclosed charge over ${\epsilon }_0$ ke barabar hai, toh totals ek $Q_{\text{enc}}$ mein combine ho jaate hain.

Enclosed charge. Cylinder wire ki length $L$ plus poora point charge trap karta hai: $Q_{\text{enc}}=\lambda L+q=(1.0\times 10^{-8})(0.30)+2.0\times 10^{-9}=3.0\times 10^{-9}+2.0\times 10^{-9}=5.0\times 10^{-9}\text{C}.$

Total flux. $\Phi =\frac{Q_{\text{enc}}}{{\epsilon }_0}=\frac{5.0\times 10^{-9}}{8.85\times 10^{-12}}\approx 565{\text{N⋅m}}^2/\text{C}.$ Note karo ki humne flux find kiya, jabki combined field itna symmetric nahi hai ki $E$ ko integral se bahar kheencha ja sake — flux ko sirf enclosed charge chahiye, detailed field kabhi nahi.

Enclosed charge, radius $r$ ki ball par $\rho$ ka integral hai. Thickness $d{r}^{\prime }$ ke shells (volume $4\pi r^{\prime 2}d{r}^{\prime }$) use karte hue: $Q_{\text{enc}}={\int }_0^r\alpha r^{\prime }(4\pi r^{\prime 2})d{r}^{\prime }=4\pi \alpha {\int }_0^r{r}^{\prime 3}d{r}^{\prime }=4\pi \alpha \frac{r^4}{4}=\pi \alpha r^4.$ Phir radius $r$ ki sphere par Gauss se: $E(4\pi r^2)=\frac{\pi \alpha r^4}{{\epsilon }_0}\Rightarrow E=\frac{\alpha r^2}{4{\epsilon }_0}.$ Plug in karo: $E=\frac{(3.0\times 10^{-5})(0.05{)}^2}{4(8.85\times 10^{-12})}=\frac{7.5\times 10^{-8}}{3.54\times 10^{-11}}\approx 2119\text{N/C}.$ Sanity check: yahan $E\propto r^2$ hai (uniform-density $E\propto r$ se steeper) kyunki charge baaharon pile up karta hai.

Conductor ke liye, metal ke andar field zero hoti hai (dekho Electric field of conductors). Ek pillbox draw karo jiska ek face metal ke andar ho (wahan flux $=0$) aur ek face bilkul baaharon (flux $=EA$). Sirf ek face leak karta hai: $EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{{\epsilon }_0}=\frac{1.0\times 10^{-6}}{8.85\times 10^{-12}}\approx 1.13\times 10^5\text{N/C}.$ Isolated sheet dono faces se leak karta hai → $2EA=\sigma A/{\epsilon }_0\Rightarrow E=\sigma /2{\epsilon }_0$. Same law, leaking faces ki alag count — bas itna hi fark hai.

• Jab $r\to 0^{+}$: $E\to \infty$. Idealized line apna saara charge zero-thickness filament par cramm karti hai, toh usse chipakne par unbounded field milti hai. (Real wires ki finite radius hoti hai, jo isko cap kar deti hai.)
• Jab $r\to \infty$: $E\to 0$, lekin sirf $1/r$ ki tarah — dheere-dheere.
• Contrast: point charge $1/r^2$ ki tarah fall off karta hai kyunki uski lines ek sphere par spread hoti hain ($\propto r^2$). Ek line ki lines cylinder par spread hoti hain ($\propto r$), ek dimension kam, toh woh zyada gently thin out hoti hain — $1/r$. Field lines ke dilute hone ki geometry hi exponent hai.

1. Symmetry identify karo → sphere / cylinder / pillbox choose karo, $d\vec{A}$ baaharon point karte hue.
2. Surface ko faces mein split karo; har flux ko $EA$, $0$, ya $2EA$ mark karo.
3. Sirf enclosed charge count karo, charges ko unke signs ke saath add karte hue (non-uniform ho toh $\rho$ integrate karo); Gaussian surface kisi charge se kabhi nahi guzarni chahiye.
4. Field solve karo: $E\times (\text{leaking area})=Q_{\text{enc}}/{\epsilon }_0$ set karo aur leaking area se divide karke $E$ pao.
5. Limits aur signs check karo — net negative $Q_{\text{enc}}$ negative flux aur inward field deta hai.