Applications — sphere, cylinder, infinite plane: Worked examples

1.8.7 · D3 · Physics › Electromagnetism › Applications — sphere, cylinder, infinite plane

Yeh page Applications of Gauss's Law ka drill ground hai. Parent note ne teen formulas build ki thi; yahan hum har woh case cover karte hain jisme woh aa sakti hain — inside, outside, surface pe, zero-charge regions, limiting distances, negative charge, real-world word problems, aur ek exam-style twist.

Shuru karne se pehle, ek reminder un vocabulary ki jo hum use karte rahenge, taaki koi symbol unexplained na lage:

Neeche har formula Gauss's Law se aata hai, sahi symmetry ke saath apply karke; flux ka idea khud Electric Flux mein hai.

The scenario matrix

Har row ek case class hai jo yeh topic tumhare saamne rakh sakta hai. Last column us worked example ka naam deta hai jo use cover karta hai.

#	Case class	Kya tricky banata hai	Covered by
A	Sphere, point outside ( $r > R$ )	point charge jaisa behave karta hai	Ex 1
B	Conducting shell, point inside ( $r < R$ )	$E = 0$ , degenerate	Ex 2
C	Solid insulator, point inside ( $r < R$ )	$E \propto r$ , zero se badhta hai	Ex 3
D	Negative charge (sign case)	arrows reverse, magnitude same	Ex 4
E	Line / cylinder, radial field	$1/ s$ fall-off	Ex 5
F	Infinite sheet + conductor surface	$/2 ε_{0}$ vs $/ ε_{0}$	Ex 6
G	Two-plate capacitor (superposition)	andar add, bahar cancel	Ex 7
H	Limiting behaviour ( $r \to R$ , $r \to \infty$ )	continuity & decay checks	Ex 8
I	Real-world word problem	words ko translate karo → $λ, σ, Q$	Ex 9
J	Exam twist: concentric shells	surface choose karo, $Q_{e n c}$ count karo	Ex 10

Example 1 — Sphere, point outside (Cell A)

Worked example Statement

$R = 0.10 m$ radius ki ek metal sphere $Q = + 3 μ C$ carry karti hai. $r = 0.30 m$ pe centre se $E$ find karo.

Forecast: compute karne se pehle guess karo — kya yeh surface pe field se bada hoga ya chota? (Chota hona chahiye: tum door chale gaye.)

Region pehchano. $r = 0.30 > R = 0.10$ , toh hum outside hain. Yeh step kyun? Jo formula tum use karte ho woh bilkul isi baat par depend karta hai ki tum $R$ ke kis taraf khade ho — parent note ne do alag results diye the.
Outside law use karo $E_{o u t} = \frac{k Q}{r ^{2}}$ . Yeh step kyun? Bahar se, saari charge centre pe ek point jaisi "lagti" hai, toh field point-charge field hai.
Plug in karo: $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{( 0.30 ) ^{2}} = \frac{27000}{0.09} = 3.0 \times 1 0^{5} N/C$ , radially outward point karta hua (charge positive hai).

Verify: Units: $\frac{( N\cdotm ^{2} / C ^{2} ) ( C )}{m ^{2}} = N/C$ ✓. Sanity: surface pe $r = 0.10$ pe field hogi $\frac{27000}{0.01} = 2.7 \times 1 0^{6}$ , naun guna bada — $1/ r^{2}$ aur hamare forecast (door se chota) ke saath consistent. ✓

Example 2 — Conducting shell, point inside (Cell B, degenerate)

Worked example Statement

Ex 1 wali hi sphere ( $Q = + 3 μ C$ , $R = 0.10 m$ ), lekin yeh hollow conducting shell hai. $r = 0.05 m$ pe $E$ find karo.

Forecast: trap alert — kya paas mein zyada charge andar field bada kardega? (Nahi.)

Gaussian sphere $r = 0.05$ pe draw karo (shell ke andar). Yeh step kyun? Gauss sirf is bag ke andar charge ki parwah karta hai.
$Q_{e n c}$ count karo. Conductor pe saari charge outer surface pe hoti hai, $r = R$ pe. Hamara bag $r = 0.05$ pe kuch bhi enclose nahi karta: $Q_{e n c} = 0$ . Yeh step kyun? Conductors saari charge ko surface pe push kar dete hain, interior khaali rehta hai.
Gauss apply karo: $E (4 π r^{2}) = \frac{0}{ε _{0}} = 0 \Rightarrow E = 0$ .

