Electric field of point charge, dipole, ring, disk, line charge (Gauss's law): Worked examples

1.8.5 · D3 · Physics › Electromagnetism › Electric field of point charge, dipole, ring, disk, line cha

Poore note mein, $k \equiv \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ aur $ε_{0} \approx 8.85 \times 1 0^{- 12} C^{2} / (N\cdotm^{2})$ .

Scenario matrix

Is topic ke har problem ka seedha connection inhi cells se hai. Neeche ke har worked example par uska cell tag laga hua hai.

#	Cell (scenario)	Kaun sa formula	Example
A	Point charge, positive, seedhi distance	$k q / r^{2}$	Ex 1
B	Point charge, negative — direction/sign	$k q / r^{2}$ , direction	Ex 2
C	Do point charges — vector superposition, ek component cancel hota hai	$\int d E$	Ex 3
D	Dipole, axial vs equatorial compare karna	$2 k p / r^{3}$ , $k p / r^{3}$	Ex 4
E	Ring, general point + do limits ( $x = 0$ , $x ≫ R$ ) + max	$k Q x / (R^{2} + x^{2})^{3/2}$	Ex 5
F	Disk general, aur uska degenerate infinite-sheet limit	$\frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$	Ex 6
G	Infinite line — real-world word problem	$2 k λ / r$	Ex 7
H	Exam twist: do sheets (capacitor-jaisa), fields add vs cancel	$σ /2 ε_{0}$ superposed	Ex 8

Example 1 — Cell A: ek single positive point charge

Worked example Statement

Origin par ek charge $q = + 5.0 nC$ rakha hai. $r = 0.20 m$ par field magnitude nikalo.

Forecast: chhota charge, thodi si distance — kya tumhe hajaaron N/C expect karne chahiye, ya laakhon? Compute karne se pehle power of ten guess karo.

Step 1. Point-charge law use karo: $E = \frac{k q}{r ^{2}}$ . Yeh step kyun? Ek single point charge "atom" hai — koi integration nahi, bas plug in karo. Yahan sphere symmetry toot nahi rahi, isliye formula directly apply hota hai.

Step 2. Substitute karo: $q = 5.0 \times 1 0^{- 9} C$ , $r = 0.20 m$ , $r^{2} = 0.040 m^{2}$ . $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 5.0 \times 1 0 ^{- 9} )}{0.040} = \frac{44.95}{0.040} \approx 1124 N/C .$ Yeh step kyun? Sirf arithmetic hai — lekin $1 0^{9}$ aur $1 0^{- 9}$ ko visible rakhne se pata chalta hai ki woh cancel ho jaate hain, isliye answer $1 0^{3}$ ke paas aata hai, jo sensible forecast se match karta hai.

Step 3. Direction: radially outward, kyunki $q > 0$ . Yeh step kyun? $q$ ka sign direction set karta hai; sirf magnitude ek vector field ka adhura jawab hai.

Verify: Units: $\frac{N\cdotm ^{2} / C ^{2} \cdot C}{m ^{2}} = N/C$ . ✓ Order of magnitude $\sim 1 0^{3} N/C$ forecast se match karta hai.

Example 2 — Cell B: ek negative charge (sign & direction)

Worked example Statement

Charge $q = - 3.0 nC$ origin par hai. Point $P$ jo origin se $0.10 m$ right ( $+ x$ ) par hai, wahan ka field kya hai? Magnitude aur direction dono do.

Forecast: magnitude sirf $∣ q ∣$ par depend karta hai — lekin $P$ par arrow kis taraf point karega?

Step 1. Magnitude ke liye $∣ q ∣$ use karo: $E = \frac{k ∣ q ∣}{r ^{2}}$ . Yeh step kyun? Formula ka $q$ sign carry karta hai, lekin "kitna strong" hai yeh size se tay hota hai, sign se nahi. Size aur direction ko alag rakhne se arithmetic mein minus sign nahi khoega. $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 9} )}{( 0.10 ) ^{2}} = \frac{26.97}{0.010} \approx 2697 N/C .$

Step 2. Direction: negative charge ka field uski taraf point karta hai. Yeh step kyun? Field lines negative charge par end hoti hain. $P$ , $+ x$ side par hai, isliye wahan field origin ki taraf yani $- x$ direction mein point karega.

Verify: Agar hum point-charge test-charge idea use karte — $P$ par ek hypothetical $+ 1 C$ negative charge ki taraf pull hoga — force (isliye field) $- x$ hai. ✓ Magnitude $\approx 2.7 \times 1 0^{3} N/C$ .

