Electric field — definition, field lines, superposition: Worked examples

1.8.4 · D3 · Physics › Electromagnetism › Electric field — definition, field lines, superposition

Shuru karne se pehle, do symbols jo poore page pe kaam aayenge:

The scenario matrix

Is topic ka har problem inhi cells mein se ek hai. Neeche ke examples mein tag hai ki woh kaun si cell cover karte hain.

#	Cell (case class)	Kya different hai	Covered by
A	Single positive charge	field door point karta hai	Ex 1
B	Single negative charge	field taraf point karta hai; sign $\overset{r}{^}$ ko flip karta hai	Ex 2
C	Two like charges, symmetric point	horizontal parts cancel, vertical add	Ex 3
D	Two unlike (dipole), symmetric point	vertical cancel, horizontal add	Ex 4
E	Limiting behaviour — far away	C aur D dono ka far-field check karo	Ex 3, 4
F	Degenerate / zero-field point	fields exactly $0$ tak cancel	Ex 5
G	Symmetry gives zero (ring centre)	continuous charge, sab cancel	Ex 6
H	Real-world word problem	words → numbers translate karo	Ex 7
I	Exam twist — third charge kahan rakhein ki use force na lage	$E = 0$ solve karo	Ex 8

Prerequisite tools jo use hote dikhenge: Coulomb's Law, Superposition Principle, aur dipole ke liye Electric Dipole.

Ex 1 — Cell A: single positive charge

Worked example Charge

$+ 2 nC$ origin par rakha hai. $E$ nikalo us point par jo x-axis par $3 cm$ dayi taraf hai. Forecast: Compute karne se pehle andaza lagao — field left point karega ya right, aur roughly kitna bada hoga ( $1 0^{4} N/C$ se bada ya chhota)?

Step 1 — SI units mein knowns likhो. $q = 2 \times 1 0^{- 9} C$ , $r = 3 \times 1 0^{- 2} m$ . Yeh step kyun? $k$ metres aur coulombs mein hai; nC ya cm mix karne se answer billions se multiply ho jaata hai silently. Pehle convert karo, hamesha.

Step 2 — point-charge formula se magnitude. $E = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} )}{( 3 \times 1 0 ^{- 2} ) ^{2}}$ Yeh step kyun? Yeh Coulomb's Law ka poora content hai test charge se divide karke — ek source ka field.

Step 3 — arithmetic. Numerator $= 17.98$ . Denominator $= 9 \times 1 0^{- 4}$ . $E = \frac{17.98}{9 \times 1 0 ^{- 4}} \approx 1.998 \times 1 0^{4} N/C \approx 2.0 \times 1 0^{4} N/C .$

Step 4 — direction. $q > 0$ , isliye $E$ $\overset{r}{^}$ ke along point karta hai, yaani charge se door — yahan, right taraf ( $+ x$ mein). Red arrow dekho.

Verify: Units: $\frac{( N m ^{2} / C ^{2} ) ( C )}{m ^{2}} = N/C$ ✓. Direction $+ q$ se door ✓, forecast se match karta hai agar "right" guess kiya tha.

Ex 2 — Cell B: single negative charge

Worked example Charge ko

$- 2 nC$ se replace karo. Same point par $E$ nikalo. Forecast: Same size number? Same direction? Guess karo.

Step 1 — magnitude $∣ q ∣$ use karta hai. Magnitude sirf kitna charge hai isse care karta hai: $E = \frac{k ∣ q ∣}{r ^{2}} = 2.0 \times 1 0^{4} N/C$ — Ex 1 jaisa hi. Yeh step kyun? Distance aur charge ki matra nahi badli; sirf sign badla, aur sign direction ki baat hai, size ki nahi.

Step 2 — sign direction flip karta hai. $E = k q \overset{r}{^} / r^{2}$ mein, negative $q$ se $E$ $\overset{r}{^}$ ke opposite point karta hai. Kyunki $\overset{r}{^}$ charge se bahar (rightward) point karta hai, $E$ ab left point karta hai, charge ki taraf. Yeh step kyun? Parent ke mistake box mein yahi sign-rule warn kiya tha — $q$ ka sign, $\overset{r}{^}$ ki direction nahi, attract/repel encode karta hai.