Verify: Agar conductor ke andar field nonzero hoti, toh free electrons ek force feel karte aur chalte rehte — lekin electrostatics ka matlab hai kuch nahi hilta. Toh $E = 0$ andar forced hai. ✓ Yeh degenerate / zero cell hai.

Example 3 — Solid insulator, point inside (Cell C)

Worked example Statement

Ek uniformly charged insulating ball mein $Q = + 3 μ C$ uski volume mein spread hai, $R = 0.10 m$ . $r = 0.05 m$ pe $E$ find karo.

Forecast: conductor ke unlike, charge ab andar spread hai — toh kuch charge enclosed hogi. Field nonzero hogi lekin surface pe field se kam.

Sirf enclosed charge count hoti hai. $r = 0.05$ pe bag ek choti ball mein charge hold karta hai. Kyunki charge evenly spread hai, enclosed fraction volume fraction $\frac{r ^{3}}{R ^{3}}$ hai. Yeh step kyun? Volume $r^{3}$ ki tarah badhta hai; uniform density ka matlab charge ∝ volume.
Inside law use karo $E_{in} = \frac{k Q r}{R ^{3}}$ . Yeh step kyun? Enclosed charge $\propto r^{3}$ hai lekin sphere ka area $\propto r^{2}$ , toh $r$ ki ek power bachti hai → $E \propto r$ , zero se linear.
Plug in karo: $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} ) ( 0.05 )}{( 0.10 ) ^{3}} = \frac{1350}{0.001} = 1.35 \times 1 0^{6} N/C$ , radially outward.

Verify: Surface pe $r = R = 0.10$ par yeh $\frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} ) ( 0.1 )}{0.001} = 2.7 \times 1 0^{6}$ deta hai, exactly bahar wali value $r = R$ pe — dono formulas boundary pe agree karte hain ✓. Aur $1.35 \times 1 0^{6} < 2.7 \times 1 0^{6}$ jaise forecast tha.

Figure — Applications — sphere, cylinder, infinite plane

Graph dekho: red curve centre se seedha upar uthti hai (Ex 3, inside insulator), blue curve bahar $1/ r^{2}$ ki tarah girta hai, aur dono $r = R$ pe touch karte hain — woh touching point wahi check hai jo humne abhi kiya.

Example 4 — Negative charge, sign case (Cell D)

Worked example Statement

Ek point-like sphere $Q = - 4 μ C$ carry karti hai. $r = 0.20 m$ pe field find karo, aur uski direction batao.

Forecast: magnitude same formula use karti hai; sirf direction alag hogi.

Magnitude compute karo $∣ Q ∣$ se: $E = \frac{( 9 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} = \frac{36000}{0.04} = 9.0 \times 1 0^{5} N/C$ . Yeh step kyun? Formula mein $Q$ field ki size ke liye magnitude ki tarah enter karta hai; sign direction se handle hoti hai.
Sign se direction fix karo. $Q < 0$ ⇒ field arrows inward point karte hain, charge ki taraf. Yeh step kyun? $E$ ki definition: yeh us taraf point karta hai jis taraf ek positive test charge push hoga — aur positive charge ek negative source ki taraf pulled hota hai.

Verify: Negative charge ke aaspaas ek bag se flux $\frac{Q _{e n c}}{ε _{0}} < 0$ hai — negative flux ka matlab arrows bag ke andar point karte hain, "inward" se match karta hai. ✓ Magnitude units N/C ✓.

Example 5 — Line / cylinder, $1/ s$ fall-off (Cell E)

Worked example Statement

Ek lambi seedhi wire mein $λ = + 2 \times 1 0^{- 7} C/m$ hai. $s = 0.05 m$ pe $E$ find karo.

Forecast: yeh $1/ s$ ki tarah fade hota hai, point ke $1/ s^{2}$ se slow.

Coaxial cylinder choose karo radius $s$ , length $L$ ka. Yeh step kyun? Axial symmetry $E$ ko radial aur curved side pe constant banata hai; flat end caps kuch contribute nahi karte.
Line-charge law $E = \frac{λ}{2 π ε _{0} s}$ . Yeh step kyun? Length $L$ cancel ho jaata hai — proof ki infinite-line result is baat pe depend nahi karta ki tumhara bag kitna tall hai.
Plug in karo: $E = \frac{2 \times 1 0 ^{- 7}}{2 π ( 8.85 \times 1 0 ^{- 12} ) ( 0.05 )}$ . Denominator $= 2 π (8.85 \times 1 0^{- 12}) (0.05) = 2.780 \times 1 0^{- 12}$ , toh $E = \frac{2 \times 1 0 ^{- 7}}{2.780 \times 1 0 ^{- 12}} \approx 7.19 \times 1 0^{4} N/C$ , radially outward.

