Exercises — Applications — sphere, cylinder, infinite plane

Symmetry: spherical. Surface: $r=0.15\text{m}$ radius ka ek sphere, jo $R=0.05\text{m}$ se bahar hai. Flux yahan kyun collapse hota hai: is concentric sphere par, spherical symmetry force karti hai ki $\vec{E}$ ka magnitude har jagah same ho aur seedha bahar point kare — bilkul har tile ke $d\vec{A}$ ke saath. Toh har tile contribute karta hai $E\times (\text{tile area})$, aur tile areas ka sum sphere ka area $4\pi r^2$ deta hai. Isi liye $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2)$ hota hai aur integral plain multiplication ban jaata hai. Kyunki hum bahar hain, sara charge enclosed hai aur woh ek point charge ki tarah act karta hai: $E_{out}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}=\frac{(9\times 10^9)(3\times 10^{-6})}{(0.15{)}^2}$ Numerator $(9\times 10^9)(3\times 10^{-6})=2.7\times 10^4$; denominator $(0.15{)}^2=0.0225$. Upar ko neeche se divide karne par (yeh bas $E=\text{numerator}/r^2$ hai) milta hai: $E=\frac{2.7\times 10^4}{0.0225}=1.2\times 10^6\text{N/C}.$ Direction: radially outward (positive charge test charge ko door push karta hai).

Symmetry: planar. Flux yahan kyun collapse hota hai: ek pillbox jo sheet ko pierce karta hai, uspar $\vec{E}$ sheet ke perpendicular hai, isliye dono flat faces par woh $d\vec{A}$ ke parallel hai (har ek $EA$ contribute karta hai) lekin side walls par $d\vec{A}$ ke perpendicular hai (woh tiles sideways skim karti hain → zero). Isi liye sirf dono faces count karti hain aur $\oint \vec{E}\cdot d\vec{A}=2EA$ hota hai. Isolated-sheet ka result distance-independent hai: $E=\frac{\sigma }{2{\epsilon }_0}=\frac{2\times 10^{-6}}{2(8.85\times 10^{-12})}=1.13\times 10^5\text{N/C}.$ Same value $4\text{cm}$ aur $40\text{cm}$ par — yahi poora point hai: ek infinite sheet ka field distance ke saath kabhi weak nahi hota. Direction: sheet positive hai, toh field sheet se dono sides door point karta hai — left face par left aur right face par right, har jagah sheet ke perpendicular.

Symmetry: axial (cylindrical). Flux yahan kyun collapse hota hai: ek coaxial cylinder par, $\vec{E}$ radially outward point karta hai, toh curved side par woh $d\vec{A}$ ke parallel aur constant hai (contributing $E\times 2\pi sL$), jabki dono flat end caps par $\vec{E}$, $d\vec{A}$ ke perpendicular hai → zero. Isi liye $\oint \vec{E}\cdot d\vec{A}=E(2\pi sL)$ hota hai. Line-charge result: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}=\frac{8\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.10)}=1.44\times 10^4\text{N/C}.$ Direction: wire se radially door.

Symmetry: spherical. Surface: $r=0.04\text{m}<R$ radius ka sphere, toh hum inside hain. Sirf radius $r$ ke andar ka charge count hota hai.

Gauss's law explicitly set up karo. Hamare sphere par $E$ constant hai aur radially point karta hai, toh har tile par $\vec{E}$, $d\vec{A}$ ke parallel hai (definition box yaad karo: $d\vec{A}$ har tile ka outward arrow hai). Isliye har tile deta hai $E\times (\text{tile area})$ aur tiles ka sum sphere ka area hai: $\oint \vec{E}\cdot d\vec{A}=E(4\pi r^2).$ Kyun: $E$ factor out hota hai (surface par constant), aur $\oint dA$ bas sphere ka area $4\pi r^2$ hai.

Kitna charge enclosed hai? $Q$ sara nahi — sirf radius $r$ ke andar ka fraction. Uniform density $\rho$ ke saath, woh density times chhote ball ka volume hai: $Q_{enc}=\rho \cdot \frac{4}{3}\pi r^3.$ Kyun: charge volume mein evenly spread hai, toh enclosed charge = density × enclosed volume.