Verify: Wahan rakha ek tiny positive test charge negative source ki taraf khicha jaayega — leftward — aur $F = q_{0} E$ $E$ ke along point karta hai. Consistent ✓.

Ex 3 — Cell C + E: do equal $+ q$ , symmetric point, aur far limit

Worked example Do charges

$+ q = + 3 nC$ $(\pm a, 0)$ par hain, $a = 4 cm$ . $E$ nikalo $(0, y)$ par jahan $y = 3 cm$ . Forecast: Answer up, down, sideways point karega, ya zero hoga?

Step 1 — har charge se point tak distance. Pythagoras se, $r = a^{2} + y^{2} = 4^{2} + 3^{2} cm = 5 cm = 0.05 m .$ Yeh step kyun? Har charge ek right triangle ke corner par hai field point ke saath; hypotenuse woh separation hai jo $1/ r^{2}$ ko chahiye.

Step 2 — ek charge se magnitude. $E_{1} = \frac{k q}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 3 \times 1 0 ^{- 9} )}{( 0.05 ) ^{2}} = \frac{26.97}{0.0025} \approx 1.079 \times 1 0^{4} N/C .$

Step 3 — symmetry se horizontal parts khatam karo. Left charge upar-aur-right push karta hai, right charge upar-aur-left push karta hai. Unke $x$ -components equal aur opposite hain → cancel. Sirf upward ( $+ y$ ) parts bachte hain. Yeh step kyun? Y-axis ke baare mein mirror symmetry: reflect karne se do charges swap ho jaate hain par point fix rehta hai, isliye $E_{x}$ apne negative ke barabar hona chahiye, yaani $E_{x} = 0$ .

Step 4 — vertical fraction. Har field ka upward slice $E_{1} cos θ$ hai, jahan $cos θ = y / r = 3/5 = 0.6$ . Do charges: $E_{y} = 2 E_{1} cos θ = \frac{2 k q y}{( a ^{2} + y ^{2} ) ^{3/2}} = 2 (1.079 \times 1 0^{4}) (0.6) \approx 1.295 \times 1 0^{4} N/C (upward) .$ $3/2$ power kyun? Coulomb se $r^{2}$ ka ek power, $cos θ = y / r$ se $y$ par project karne mein aadha power aur.

Step 5 — far limit ( $y ≫ a$ ). $y ≫ a$ ke liye, $(a^{2} + y^{2})^{3/2} \to y^{3}$ , isliye $E_{y} \to \frac{2 k q}{y ^{2}} .$ Yeh check kyun? Door se pair ek single charge $2 q$ ki tarah dikhta hai — aur waqai hum point-charge $1/ y^{2}$ law recover kar lete hain. Sanity confirmed.

Verify: Seedha upar point karta hai ✓ (koi sideways nahi, symmetry forecast se match). Number VERIFY mein check kiya.

Ex 4 — Cell D + E: dipole, symmetric point, aur uska far limit

Worked example Ab right charge flip karo:

$+ q$ at $(- a, 0)$ replace ho gaya, ab $+ 3 nC$ at $(+ a, 0)$ aur $- 3 nC$ at $(- a, 0)$ hai, same $a = 4 cm$ , $y = 3 cm$ . $E$ nikalo. Forecast: Is baar kaun se parts cancel hongे — ups ya sideways wale?

Step 1 — same $r$ , same $E_{1}$ . Geometry nahi badli: $r = 0.05 m$ , $E_{1} \approx 1.079 \times 1 0^{4} N/C$ .

Step 2 — ab kaun se components cancel hain. $+ q$ (right) ka field us se door point karta hai — hamare point par upar-aur-left. $- q$ (left) ka field us ki taraf point karta hai — neeche-aur-left. Vertical parts ab opposite hain → cancel; horizontal ( $- x$ ) parts same direction mein hain → add. Yeh step kyun? Ek charge ka sign swap karna us charge ke poore field vector ko flip karta hai, cancelling pair ko adding pair mein convert karta hai. Yahi Electric Dipole ka dil hai.