Verify: $s$ ko double karke $0.10$ karo toh field half honi chahiye (quarter nahi). Formula deta hai $\approx 3.60 \times 1 0^{4}$ , exactly half. ✓ Point charge se compare karo, jo quarter ho jaata — yeh $1/ s$ law confirm karta hai.

Example 6 — Sheet vs conductor surface (Cell F)

Worked example Statement

Ek flat sheet mein $σ = + 1.77 \times 1 0^{- 8} C/m^{2}$ hai. Dono do: (a) isolated sheet ka field, aur (b) ek conductor ke bilkul bahar ka field jiska surface yahi $σ$ hold karta ho.

Forecast: conductor ki value sheet ki value se exactly double honi chahiye.

Isolated sheet $E = \frac{σ}{2 ε _{0}} = \frac{1.77 \times 1 0 ^{- 8}}{2 ( 8.85 \times 1 0 ^{- 12} )} = \frac{1.77 \times 1 0 ^{- 8}}{1.77 \times 1 0 ^{- 11}} = 1.0 \times 1 0^{3} N/C$ . Yeh step kyun? Ek free sheet dono taraf radiate karti hai, toh pillbox ke do faces hain: $2 E A = σ A / ε_{0}$ .
Conductor surface $E = \frac{σ}{ε _{0}} = \frac{1.77 \times 1 0 ^{- 8}}{8.85 \times 1 0 ^{- 12}} = 2.0 \times 1 0^{3} N/C$ . Yeh step kyun? Conductor ke andar $E = 0$ , toh saara flux sirf outer face se nikalti hai — ek face, do nahi, toh 2 ka factor nahi.

Verify: Ratio $\frac{2.0 \times 1 0 ^{3}}{1.0 \times 1 0 ^{3}} = 2$ exactly ✓ — woh classic factor-of-two jiske baare mein parent note ne warn kiya tha.

Example 7 — Two-plate capacitor, superposition (Cell G)

Worked example Statement

Do bade parallel plates $+ σ$ aur $- σ$ carry karte hain jahan $σ = 1.77 \times 1 0^{- 8} C/m^{2}$ hai. $E$ plates ke beech aur bahar find karo. (Yeh parallel-plate result hai.)

Forecast: plates ke beech dono single-sheet fields ek hi taraf point karti hain aur add hoti hain; bahar woh oppose karti hain aur cancel hoti hain.

Har sheet akele ka field: $\frac{σ}{2 ε _{0}} = 1.0 \times 1 0^{3} N/C$ (Ex 6 se). Yeh step kyun? Superposition — har plate ko isolated sheet maano, phir vectors add karo.
Beech mein: dono $+$ se $-$ ki taraf point karte hain, toh $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}} = 2.0 \times 1 0^{3} N/C$ .
Bahar: dono fields opposite directions mein hain aur cancel hoti hain → $E = 0$ . Yeh step kyun? Direction sab kuch hai: same direction → add, opposite → subtract.

Verify: $\frac{σ}{ε _{0}} = \frac{1.77 \times 1 0 ^{- 8}}{8.85 \times 1 0 ^{- 12}} = 2.0 \times 1 0^{3}$ ✓, aur "beech mein" value Ex 6 ki single-conductor value ke barabar hai — sense banata hai: capacitor gap ek conductor ke bahar jaisa behave karta hai. ✓

Arrows seedha dikhate hain: gap ke andar blue aur yellow arrows stack hote hain (double strength); bahar, woh ek doosre ke against point karte hain aur gayab ho jaate hain.

Example 8 — Limiting behaviour (Cell H)

Worked example Statement

Ex 3 ki solid insulating ball ke liye ( $Q = + 3 μ C$ , $R = 0.10 m$ ), do limits check karo: (a) andar aur bahar se surface pe aana, (b) bahut door jaana.

Forecast: field $r = R$ pe continuous honi chahiye aur $r \to \infty$ pe vanish karni chahiye.

Andar se, $r \to R$ : $E_{in} (R) = \frac{k QR}{R ^{3}} = \frac{k Q}{R ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{0.01} = 2.7 \times 1 0^{6} N/C$ . Yeh step kyun? $r = R$ ko inside law mein plug karo.
Bahar se, $r \to R$ : $E_{o u t} (R) = \frac{k Q}{R ^{2}} = 2.7 \times 1 0^{6} N/C$ — identical. Yeh step kyun? Wahi $r = R$ outside law mein; matching values ka matlab graph mein koi jump nahi.
Door, $r = 10 m$ pe: $E = \frac{( 9 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 6} )}{100} = 270 N/C$ , aur $\to 0$ as $r \to \infty$ . Yeh step kyun? $1/ r^{2}$ decay confirm karta hai ki ek finite ball door se ek fading point jaisi lagti hai.

