Gauss's law — integral form, choosing Gaussian surfaces

1.8.6 · Physics › Electromagnetism

Gauss's Law KYA hai?

Electric flux ${\Phi }_E=\oint \vec{E}\cdot d\vec{A}$ measure karta hai ki kitna field surface se "flow" kar raha hai. Dot product $\vec{E}$ ka outward normal ke saath wala component pick karta hai — sideways field kuch contribute nahi karta.

YEH SACH KYU HAI? (Coulomb's law se derive karo)

Hum ise scratch se, step by step, banate hain.

Step 1 — Ek single point charge ka flux ek sphere se. Charge $q$ ko radius $r$ wale sphere ke centre pe rakho. Coulomb's law se, $\vec{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\hat{r}.$ Yeh step kyun? Hum sirf us field law se shuru karte hain jis par hum already trust karte hain (Coulomb), sabse simple symmetric case mein.

Sphere par $\vec{E}\parallel d\vec{A}$ hai (dono radial hain), aur $|\vec{E}|$ constant hai. Isliye $\Phi =\oint EdA=E\oint dA=\frac{q}{4\pi {\epsilon }_0{r}^2}\cdot (4\pi r^2)=\frac{q}{{\epsilon }_0}.$ Yeh step kyun? Coulomb ke denominator mein $r^2$ cancel ho jaata hai sphere ki area ke $r^2$ se. Radius gayab ho gayi — flux kisi bhi sphere ke liye same hai. Yahi cancellation Gauss's law ka secret heart hai.

Step 2 — Koi bhi closed shape, charge andar. $q$ se ek field line jo ek tilted patch se exit karti hai woh area $dA$ cross karti hai, lekin sirf perpendicular part $dA\cos \theta$ lines pakadta hai, jabki patch exactly utne factor se door hai jitna $r^2$ badhata hai. Dono effects cancel ho jaate hain, isliye solid angle hi matter karta hai: $d\Phi =\frac{q}{4\pi {\epsilon }_0}d\Omega ,\oint d\Omega =4\pi \Rightarrow \Phi =\frac{q}{{\epsilon }_0}.$ Yeh step kyun? Yeh dikhata hai ki sphere special nahi tha — koi bhi enclosing surface $q/{\epsilon }_0$ deta hai.

Step 3 — Charge bahar. Ek field line jo surface ke ek side se enter karti hai, doosri side se zaroor nikalti hai: yeh even number of times pierce karti hai, pehle $+$ phir $-$ contribute karti hai. Bahar ke charges ke liye net $\Phi =0$. Yeh step kyun? Yeh prove karta hai ki bahar ke charges $Q_{\text{enc}}$ mein nahi aate.

Step 4 — Superposition. Real fields add hoti hain: $\vec{E}=\sum {\vec{E}}_i$. Flux linear hai, isliye $\Phi ={\sum }_i\frac{q_i^{\text{(inside)}}}{{\epsilon }_0}=\frac{Q_{\text{enc}}}{{\epsilon }_0}.■$

KAISE use karein: Gaussian surface choose karna

Law hamesha true hai lekin solvable sirf tab hota hai jab tum $E$ ko integral se bahar nikal sako. Aisi surface choose karo jis par har face par ya toh:

1. $\vec{E}$ magnitude mein constant ho aur $d\vec{A}$ ke parallel ho ($\theta =0$), to $\int EdA=EA$, ya
2. $\vec{E}\perp d\vec{A}$ ($\theta =90°$), to flux zero hoga.

Worked examples

Example 1 — Point charge ka field

Charge $q$ se distance $r$ par $\vec{E}$ nikalo.

• Surface: sphere radius $r$. Kyun? Spherical symmetry → $E$ sirf $r$ par depend karta hai aur radially point karta hai.
• Sphere par: $\Phi =E(4\pi r^2)$. Kyun? $E$ constant hai aur normal ke parallel hai.
• Gauss: $E4\pi r^2=q/{\epsilon }_0\Rightarrow E=\frac{q}{4\pi {\epsilon }_0{r}^2}$. Humne Coulomb recover kar liya — consistency check pass ho gaya.

Example 2 — Infinite line charge, linear density $\lambda$

• Surface: cylinder radius $r$, length $L$, coaxial.
• Curved face: $\vec{E}$ radial hai, $\parallel d\vec{A}$, constant → flux $=E(2\pi rL)$.
• Flat end caps: $\vec{E}\perp d\vec{A}$ → flux $=0$. Kyun? Field radially point karta hai, caps axis ke along face karte hain.
• $Q_{\text{enc}}=\lambda L$. $E(2\pi rL)=\frac{\lambda L}{{\epsilon }_0}\Rightarrow \boxed{E=\frac{\lambda }{2\pi {\epsilon }_{0}r}}$ Length $L$ kyun cancel hoti hai? Trapped charge aur curved area dono $L$ ke saath scale karte hain.

Example 3 — Infinite charged plane, surface density $\sigma$

• Surface: pillbox, cross-section area $A$, sheet ko symmetrically straddle karte hue.
• Do flat faces: $\vec{E}\parallel d\vec{A}$, total flux $=2EA$ (field dono sides se exit karta hai).
• Side wall: $\vec{E}\perp d\vec{A}$ → 0.
• $Q_{\text{enc}}=\sigma A$. $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow \boxed{E=\frac{\sigma }{2{\epsilon }_0}}$ Factor 2 kyun? Flux dono faces se escape karta hai. $E$ distance se independent kyun? Infinite plane = constant solid-angle coverage; field uniform hai.

Example 4 — Uniformly charged solid sphere, radius $R$, total $Q$, inside ($r<R$)

• Surface: sphere radius $r<R$.
• $Q_{\text{enc}}=Q\frac{r^3}{R^3}$ (charge enclosed volume ke saath scale karta hai). $E4\pi r^2=\frac{Q{r}^3/R^3}{{\epsilon }_0}\Rightarrow \boxed{E=\frac{Qr}{4\pi {\epsilon }_0{R}^3}}$ Andar $r$ mein linear kyun? Jitna deep jaao, utna kam charge enclosed hota hai; centre par $E\to 0$.

Common mistakes (steel-manned)

Flashcards

Connections

• Coulomb's law — Gauss's law isi se derive hoti hai; statics ke liye dono equivalent hain.
• Electric flux — woh quantity jise Gauss's law constrain karta hai.
• Electric field of conductors — andar field = 0, surface field $\sigma /{\epsilon }_0$ ek pillbox ke through.
• Divergence theorem — integral form ko differential form $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$ mein convert karta hai.
• Maxwell's equations — Gauss's law charon mein se pehla hai.
• Symmetry in physics — isliye surface choice kaam karta hai.

Concept Map