Step 3 — horizontal fraction. Yahan relevant slice $cos ϕ = a / r = 4/5 = 0.8$ use karta hai: $E = 2 E_{1} cos ϕ = \frac{2 k q a}{( a ^{2} + y ^{2} ) ^{3/2}} = 2 (1.079 \times 1 0^{4}) (0.8) \approx 1.726 \times 1 0^{4} N/C (in - x) .$

Step 4 — far limit ( $y ≫ a$ ). $(a^{2} + y^{2})^{3/2} \to y^{3}$ , isliye dipole moment $p = 2 q a$ ke saath: $E \to \frac{2 k q a}{y ^{3}} = \frac{k p}{y ^{3}} .$ Point charge se steeper ( $1/ y^{3}$ ) kyun? Door se $+$ aur $-$ almost cancel ho jaate hain; sirf unka tiny offset $2 a$ bachta hai, aur woh extra "smallness" ek aur power of distance ka kharcha leti hai.

Verify: Sideways point karta hai ( $- x$ mein) ✓, like-charge case ke vertical answer ke opposite — do symmetries complementary hain. Number VERIFY mein.

Ex 5 — Cell F: exact zero-field point

Worked example Do charges:

$+ 4 nC$ at $x = 0$ aur $+ 1 nC$ at $x = 0.30 m$ . Unhe join karne wali line par, $E = 0$ kahan hai? Forecast: Bade charge ke paas ya chhote ke? Solve karne se pehle guess karo.

Step 1 — kahan cancel ho sakta hai? Do positive charges ke beech, do fields sirf unke darmiyan ki gap mein opposite directions mein point karte hain. Bahar, dono same taraf point karte hain aur cancel nahi ho sakte. Yeh step kyun? Cancellation ke liye opposition chahiye; sirf interior region yeh deta hai.

Step 2 — magnitudes equal set karo. Null point ko $+ 4 nC$ charge se $x$ distance par mano, toh $+ 1 nC$ se $0.30 - x$ : $\frac{k ( 4 \times 1 0 ^{- 9} )}{x ^{2}} = \frac{k ( 1 \times 1 0 ^{- 9} )}{( 0.30 - x ) ^{2}} .$ Yeh step kyun? $E = 0$ ka matlab hai dono magnitudes match karein (directions already oppose kar rahe hain). $k$ aur $1 0^{- 9}$ cancel ho jaate hain.

Step 3 — solve karo. $4 (0.30 - x)^{2} = x^{2} \Rightarrow 2 (0.30 - x) = x$ (positive root lo): $0.60 - 2 x = x \Rightarrow x = 0.20 m .$ Toh null point bade charge se $0.20 m$ aur chhote se $0.10 m$ dur hai.

Verify: Yeh chhote charge ke paas hai ✓ (weak charge ko stronger se match karne ke liye paas hona padta hai). Distance ratio $0.20/0.10 = 2 = 4/1$ ✓ — exactly $q_{1} / q_{2}$ .

Ex 6 — Cell G: symmetry ⇒ zero (ring centre)

Worked example Ek thin ring of radius

$R$ par total charge $Q$ uniformly spread hai. Continuous Charge Distributions use karke dikhao ki centre par $E$ zero hai. Forecast: Jab charge chaaro taraf ho, toh centre field bada hoga ya zero?

Step 1 — ring ko tiny pieces $d q$ mein kaat do. Har chhota arc almost ek point charge hai; yeh $d E = \frac{k d q}{R ^{2}}$ banata hai jo us piece se centre ki taraf point karta hai. Yeh step kyun? Hum point-charge formula directly ek smear par apply nahi kar sakte, isliye hum infinitesimal points par superpose karte hain aur integrate karte hain.

Step 2 — opposite pieces pair karo. Angle $θ$ par har piece ke liye directly across $θ + 18 0^{\circ}$ par ek identical piece hai. Unke do $d E$ size mein equal aur direction mein exactly opposite hain.

Step 3 — sum collapse ho jaata hai. Har pair cancel ho jaata hai, isliye $E_{centre} = \oint d E = 0 .$ Yeh step kyun? Perfectly balanced opposite vectors ka integral zero hai — symmetry dekhne ke baad koi arithmetic nahi chahiye.

Verify: $Q$ aur $R$ se independent hai — jab tak ring uniform hai, centre hamesha null rehta hai. Yahi reasoning Gauss's Law exploit karta hai: symmetry kaam karti hai.