Example 3 — Cell C: do charges, ek component cancel hota hai

Figure — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

Worked example Statement

Do charges $+ q$ aur $+ q$ ( $q = 4.0 nC$ ) $x$ -axis par $x = \pm 0.30 m$ par hain. $y$ -axis par $P = (0, 0.40) m$ par field nikalo.

Forecast: symmetry se, $P$ par field $x$ ke along point karega, $y$ ke along, ya diagonal? Compute karne se pehle decide karo.

Step 1. Har charge $P$ se $r = 0.3 0^{2} + 0.4 0^{2} = 0.50 m$ door hai (3-4-5 triangle — figure dekho). Yeh step kyun? Distance se $k q / r^{2}$ ke zariye har individual magnitude milti hai; clean 3-4-5 arithmetic ko honest rakhta hai.

Step 2. Har ek ki magnitude: $E_{1} = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4.0 \times 1 0 ^{- 9} )}{0.25} = 143.8 N/C .$ Yeh step kyun? Dono charges equal aur equidistant hain, isliye $E_{1} = E_{2}$ — yahi equality hai jo ek component cancel karwati hai.

Step 3. Resolve karo. $x$ -components equal aur opposite hain → cancel. $y$ -components add hote hain. $y$ -fraction hai $cos θ = \frac{0.40}{0.50} = 0.80$ (figure mein red arrow ka vertical shadow dekho). Yeh step kyun? Yahi superposition ka "silent hero" $\overset{r}{^}$ hai: woh axis chuno jahan symmetry kaam kare, taaki sirf ek component bache. $E_{P} = 2 E_{1} cos θ = 2 (143.8) (0.80) = 230.1 N/C, along + y .$

Verify: Direction seedha upar ( $+ y$ ) hai — "mirror symmetry across the $y$ -axis se horizontal part khatam" forecast se match karta hai. Units N/C. ✓

Example 4 — Cell D: dipole axial vs equatorial

Worked example Statement

Ek dipole ka $p = 6.0 \times 1 0^{- 30} C\cdotm$ hai (water molecule scale ki value). $r = 1.0 \times 1 0^{- 9} m$ par far field nikalo — axis par aur equator par — aur confirm karo ki axial field bilkul exactly equatorial se double hai.

Forecast: kaun sa bada hoga — axial ya equatorial? Kitne se?

Step 1. Axial: $E_{a x} = \frac{2 k p}{r ^{3}}$ . Yeh step kyun? Axis par paas wala charge dominate karta hai aur dono fields ek hi direction mein point karti hain, isliye factor $2$ hai. Hum $r^{3}$ use karte hain ( $r^{2}$ nahi) kyunki dono nearly-equal charges apna leading $1/ r^{2}$ term cancel kar dete hain — parent note ka key dipole fact. $E_{a x} = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 6.0 \times 1 0 ^{- 30} )}{( 1.0 \times 1 0 ^{- 9} ) ^{3}} = \frac{1.0788 \times 1 0 ^{- 19}}{1.0 \times 1 0 ^{- 27}} \approx 1.079 \times 1 0^{8} N/C .$

Step 2. Equatorial: $E_{e q} = \frac{k p}{r ^{3}}$ (axial ka aadha). Yeh step kyun? Bisector par axial components add hote hain lekin sirf factor $1$ carry karte hain ( $2$ nahi), aur yeh $p$ ke opposite direction mein point karte hain (figure dekho). Same $r^{3}$ law. $E_{e q} = \frac{( 8.99 \times 1 0 ^{9} ) ( 6.0 \times 1 0 ^{- 30} )}{1.0 \times 1 0 ^{- 27}} \approx 5.394 \times 1 0^{7} N/C .$

Step 3. Ratio $E_{a x} / E_{e q} = 2$ exactly. Yeh step kyun? Structural fact confirm hota hai — yeh inhi numbers ka coincidence nahi hai, yeh formulas mein $2$ vs $1$ hai.

Verify: $1.079 \times 1 0^{8} /5.394 \times 1 0^{7} = 2.00$ . ✓ Axial bada nikla, forecast sahi tha.

Example 5 — Cell E: ring — general point, dono limits, aur maximum

Worked example Statement

$R = 0.10 m$ radius ki ring par $Q = 8.0 nC$ charge hai. $E$ nikalo (a) axis par $x = 0.20 m$ par, (b) centre par $x = 0$ par, (c) jab $x \to \infty$ (dikhao ki yeh point charge ban jaata hai), aur (d) maximum field ki axial position locate karo.