Verify: Inside-limit = outside-limit at $R$ (dono $2.7 \times 1 0^{6}$ ) ✓ — continuity confirmed, wahi touching point jo Ex 3 figure mein draw hua tha. Aur $E (10 m) = 270 ≪ E (R)$ ✓.

Example 9 — Real-world word problem (Cell I)

Worked example Statement

Ek charged raincloud base ko ek badi flat sheet ke roop mein model kiya gaya hai jisme total charge $- 25 C$ hai jo $10 km^{2}$ area pe spread hai. Uske bilkul neeche electric field estimate karo.

Forecast: hume pehle "total charge over an area" ko $σ$ mein convert karna hoga, phir sheet formula use karna hoga.

$σ$ mein convert karo. $A = 10 km^{2} = 10 \times 1 0^{6} m^{2} = 1.0 \times 1 0^{7} m^{2}$ . Toh $σ = \frac{∣ Q ∣}{A} = \frac{25}{1.0 \times 1 0 ^{7}} = 2.5 \times 1 0^{- 6} C/m^{2}$ . Yeh step kyun? Formula ko charge per area chahiye, total charge nahi.
Isolated-sheet field $E = \frac{σ}{2 ε _{0}} = \frac{2.5 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )}$ . Yeh step kyun? Cloud base bilkul neeche se dekhi jaaye toh effectively infinite sheet hai.
Compute karo: $E = \frac{2.5 \times 1 0 ^{- 6}}{1.77 \times 1 0 ^{- 11}} \approx 1.41 \times 1 0^{5} N/C$ , cloud ki taraf point karta hua (charge negative hai).

Verify: Units $\frac{C/m ^{2}}{C ^{2} / ( N\cdotm ^{2} )} = N/C$ ✓. Air break down hoti hai (sparks) around $3 \times 1 0^{6} N/C$ pe; hamaara $1.4 \times 1 0^{5}$ usse kam hai lekin large — pre-lightning cloud ke liye plausible. ✓

Example 10 — Exam twist: two concentric shells (Cell J)

Worked example Statement

Ek inner conducting sphere ( $R_{1} = 0.05 m$ , charge $+ 2 μ C$ ) ek concentric conducting shell ke andar baitha hai ( $R_{2} = 0.10 m$ , charge $- 2 μ C$ ). $E$ find karo (a) $r = 0.07 m$ pe (dono ke beech) aur (b) $r = 0.20 m$ pe (dono ke bahar).

Forecast: bahar ka field zero hona chahiye — charges cancel ho jaate hain. Beech mein, sirf inner charge enclosed hai.

Beech mein ( $r = 0.07$ ): bag sirf inner $+ 2 μ C$ enclose karta hai. $E = \frac{k Q _{e n c}}{r ^{2}} = \frac{( 9 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 6} )}{( 0.07 ) ^{2}} = \frac{18000}{0.0049} \approx 3.67 \times 1 0^{6} N/C$ , outward. Yeh step kyun? $Q_{e n c}$ sirf tumhari surface ke andar wali charge count karta hai — outer shell is bag ke bahar hai.
Bahar ( $r = 0.20$ ): bag ab $+ 2 μ C - 2 μ C = 0$ enclose karta hai. Toh $E = 0$ . Yeh step kyun? Net enclosed charge zero hai → flux zero → field zero.

Verify: Beech wali value sirf $+ 2 μ C$ use karti hai; check $\frac{18000}{0.0049} = 3.673 \times 1 0^{6}$ ✓. Bahar, $Q_{e n c} = 0 \Rightarrow E = 0$ — yahi wajah hai ki coaxial cables aur shields ka koi external field nahi hota. ✓

Recall Quick self-test

$σ$ surface density wale conductor ke bilkul bahar field ::: $E = σ / ε_{0}$ Hollow conducting shell ke andar field ::: $E = 0$ Uniformly charged insulating ball ke andar, $E$ badhta hai ::: $E \propto r$ (linearly) Do equal-and-opposite concentric shells ke bahar net field ::: $0$ $\pm σ$ wale capacitor plates ke beech ::: $E = σ / ε_{0}$