Equal karo aur solve karo. $E(4\pi r^2)=\frac{\rho \frac{4}{3}\pi r^3}{{\epsilon }_0}.$ Dono sides se $4\pi$ cancel karo aur right side ke $r^3$ se $r^2$ ka ek power cancel karo (isi liye field linear end up hota hai — area $r^2$ ki tarah grow karta hai lekin enclosed charge $r^3$ ki tarah, ek power of $r$ bach jaata hai): $E_{in}=\frac{\rho r}{3{\epsilon }_0}.$

Numbers. Volume density: $\rho =\frac{Q}{\frac{4}{3}\pi R^3}=\frac{6\times 10^{-6}}{\frac{4}{3}\pi (0.10{)}^3}=1.43\times 10^{-3}{\text{C/m}}^3.$ $E_{in}=\frac{\rho r}{3{\epsilon }_0}=\frac{(1.43\times 10^{-3})(0.04)}{3(8.85\times 10^{-12})}=2.16\times 10^6\text{N/C}.$ Equivalent form ($\rho$ calculate karne se bachne ke liye; $\rho =Q/\frac{4}{3}\pi R^3$ ko $E_{in}$ mein substitute karo): $E_{in}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}=\frac{(9\times 10^9)(6\times 10^{-6})(0.04)}{(0.10{)}^3}=2.16\times 10^6\text{N/C}.$ ✓

Conductor surface, isolated sheet nahi, toh $\sigma /{\epsilon }_0$ version use karo (sara flux bahar jaata hai; metal ke andar field zero hai — dekho Conductors in Electrostatics): $E=\frac{\sigma }{{\epsilon }_0}=\frac{5\times 10^{-6}}{8.85\times 10^{-12}}=5.65\times 10^5\text{N/C}.$

Do isolated sheets superpose karo (Parallel Plate Capacitor). Har ek $\sigma /2{\epsilon }_0$ deta hai. Beech mein: dono fields ek hi direction mein point karti hain, toh add ho jaati hain: $E_{between}=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{1.0\times 10^{-6}}{8.85\times 10^{-12}}=1.13\times 10^5\text{N/C}.$ Bahar: dono fields opposite directions mein point karti hain aur cancel ho jaati hain: $E_{outside}=0$.

(i) Inner wall. Shell ke metal ke andar, $E=0$ hai. Metal ke andar ek Gaussian sphere banao ($a<r<b$). Kyunki wahan $E=0$ hai, enclosed charge zero hona chahiye. Centre charge $+4\mu \text{C}$ hai, toh inner wall ko usse cancel karne ke liye $-4\mu \text{C}$ carry karna hoga: $Q_{inner}=-4\mu \text{C}$. (ii) Outer wall. Shell overall neutral hai, toh uski dono walls ka sum zero hai: $Q_{inner}+Q_{outer}=0\Rightarrow Q_{outer}=+4\mu \text{C}$. (iii) $r=0.20\text{m}$ par field (sab kuch se bahar). $r=0.20$ par Gaussian sphere $q$ plus dono walls enclose karta hai: $+4-4+4=+4\mu \text{C}$ net. $E=\frac{(9\times 10^9)(4\times 10^{-6})}{(0.20{)}^2}=9.0\times 10^5\text{N/C, radially outward.}$

Andar ($r\le R$): $E_{in}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}$ — linearly rise karta hai, $r=R$ par sabse bada. Bahar ($r\ge R$): $E_{out}=\frac{Q}{4\pi {\epsilon }_0{r}^2}$ — girta hai, $r=R$ par sabse bada. Dono branches exactly surface par milti hain aur peak karti hain, $r=R$: $E_{max}=\frac{Q}{4\pi {\epsilon }_0{R}^2}.$ Field wahan continuous hai (ek acha sanity check): $r=R$ kisi bhi formula mein daalo, same result milega.

Potential woh work per unit charge hai jo ek test charge ko infinity se laane mein lagta hai: $V(R)=-{\int }_{\infty }^R{E}_{out}dr={\int }_R^{\infty }\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}dr=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}.$ Yahan integral kyun? $E$ tumhe har point par force per charge batata hai; potential puri path par accumulated push hai, jo exactly woh hai jo integral sum karta hai. Numerically: $V(R)=\frac{(9\times 10^9)(5\times 10^{-6})}{0.10}=4.5\times 10^5\text{V}.$

Surface par inside aur outside formulas agree karti hain; $E(R)=\frac{\rho R}{3{\epsilon }_0}$ use karo aur $\rho$ ke liye solve karo: $\rho =\frac{3{\epsilon }_{0}E(R)}{R}=\frac{3(8.85\times 10^{-12})(3.0\times 10^5)}{0.10}=7.97\times 10^{-5}{\text{C/m}}^3.$

Beech mein ($s=0.02$): Gaussian cylinder sirf inner wire ka $\lambda$ enclose karta hai: $E=\frac{\lambda }{2\pi {\epsilon }_{0}s}=\frac{3\times 10^{-8}}{2\pi (8.85\times 10^{-12})(0.02)}=2.70\times 10^4\text{N/C, outward.}$ Bahar ($s=0.5$): length per enclosed charge $+\lambda -\lambda =0$ hai, toh $E=0$ hai. Cable externally "silent" hai — isi liye coax shield karta hai.

