Worked examples — Gauss's law — integral form, choosing Gaussian surfaces

1.8.6 · D3 · Physics › Electromagnetism › Gauss's law — integral form, choosing Gaussian surfaces

Yeh page parent Gauss's law note ke liye exercise gym hai. Yahan koi nayi theory nahi padhate — hum har tarah ki situation ko grind karte hain jo yeh law de sakta hai, taaki jab exam koi "naya" problem banaye, tum pehchaan sako ki woh matrix ke kaunse cell mein hai.

Neeche sab kuch ek boxed law par tika hai: ${\oint }_S\vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Agar koi symbol yahan unfamiliar lage, ruko aur parent note dobara padho — yeh page assume karta hai ki tum pehle se Electric flux ("kitni field lines bahar nikal rahi hain" wali quantity), Coulomb's law (point charge ka $1/r^2$ field), aur symmetry-matching rule (surfaces choose karne ka) se mil chuke ho.

Scenario matrix

Examples karne se pehle, har case class ka naam rakho jo yeh topic produce kar sakta hai. Har cell ek distinct tarah ka trap hai. Baad ke examples un cell(s) ke label ke saath hain jinhe woh cover karte hain.

Hum har row cover karte hain. Chalo shuru karte hain.

Example 1 — Positive point charge (cell A)

Forecast: Padhne se pehle andaza lagao — kya flux us radius par depend karega jo tum choose karo? Positive hoga ya negative?

1. Radius $r$ ki concentric sphere lo. Yeh step kyun? Charge mein spherical symmetry hai, isliye $E$ sirf $r$ par depend kar sakta hai aur radially point karna chahiye (figure mein lavender arrows dekho — dashed circle par sab same length ke hain). Yahi ek aisi surface hai jis par $E$ constant aur $d\vec{A}$ ke parallel ho.
2. Flux $=E\cdot (4\pi r^2)$ likho. Yeh step kyun? Kyunki $\vec{E}\parallel d\vec{A}$ har jagah ($\theta =0$, isliye $\cos \theta =1$) aur $E$ constant hai, $\oint EdA=E\oint dA=E\cdot A_{\text{sphere}}$, jahan $A_{\text{sphere}}=4\pi r^2$ poori $S$ ki area hai.
3. $q/{\epsilon }_0$ se barabar rakho aur solve karo. $E\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}\Rightarrow E=\frac{q}{4\pi {\epsilon }_0{r}^2}=\frac{(9.0\times 10^9)(2.0\times 10^{-9})}{(0.30{)}^2}=200\text{N/C}.$ Yeh step kyun? $1/(4\pi {\epsilon }_0)=9.0\times 10^9$ Coulomb constant $k$ hai — humne Coulomb's law wapas nikaal liya.
4. Total flux $=q/{\epsilon }_0=\frac{2.0\times 10^{-9}}{8.85\times 10^{-12}}=226{\text{N⋅m}}^2/\text{C}$. Yeh step kyun? Total flux Gauss's law ki right-hand side seedha padhne se milta hai: poora surface integral bas $Q_{\text{enc}}/{\epsilon }_0$ equals karta hai, isliye humein $E$ ki bhi zaroorat nahi — flux sirf trapped charge par depend karta hai.

Example 2 — Negative charge, aur ek charge outside (cells B & C)

Forecast: Inn mein se kaun sa zero flux deta hai? Kaun sa negative flux?

Part (a):

1. Flux $=q/{\epsilon }_0$. Kyun? Gauss's law sign ki parwah nahi karta — signed charge seedha daalo. $\Phi =\frac{-3.0\times 10^{-9}}{8.85\times 10^{-12}}=-339{\text{N⋅m}}^2/\text{C}.$
2. Minus ka matlab samjho. Yeh step kyun? Negative flux matlab field lines surface ke andar point kar rahi hain — figure mein coral inward arrows. Ek negative charge field lines ko nigal leta hai.

Part (b):

1. $Q_{\text{enc}}$ identify karo. Kyun? Gauss sirf closed surface $S$ ke andar charge count karta hai. $+5$ nC bahar hai, isliye $Q_{\text{enc}}=0$.
2. Conclude $\Phi =0$. $\Phi =\frac{0}{{\epsilon }_0}=0.$ Yeh step kyun? Right panel dekho: jo field line ek taraf se surface mein enter karti hai woh doosri taraf se nikl jaati hai. Enter $=-$, leave $=+$, yeh pairs mein cancel ho jaate hain. Surface par $\vec{E}$ zero nahi hai — lekin net flux zero hai.

Example 3 — Infinite line charge, near and far (cell D)

Forecast: Distance double karne par — $E$ factor of 2 se girega ya 4 se?

