Electric field — definition, field lines, superposition: Exercises

1.8.4 · D4 · Physics › Electromagnetism › Electric field — definition, field lines, superposition

Level 1 — Recognition

Kya tum sahi formula pick kar sakte ho aur signs/directions padhh sakte ho?

Recall Solution L1·1

WHAT hum use karte hain: point charge ka field, $E = \frac{k ∣ q ∣}{r ^{2}}$ . WHY yeh formula: ek akela isolated point charge bilkul wahi case hai jiske liye Coulomb's law banaya gaya tha; Coulomb force ko test charge se divide karne par $E = k ∣ q ∣/ r^{2}$ mila.

Plug in karo ( $3.0 nC = 3.0 \times 1 0^{- 9} C$ , $2.0 cm = 0.020 m$ ): $E = \frac{( 8.99 \times 1 0 ^{9} ) ( 3.0 \times 1 0 ^{- 9} )}{( 0.020 ) ^{2}} = \frac{26.97}{4.0 \times 1 0 ^{- 4}} \approx 6.74 \times 1 0^{4} N/C .$ Direction: $q$ positive hai, isliye $E$ radially outward, origin se door point karta hai.

Recall Solution L1·2

WHAT: $E = \frac{k q}{r ^{2}} \overset{r}{^}$ , jahan $\overset{r}{^}$ source se field point ki taraf point karta hai — yahan woh $+ x$ direction hai. WHY sign matter karta hai: $q$ negative hone par, $\frac{k q}{r ^{2}}$ ek negative number times $\overset{r}{^}$ hai, isliye $E$ $\overset{r}{^}$ ke opposite point karta hai — yaani charge ki taraf wapas, $- x$ direction mein. Yeh kaisa dikhta hai: field lines ek negative charge mein enter karti hain, isliye aas-paas ke kisi bhi point par arrow andar ki taraf point karta hai. ✓

Level 2 — Application

Numbers plug karo ek multi-step lekin phir bhi single-idea calculation mein.

Recall Solution L2·1

WHAT: jab $E$ pata ho, toh kisi bhi charge par force hota hai $F = Q E$ . WHY yeh ek line hai: field ne sources ke baare mein sab kuch already package kar diya hai; charge bas respond karta hai. $∣ F ∣ = ∣ - e ∣ E = (1.6 \times 1 0^{- 19}) (2.0 \times 1 0^{4}) = 3.2 \times 1 0^{- 15} N .$ Direction: $Q = - e$ negative hai, isliye $F$ $E$ ke opposite point karta hai → $- x$ direction mein.

Recall Solution L2·2

WHAT chahiye: ek aisa point jahan do fields ki equal magnitude aur opposite direction ho taaki woh cancel ho jayein. WHY unke beech: do positive charges ke beech, left wale ka field right point karta hai, right wale ka left — sirf yahan woh oppose karke cancel ho sakte hain.

Maan lo null point $+ q$ se $x$ distance par hai, toh woh $+ 4 q$ se $d - x$ door hai. Magnitudes equal set karo: $\frac{k q}{x ^{2}} = \frac{k ( 4 q )}{( d - x ) ^{2}} \Rightarrow \frac{1}{x ^{2}} = \frac{4}{( d - x ) ^{2}} .$ Square root lo (dono sides positive hain): $\frac{1}{x} = \frac{2}{d - x} \Rightarrow d - x = 2 x \Rightarrow x = \frac{d}{3} .$ Yeh kaisa dikhta hai: null point chhote charge ke kareeb hota hai — yeh sense banata hai, tumhe weak charge ke paas khada hona padta hai taaki uska field strong wale se compete kar sake.

Level 3 — Analysis

Ab geometry kaam karta hai — components aur symmetry.

Figure — Electric field — definition, field lines, superposition

Recall Solution L3·1

Step 1 — distance (figure mein triangle dekho). Har charge field point se $r = a^{2} + y^{2}$ door hai (Pythagoras legs $a$ aur $y$ wale right triangle par). Toh har field ka magnitude hai $E_{1} = \frac{k q}{a ^{2} + y ^{2}}$ .

Step 2 — components (WHY symmetry help karta hai). Do charges $y$ -axis ke across ek doosre ke mirror hain. Unke horizontal parts $E_{x}$ opposite directions mein point karte hain aur cancel ho jaate hain; unke vertical parts dono $+ y$ point karte hain aur add hote hain. Vertical fraction hai $cos θ = \frac{y}{a ^{2} + y ^{2}}$ — usi triangle ka "adjacent over hypotenuse."

Step 3 — total. $E_{y} = 2 E_{1} cos θ = \frac{2 k q}{a ^{2} + y ^{2}} \cdot \frac{y}{a ^{2} + y ^{2}} = \frac{2 k q y}{( a ^{2} + y ^{2} ) ^{3/2}} .$ $3/2$ power: Coulomb se ek pura power $r^{2}$ , aur projection $cos θ$ se aadha power aur.