Ex 7 — Cell H: real-world word problem

Worked example Ek tiny paint droplet charge

$- 5.0 nC$ aur mass $2.0 \times 1 0^{- 8} kg$ carry karta hai. Yeh hawa mein motionless hover karta hai. Kaun sa electric field (size aur direction) ise uthaye rakha hai? ( $g = 9.8 m/ s^{2}$ .) Forecast: Charge negative hai toh field upar point karega ya neeche?

Step 1 — force balance. Hovering ka matlab net force zero: upar ki electric force neeche ki gravity ke barabar honi chahiye. $∣ q ∣ E = m g .$ Yeh step kyun? "Motionless" ka physics translation $\sum F = 0$ hai; yahan energy ideas nahi chahiye — seedha Newton.

Step 2 — $E$ ke liye solve karo. $E = \frac{m g}{∣ q ∣} = \frac{( 2.0 \times 1 0 ^{- 8} ) ( 9.8 )}{5.0 \times 1 0 ^{- 9}} = \frac{1.96 \times 1 0 ^{- 7}}{5.0 \times 1 0 ^{- 9}} = 39.2 N/C .$

Step 3 — direction. Droplet par force upward hona chahiye. Charge par force $F = q E$ hai; $q < 0$ ke saath, $F$ $E$ ke opposite hoti hai. Force upar paane ke liye, $E$ neeche point karna chahiye. Yeh step kyun? Ex 2 jaisi hi sign logic — jab charge negative ho toh kabhi sirf force se direction mat padho.

Verify: Units $\frac{kg \cdot m/ s ^{2}}{C} = N/C$ ✓. $E$ downward, force $q E$ upward, gravity balance karta hai ✓.

Ex 8 — Cell I: exam twist (third charge ko force-free kahan rakkhein?)

Worked example Charges

$+ q$ at $x = 0$ aur $+ 9 q$ at $x = 1.0 m$ . Ek third charge $Q$ ko line par rakhna hai taki use koi force na lage. Kahan — aur kya answer $Q$ ke sign ya size par depend karta hai? Forecast: $+ q$ ke paas ya $+ 9 q$ ke paas?

Step 1 — "no force" ka matlab "us jagah $E = 0$ ". Kyunki $F = Q E$ , force bilkul wahan zero hogi jahan baaki dono ka net field zero hoga — $Q$ se independent. Yeh step kyun? Yahi key twist hai: yeh ek force problem ko Ex-5 wale field-null problem mein reduce karta hai. $Q$ ka apna field kabhi khud par act nahi karta (parent mistake box).

Step 2 — field magnitudes equal set karo (gap ke interior mein). Mano yeh $+ q$ se $x$ distance par hai: $\frac{k q}{x ^{2}} = \frac{k ( 9 q )}{( 1 - x ) ^{2}} \Rightarrow (1 - x)^{2} = 9 x^{2} \Rightarrow 1 - x = 3 x .$

Step 3 — solve karo. $1 = 4 x \Rightarrow x = 0.25 m$ $+ q$ charge se.

Verify: Chhote charge $+ q$ ke paas ✓. Distance ratio $0.25 : 0.75 = 1 : 3 = 1 : 9$ ✓. Answer $Q$ ke sign aur size se independent ✓ — question mein yahi trap tha.

Recall Quick self-test

Do equal $+ q$ at $\pm a$ , symmetry axis par field kis taraf point karta hai? ::: Axis ke along (charges se door), horizontal parts cancel ho jaate hain. Perpendicular bisector par dipole kis taraf point karta hai? ::: Dipole ke antiparallel (horizontally $+$ se $-$ side tak), vertical parts cancel ho jaate hain. Do unequal like charges ke beech, null bade ke paas hai ya chhote ke? ::: Chhote ke paas; distance ratio $q_{1} / q_{2}$ hai. Ring ka centre field zero kyun hai? ::: Opposite pieces symmetry se cancel ho jaate hain, kisi bhi uniform $Q$ aur $R$ ke liye. Ek negative charge hover karta hai; $E$ kis taraf point karta hai? ::: Neeche — force $q E$ $E$ ke opposite hai, isliye upward force ke liye downward field chahiye.

Parent: topic note.