Pehle insulator ka total charge: $Q_1=\rho \cdot \frac{4}{3}\pi R^3=(2\times 10^{-5})\frac{4}{3}\pi (0.10{)}^3=8.38\times 10^{-8}\text{C}$. (a) $r=0.05\text{m}$ (insulator ke andar): sirf $r$ ke andar ka charge count hota hai. $E=\frac{\rho r}{3{\epsilon }_0}=\frac{(2\times 10^{-5})(0.05)}{3(8.85\times 10^{-12})}=3.77\times 10^4\text{N/C.}$ (b) $r=0.15\text{m}$ (insulator aur shell ke beech): is radius par Gaussian sphere $Q_1$ poora enclose karta hai; conducting shell $R_2=0.20\text{m}$ par bahar baitha hai, toh jo enclosed hai usme kuch contribute nahi karta. $Q_1$ ko centre par point charge maano: $E=\frac{(9\times 10^9)(8.38\times 10^{-8})}{(0.15{)}^2}=3.35\times 10^4\text{N/C.}$ (c) $r=0.30\text{m}$ (sab kuch se bahar): ab Gaussian sphere insulator ka $Q_1$ aur shell ka $Q_2$ dono enclose karta hai: $Q_{enc}=Q_1+Q_2=8.38\times 10^{-8}+1\times 10^{-6}=1.084\times 10^{-6}\text{C}.$ $E=\frac{(9\times 10^9)(1.084\times 10^{-6})}{(0.30{)}^2}=1.08\times 10^5\text{N/C.}$

Har isolated sheet $E=\sigma /2{\epsilon }_0$ banati hai apni dono sides par door point karti hui. Rightward ko positive maano. Field units mein, ${\sigma }_A/2{\epsilon }_0=2.26\times 10^5$ aur ${\sigma }_B/2{\epsilon }_0=5.65\times 10^4\text{N/C}$ hai. Dono ke left mein: dono left point karti hain (door, − ki taraf). $E=-(2.26\times 10^5+5.65\times 10^4)=-2.82\times 10^5\text{N/C}$ → magnitude $2.82\times 10^5$, left ki taraf. Beech mein (A left par, B right par): A right push karta hai (+), B left push karta hai (−). $E=2.26\times 10^5-5.65\times 10^4=1.69\times 10^5\text{N/C}$ → magnitude $1.69\times 10^5$, right ki taraf. Dono ke right mein: dono right point karti hain (+). $E=2.26\times 10^5+5.65\times 10^4=2.82\times 10^5\text{N/C}$ → magnitude $2.82\times 10^5$, right ki taraf.

Gauss's law $E=\lambda /2\pi {\epsilon }_{0}s$ deta hai sirf tabhi jab field purely radial aur axis ke saath constant ho — jo require karta hai ki source aise dikhe jaise tum uske saath slide karo (translational/axial symmetry), taaki Gaussian cylinder ke end caps zero flux carry karen.

Paas se ($s\ll L$). Jab tumhari distance $s$ rod ki length $L$ se bahut chhoti hai, toh dono ends bahut dur hain aur practically invisible hain. Locally rod "infinite" lagti hai: field essentially radial hai, end-cap flux negligible hai, aur derivation ka assumption hold karta hai. Toh $1/s$ law yahan ek excellent approximation hai.

Door se ($s\gg L$). Ab poori rod ek chhota sa angle subtend karti hai — woh ek single dot of total charge $Q=\lambda L$ ki tarah lagti hai. Point charge ki field $1/r^2$ ki tarah fall off hoti hai, toh field $1/s$ (paas) se $1/r^2$ (door) mein cross over karni chahiye. In dono regimes ke beech ek smooth transition hoti hai.

Kaun sa assumption fail hota hai? Axial / translational symmetry (axis ke saath invariance). Ek finite rod aise nahi hai jaise tum uske saath slide karo — ends ke paas field tilt ho jaati hai aur purely radial nahi rehti, toh Gaussian cylinder ke end caps ab zero flux carry nahi karte. Jab end-cap flux matter karta hai, toh clean $E(2\pi sL)=\lambda L/{\epsilon }_0$ step invalid ho jaata hai. Takeaway: tidy $1/s$ formula sirf idealised infinite line ke liye exact hai; finite rod ke liye woh ek near-field approximation hai jo door jaate jaate point-charge $1/r^2$ mein decay ho jaati hai.

Recognition ::: symmetry ka naam lo aur $1/r^2$, $1/s$, ya constant pick karo Application ::: inside vs outside decide karo, phir ek baar plug karo Analysis ::: $E=0$ in metal + conservation use karo induced charge place karne ke liye Synthesis ::: integrate karke $E$ ko $V$ se link karo, ya do geometries combine karo Mastery ::: vectors ko sahi signs aur enclosed-charge bookkeeping ke saath superpose karo