Numbers ( $q = 2.0 \times 1 0^{- 9}$ , $a = 0.03$ , $y = 0.04$ , toh $a^{2} + y^{2} = 0.0025 m^{2}$ , $(a^{2} + y^{2})^{3/2} = 1.25 \times 1 0^{- 4}$ ): $E_{y} = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 2.0 \times 1 0 ^{- 9} ) ( 0.04 )}{1.25 \times 1 0 ^{- 4}} \approx 1.15 \times 1 0^{4} N/C, pointing + y .$

Recall Solution L3·2

Kya badla: ek charge negative hone par, vertical parts cancel ho jaate hain aur horizontal parts add hote hain (dono $- x$ mein point karte hain). Hum $cos ϕ = \frac{a}{a ^{2} + y ^{2}}$ se project karte hain (triangle ki doosri leg). $E = 2 E_{1} cos ϕ = \frac{2 k q}{a ^{2} + y ^{2}} \cdot \frac{a}{a ^{2} + y ^{2}} = \frac{2 k q a}{( a ^{2} + y ^{2} ) ^{3/2}}, along - x .$ Yeh Electric Dipole result hai; $p = 2 q a$ define karke, door jaane par yeh $E \to \frac{k p}{y ^{3}}$ ban jaata hai.

Level 4 — Synthesis

Superposition, limits, aur physical reasoning combine karo.

Recall Solution L4·1

WHAT limit karta hai: jab $y ≫ a$ , toh $(a^{2} + y^{2})^{3/2}$ ke andar $a^{2}$ negligible ho jaata hai $y^{2}$ ke muqable mein, isliye $(a^{2} + y^{2})^{3/2} \to (y^{2})^{3/2} = y^{3}$ . $E_{y} \to \frac{2 k q y}{y ^{3}} = \frac{2 k q}{y ^{2}} = \frac{k ( 2 q )}{y ^{2}} .$ WHY yeh bilkul sahi hai: door se do charges ek total charge $2 q$ ke blob mein blur ho jaate hain — tum ab $\pm a$ separation resolve nahi kar sakte. Ek single charge ka inverse-square law phir se ubhar aata hai. Number at $y = 1.0 m$ : $E \approx \frac{( 8.99 \times 1 0 ^{9} ) ( 4.0 \times 1 0 ^{- 9} )}{1. 0 ^{2}} \approx 36 N/C$ .

Recall Solution L4·2

$E \to \frac{2 k q a}{y ^{3}} = \frac{k p}{y ^{3}}, p = 2 q a .$ WHY steeper: ek dipole ka total charge zero hota hai. Door se $+ q$ aur $- q$ almost cancel ho jaate hain, isliye leading $1/ y^{2}$ term vanish ho jaata hai. Jo bachta hai woh sirf unki separation $2 a$ ka chhota sa effect hai — ek second-order leftover jo ek power faster decay karta hai, $1/ y^{3}$ . Yeh kaisa dikhta hai: dipole ki field lines $+$ se $-$ ki taraf tightly loop karti hain; door bahar jaate-jaate almost koi nahi bachti, kyunki $+ q$ se nikalne wali lines nearby $- q$ pakad leti hai.

Level 5 — Mastery

Ek problem jo har tool ko saath mein force karte use karti hai.

Recall Solution L5·1

Step 1 — ring ko chop karo (Continuous charge, figure dekho). Ring ko tiny pieces $d q$ mein toddo. Har piece axial point se $r = a^{2} + z^{2}$ door hai (Pythagoras: leg $a$ centre se piece tak, leg $z$ axis ke along). Har piece ek field produce karta hai $d E = \frac{k d q}{a ^{2} + z ^{2}}$ . Yahan Continuous Charge Distributions + Superposition Principle saath mein kaam kar rahe hain.

Step 2 — symmetry sideways parts ko maar deti hai. Har piece ke liye ring ke across ek opposite piece hoti hai. Unke components jo axis ke perpendicular hain woh cancel ho jaate hain; sirf axial component bachta hai. Axial fraction hai $cos θ = \frac{z}{a ^{2} + z ^{2}}$ . $d E_{z} = \frac{k d q}{a ^{2} + z ^{2}} \cdot \frac{z}{a ^{2} + z ^{2}} = \frac{k z d q}{( a ^{2} + z ^{2} ) ^{3/2}} .$

Step 3 — add up (integrate karo). $d q$ ko multiply karne wali har cheez har piece ke liye same hai (sabhi pieces same $r$ aur $z$ share karte hain), isliye $\int d q = Q$ : $E_{z} = \frac{k Q z}{( a ^{2} + z ^{2} ) ^{3/2}} axis ke along directed.$

Numbers ( $Q = 8.0 \times 1 0^{- 9}$ , $a = 0.05$ , $z = 0.12$ ; $a^{2} + z^{2} = 0.0025 + 0.0144 = 0.0169$ , $(0.0169)^{3/2} = 2.197 \times 1 0^{- 3}$ ): $E_{z} = \frac{( 8.99 \times 1 0 ^{9} ) ( 8.0 \times 1 0 ^{- 9} ) ( 0.12 )}{2.197 \times 1 0 ^{- 3}} \approx 3.93 \times 1 0^{3} N/C .$

Far-field check ( $z ≫ a$ ): $(a^{2} + z^{2})^{3/2} \to z^{3}$ , isliye $E_{z} \to \frac{k Q z}{z ^{3}} = \frac{k Q}{z ^{2}}$ — ek point charge $Q$ , jaise hona chahiye. ✓ Centre check ( $z = 0$ ): $E_{z} = 0$ — exact centre par har piece ka field uske opposite se cancel ho jaata hai. ✓

Recall Ladder ka ek-line summary

Point-charge formula pehchano → $F = Q E$ apply karo aur nulls dhundho → components aur symmetry se analyse karo → far-field limits ke zariye synthesise karo → chop-project-add se continuous ring master karo.