1. Coaxial cylinder, radius $r$, length $L$. Kyun? Line symmetry: $E$ sirf wire se doori par depend karta hai aur seedha bahar point karta hai (mint arrows). Ek cylinder woh symmetry share karta hai.
2. Curved wall: flux $=E\cdot (2\pi rL)$. Kyun? Wall par $\vec{E}\parallel d\vec{A}$ aur $E$ constant hai; yahan wall ki area $A=2\pi rL$ hai.
3. End caps: flux $=0$. Kyun? Flat caps par $\vec{E}$ (radial) aur $d\vec{A}$ (axial) perpendicular hain. $\cos 90°=0$.
4. $Q_{\text{enc}}=\lambda L$ ke saath solve karo: $E(2\pi rL)=\frac{\lambda L}{{\epsilon }_0}\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}r}.$ $L$ kyun cancel hota hai? Trapped charge aur wall area dono $L$ ke saath badhte hain.
5. Numbers daalo: $E(0.020)=\frac{4.0\times 10^{-6}}{2\pi (8.85\times 10^{-12})(0.020)}=3.60\times 10^6\text{N/C},$ $E(0.040)=1.80\times 10^6\text{N/C}.$

Example 4 — Infinite charged sheet, factor-2 trap (cell E)

Forecast: Kya $10$ cm par field kamzor hogi theek upar se?

1. Sheet ko straddle karta pillbox, cap area $A$. Kyun? Planar symmetry: field sheet ke perpendicular hai, dono sides par same. Ek symmetric box woh symmetry capture karta hai. Yahan $A$ box ki ek flat cap ki area hai.
2. Dono caps leak karte hain, side wall zero deti hai. Kyun? Field dono faces se bahar nikalti hai (butter arrows dono sides par door point karte hain) → total cap flux $=2EA$. Wall $\vec{E}$ ke parallel hai → zero.
3. $Q_{\text{enc}}=\sigma A$ ke saath solve karo: $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}=\frac{2.0\times 10^{-6}}{2(8.85\times 10^{-12})}=1.13\times 10^5\text{N/C}.$ $A$ kyun gayab hota hai? Yeh dono sides ko multiply karta hai — answer us box ke size par depend nahi kar sakta jisko humne imagine kiya.
4. $10$ cm par: phir bhi $1.13\times 10^5$ N/C. Kyun unchanged? Answer mein koi $r$ nahi — infinite plane ek uniform field deta hai.

Example 5 — Solid sphere: andar aur bahar (cell F, limiting values)

Forecast: Jab ball ke centre se bahar jaate ho, andar $E$ badhta hai ya ghatta hai?

1. Andar ($r<R$): $Q_{\text{enc}}=Q\frac{r^3}{R^3}$. Kyun? Uniform volume density $\rho$ → enclosed charge enclosed volume $\propto r^3$ ke saath scale karta hai, isliye radius $r$ ke andar ka fraction $r^3/R^3$ hai. $E\cdot 4\pi r^2=\frac{Q{r}^3/R^3}{{\epsilon }_0}\Rightarrow E_{\text{in}}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}.$ At $r=0.025$: $E_{\text{in}}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})(0.025)}{(0.050{)}^3}=1.44\times 10^4\text{N/C}.$
2. Bahar ($r>R$): $Q_{\text{enc}}=Q$ (poora). $E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}.$ At $r=0.10$: $E_{\text{out}}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.10{)}^2}=7.2\times 10^3\text{N/C}.$
3. $r=R$ par continuity check karo. Kyun? Kyunki charge purely volumetric hai — boundary par koi thin shell of surface charge nahi baitta — $r=R$ par exactly koi extra $Q_{\text{enc}}$ nahi aata, aur field jump nahi kar sakta. (Surface charge hoti toh jump hota; yahan nahi hai.) Dono formulas ko $r=R$ par evaluate karo aur actual numbers compare karo: $E_{\text{in}}(R)=\frac{QR}{4\pi {\epsilon }_0{R}^3}=\frac{Q}{4\pi {\epsilon }_0{R}^2}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.050{)}^2}=2.88\times 10^4\text{N/C},$ $E_{\text{out}}(R)=\frac{Q}{4\pi {\epsilon }_0{R}^2}=\frac{(9.0\times 10^9)(8.0\times 10^{-9})}{(0.050{)}^2}=2.88\times 10^4\text{N/C}.$ Dono same $2.88\times 10^4$ N/C dete hain, isliye surface par field continuous hai.

Example 6 — Charged spherical shell / cavity (cell G, degenerate empty inside)

Forecast: Hollow charged shell ke andar field kya hoga?