Forecast: kya centre par sabse bada field hoga (tum saare charge ke sabse paas ho) — ya zero?

Step 1 (a). $E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{3/2}}$ jahaan $R^{2} + x^{2} = 0.01 + 0.04 = 0.05$ . Yeh step kyun? Har $d q$ ka sirf axial component survive karta hai; har $d q$ ka $cos θ = x / R^{2} + x^{2}$ same hota hai, isliye woh integral se bahar aa jaata hai, aur yeh closed form milta hai. $(R^{2} + x^{2})^{3/2} = 0.0 5^{1.5} = 0.011180, E = \frac{( 8.99 \times 1 0 ^{9} ) ( 8.0 \times 1 0 ^{- 9} ) ( 0.20 )}{0.011180} \approx 1286 N/C .$

Step 2 (b). $x = 0$ par: numerator mein $x = 0$ ka factor hai, isliye $E = 0$ . Yeh step kyun? Symmetry se ring ke opposite sides se har radial pull cancel ho jaati hai — paas hona symmetry se haar jaata hai. Parent note jis galti ke baare mein warn karta hai, yahi woh hai.

Step 3 (c). $x ≫ R$ ke liye: $(R^{2} + x^{2})^{3/2} \to x^{3}$ , isliye $E \to k Q x / x^{3} = k Q / x^{2}$ — ek point charge. Yeh step kyun? Door se ring ki radius invisible ho jaati hai; field ko sab-kuch ka "atom" law reduce kar dena chahiye. Ek achha limit check.

Step 4 (d). Maximum $x = R / 2 = 0.10/ 2 \approx 0.0707 m$ par hai ( $d E / d x = 0$ set karke; figure mein peak dekho). Yeh step kyun? Field centre par $0$ se grow karta hai, peak karta hai, phir $1/ x^{2}$ ki tarah fall off hota hai — isliye max centre par nahi balki ek interior point par hota hai.

Verify (d): $x = R / 2$ plug karne par: wahan field $\approx 2769 N/C$ hai, $x = 0.20$ wali value se zyada — ring ke paas peak hone se consistent. ✓

Example 6 — Cell F: disk, general aur degenerate (infinite sheet)

Worked example Statement

$R = 0.050 m$ radius ki disk par $σ = 2.0 \times 1 0^{- 6} C/m^{2}$ hai. $E$ nikalo (a) axis par $x = 0.050 m$ par, aur (b) $x ≪ R$ ka limit (infinite-sheet value).

Forecast: $x = R$ par (disk size ke relative "aadha bahar") — kya field full sheet value $σ /2 ε_{0}$ ke karib hogi, ya kaafi kam?

Step 1 (a). $E = \frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$ . Yeh step kyun? Disk nested rings se bani hai; ring result ko integrate karne se yeh bracket milta hai. $[\dots]$ "finite-size correction" hai jo ek real disk mein hoti hai aur infinite sheet mein kho jaati hai. $\frac{σ}{2 ε _{0}} = \frac{2.0 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )} = 1.130 \times 1 0^{5} N/C .$ $\frac{x}{R ^{2} + x ^{2}} = \frac{0.050}{0.0025 + 0.0025} = \frac{0.050}{0.0707} = 0.7071.$ $E = 1.130 \times 1 0^{5} (1 - 0.7071) = 1.130 \times 1 0^{5} (0.2929) \approx 3.310 \times 1 0^{4} N/C .$

Step 2 (b). Jab $x ≪ R$ : $\frac{x}{R ^{2} + x ^{2}} \to 0$ , bracket $\to 1$ , isliye $E \to \frac{σ}{2 ε _{0}} = 1.130 \times 1 0^{5} N/C$ . Yeh step kyun? Badi disk ke paas tum usse finite nahi dekh sakte — yeh infinite lagti hai. Field constant kyun: door jaane par aur zyada sheet dikhai deti hai, jo $1/ r^{2}$ falloff ko exactly cancel kar deti hai.

Verify: $x = R$ par field ( $3.3 \times 1 0^{4}$ ) sheet value ( $1.13 \times 1 0^{5}$ ) ka sirf $\approx 29%$ hai — isliye forecast "kaafi kam" sahi hai; jab $x$ , $R$ ke comparable ho, tab edge effects matter karte hain. ✓

Example 7 — Cell G: infinite line — real-world word problem

Worked example Statement

Ek lamba straight power-line wire linear charge density $λ = 3.0 \times 1 0^{- 7} C/m$ carry karta hai. Ek chidiya $r = 2.0 m$ door baith jaati hai. Woh kis field magnitude mein hai, aur field kis direction mein point karta hai?

Forecast: point charge ke $1/ r^{2}$ ki tulna mein, line ka $1/ r$ dheere fall off karta hai. Kya kuch metres par field "point charges ke baare mein tumhari intuition" se weak lagega ya strong?

Step 1. Gauss ka cylindrical result use karo: $E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}$ . Yeh step kyun? Cylindrical symmetry $E$ ko flux integral se bahar nikalne deti hai (sirf curved wall contribute karta hai) — isliye ring ki tarah integrate kiye bina clean formula milta hai. $E = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 7} )}{2.0} = \frac{5394}{2.0} \approx 2697 N/C .$

Step 2. Direction: radially outward, wire ke perpendicular (kyunki $λ > 0$ ). Yeh step kyun? Line ka field wire se shortest (perpendicular) path ke along point karta hai — wohi direction jisme Gaussian cylinder ke curved normal point karte hain.

Verify: Distance aadhi karke $1.0 m$ karne par $E$ double ho jaana chahiye (kyunki $1/ r$ hai, $1/ r^{2}$ nahi): $2 \times 2697 = 5394 N/C$ — yeh exactly $2 k λ /1.0$ hai. ✓ Slower falloff confirm hua.

Example 8 — Cell H: exam twist — do sheets (capacitor-jaisa)

Worked example Statement

Do badi parallel sheets $+ σ$ aur $- σ$ carry karti hain, $σ = 4.0 \times 1 0^{- 6} C/m^{2}$ . Field nikalo (a) sheets ke beech aur (b) unke bahar.

Forecast: "twist" — do single-sheet fields ( $σ /2 ε_{0}$ har ek) plates ke beech add hote hain ya cancel? Bahar?

Step 1. Har isolated sheet $E_{1} = \frac{σ}{2 ε _{0}}$ banati hai jo $+ σ$ se door aur $- σ$ ki taraf point karti hai. Yeh step kyun? Ek single sheet degenerate disk limit hai; uska field uniform hai, isliye poora problem do known uniform fields ka superposition hai. $E_{1} = \frac{4.0 \times 1 0 ^{- 6}}{2 ( 8.85 \times 1 0 ^{- 12} )} = 2.260 \times 1 0^{5} N/C .$

Step 2 (a) — beech mein: dono fields ek hi direction mein point karti hain ( $+$ se $-$ ki taraf), isliye add hoti hain: $E_{in} = 2 E_{1} = \frac{σ}{ε _{0}} = 4.520 \times 1 0^{5} N/C .$ Yeh step kyun? Yahi capacitor field $σ / ε_{0}$ hai — woh factor of 2 jo parent note common mistake ke roop mein flag karta hai, yahan dono sheets count karke recover hota hai, ek ki dono faces se nahi.

Step 3 (b) — bahar: dono fields opposite directions mein point karti hain aur cancel ho jaati hain: $E_{out} = E_{1} - E_{1} = 0.$ Yeh step kyun? Bahar tum dono sheets ke far side par ho; $+$ sheet bahar push karti hai jabki $-$ sheet andar pull karti hai — equal magnitudes, opposite signs.

Verify: $E_{in} = σ / ε_{0} = 4.0 \times 1 0^{- 6} /8.85 \times 1 0^{- 12} = 4.520 \times 1 0^{5} N/C$ = $2 \times 2.260 \times 1 0^{5}$ . ✓ Aur $E_{out} = 0$ , ek ideal Parallel plate capacitor se match karta hai. ✓

Recall check

Recall Ring ka field uske centre par zero kyun hota hai lekin

$x = R / 2$ par nahi? Centre par ::: har $d q$ ka ek opposite partner hota hai jo equally doosri taraf pull karta hai, isliye saare field vectors cancel ho jaate hain — symmetry paas hone se jeet jaati hai; field phir $x = R / 2$ par maximum tak rise karta hai usse pehle ki $1/ x^{2}$ ki tarah decay kare.

Recall Do oppositely-charged sheets ke beech field

$σ / ε_{0}$ hai, bahar kya hai? Zero ::: dono single-sheet fields ( $σ /2 ε_{0}$ har ek) andar add hoti hain lekin bahar cancel ho jaati hain.

Related: Gauss's law · Coulomb's law · Flux and field lines · Electric dipole in a uniform field · Electric potential.