1. Andar ($r<R$): $Q_{\text{enc}}=0$. Kyun? Poora charge radius $R$ wali shell par hai; chhoti radius ki Gaussian sphere kuch bhi enclose nahi karti. $E\cdot 4\pi r^2=\frac{0}{{\epsilon }_0}=0\Rightarrow E_{\text{in}}=0.$ Andar har jagah exactly zero kyun? Symmetry force karta hai ki $E$ radial aur inner sphere par constant ho; zero flux + constant $E$ ⇒ $E=0$. Cavity field-free hai (figure mein lavender region).
2. Bahar ($r>R$): $Q_{\text{enc}}=Q$. $E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{(9.0\times 10^9)(6.0\times 10^{-9})}{(0.20{)}^2}=1.35\times 10^3\text{N/C}.$ Point charge jaisi form kyun? Bahar se dekho toh shell bilkul ek point charge jaisi lagti hai apne centre par.

Example 7 — Conductor surface vs insulated sheet (cell H)

Forecast: Same $\sigma$ — same field? Ya ek doosre ko double karta hai?

1. Insulated sheet: pillbox dono faces se leak karta hai. Kyun? Field dono sides par exist karta hai, isliye $2EA=\sigma A/{\epsilon }_0$ (jahan $A$ = ek cap ki area): $E_{\text{sheet}}=\frac{\sigma }{2{\epsilon }_0}=\frac{4.0\times 10^{-6}}{2(8.85\times 10^{-12})}=2.26\times 10^5\text{N/C}.$
2. Conductor surface: pillbox sirf ek face se leak karta hai. Kyun? Conductor ke andar $\vec{E}=0$ hota hai (charges tab tak rearrange hote hain jab tak yeh ho), isliye inner cap koi flux nahi pakdta. Sirf outer cap leak karta hai: $EA=\sigma A/{\epsilon }_0$: $E_{\text{cond}}=\frac{\sigma }{{\epsilon }_0}=\frac{4.0\times 10^{-6}}{8.85\times 10^{-12}}=4.52\times 10^5\text{N/C}.$
3. Compare karo. Conductor field exactly double hai sheet field se same $\sigma$ ke liye.

Example 8 — Real-world word problem (cell I)

Forecast: Ek negative speck ek positive wire ke paas — kheencha jaayega ya dhakka khayega?

1. Uss radius par line ka field nikalo. Yahan se kyun shuru karein? Force ke liye woh field chahiye jis mein speck hai; Example 3 ka line-charge result use karo. $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{1.5\times 10^{-8}}{2\pi (8.85\times 10^{-12})(1.0\times 10^{-3})}=2.70\times 10^5\text{N/C}.$
2. Force magnitude $F=|q|E$. $|q|$ kyun? Hum pehle sirf force ka size nikalte hain; $|q|$ (charge ka magnitude, sign ignore karke) times $E$ ek positive number of newtons deta hai. Direction agle step mein handle karte hain, isliye yahaan sign deliberately hata dete hain taaki double-count na ho. $F=|q|E=(2.0\times 10^{-15})(2.70\times 10^5)=5.39\times 10^{-10}\text{N}.$
3. Direction. Kyun? $\vec{E}$ positive wire se door point karta hai, lekin $q<0$ hai, isliye $\vec{F}=q\vec{E}$ (sign restore karne ke baad) wire ki taraf point karta hai — speck kheencha jaata hai. Exactly aise hi toner drum ki taraf attract hota hai.

Example 9 — Exam twist: valid par solvable nahi (cell J)

Forecast: Kya off-centre position flux badalta hai? Kya yeh $E$ solve karne ki ability badalta hai?

1. Net flux (part a). Kyun unchanged? Gauss sirf enclosed charge count karta hai, uski position nahi. Jab tak $q$ closed surface $S$ ke andar hai: $\Phi =\frac{q}{{\epsilon }_0}\text{(same as if it were centred).}$
2. $\vec{E}$ nikalna (part b). Kyun solvable nahi? Charge ka off-centre hona spherical symmetry tod deta hai: $E$ ab surface par point-to-point vary karta hai (nazdiki side par zyada, door wali side par kam). Kyunki $E$ ab $S$ par constant nahi hai, tum isse integral $\oint \vec{E}\cdot d\vec{A}$ se bahar nahi nikal sakte. Equation $\oint \vec{E}\cdot d\vec{A}=q/{\epsilon }_0$ sach rehti hai, lekin yeh ek number hai jo surface par $E$ ke infinitely many unknown values balance kar raha hai — isliye Gauss's law akele yahan $\vec{E}$ nahi de sakta.
3. Iski jagah kya karo. Kyun tools badle? Jab symmetry chali jaaye, $\vec{E}$ ke liye directly Coulomb's law / superposition use karo; Gauss sirf total flux ke liye rakho. Conclusion: flux $q/{\epsilon }_0$ par fix hai, lekin Gauss's law se bina symmetry ke field nahi milta.

Matrix coverage ka recap

Related: Divergence theorem iss integral law ko Maxwell's equations mein local form mein convert karta hai; Ex 6 ki shielding Electric field of conductors ki neenv hai.