Potential of point charge, potential from field and vice versa: Exercises

1.8.9 · D4 · Physics › Electromagnetism › Potential of point charge, potential from field and vice ver

Level 1 — Recognition

Goal: pehchano ki kaun sa relationship apply hota hai aur formula se value padhkar nikalo.

L1.1 Ek point charge $q = + 5 μ C$ origin par rakha hai. Potential $V (r)$ ko distance $r$ ke function ke roop mein likho, phir $r = 2 m$ par evaluate karo.

Recall Solution L1.1

KYA use karein: point charge ka potential, $V (r) = \frac{k q}{r}$ . KYUN: yeh ek single point charge hai — parent note ka formula seedha apply hota hai, koi integration nahi chahiye. $V (r) = \frac{( 9 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 6} )}{r} = \frac{4.5 \times 1 0 ^{4}}{r} V .$ $r = 2 m$ par: $V = \frac{4.5 \times 1 0 ^{4}}{2} = 2.25 \times 1 0^{4} V = 22500 V .$

L1.2 Ek point charge ke liye, $V$ distance ke saath kaise girta hai, aur field magnitude $E$ kaise girta hai? Kaun zyada tezi se girta hai?

Recall Solution L1.2

$V \propto \frac{1}{r}$ (neeche $r$ ki ek power hai); $E \propto \frac{1}{r ^{2}}$ (do powers). Kyunki $E$ ke liye exponent bada hai, the field drops faster potential ke mukable mein. Distance double karo: $V$ aadha ho jaata hai, lekin $E$ chauthai ho jaata hai.

Level 2 — Application

Goal: sahi se substitute karo, units aur signs bilkul honest rakhte hue.

L2.1 Ek charge $q = - 4 μ C$ kisi point par potential $V = - 1.2 \times 1 0^{4} V$ produce karta hai. Woh point charge se kitni door hai?

Recall Solution L2.1

KYA: $V = \frac{k q}{r}$ ko invert karo taaki $r = \frac{k q}{V}$ milen. KYUN: hame $V$ aur $q$ pata hai; distance hi sirf unknown hai. Negative signs rakhkar chalte hain — $q$ aur $V$ dono negative hain, isliye $r$ positive aata hai (jaisa ki distance hona chahiye). $r = \frac{k q}{V} = \frac{( 9 \times 1 0 ^{9} ) ( - 4 \times 1 0 ^{- 6} )}{- 1.2 \times 1 0 ^{4}} = \frac{- 3.6 \times 1 0 ^{4}}{- 1.2 \times 1 0 ^{4}} = 3 m .$

L2.2 Ek region mein potential $V (x) = 5 x$ volts hai ( $x$ metres mein). Field ka $x$ -component, $E_{x}$ , nikalo.

Recall Solution L2.2

KYA: $E_{x} = - \frac{d V}{d x}$ . KYUN: field, potential ke slope ka minus hota hai (parent §3). $E_{x} = - \frac{d}{d x} (5 x) = - 5 V/m .$ Constant field, $- x$ direction mein pointing (kyunki $V$ , $x$ ke saath badhta hai, aur field neeche ki taraf point karta hai).

L2.3 Ek uniform field $E = 300 V/m$ , $+ x$ direction mein point karta hai. $x_{A} = 2 m$ aur $x_{B} = 5 m$ ke beech potential difference $V_{A} - V_{B}$ nikalo.

Recall Solution L2.3

KYA: uniform field ke liye $V_{A} - V_{B} = - \int_{B}^{A} E d x = - E (x_{A} - x_{B})$ . KYUN: field constant hai, isliye integral bas $E \times (displacement)$ hai; minus sign potential ki definition se aata hai. $V_{A} - V_{B} = - 300 (2 - 5) = - 300 \times (- 3) = 900 V .$ $A$ (chhota $x$ ) zyada potential par hai — field ke against move karne se $V$ badhta hai.

Level 3 — Analysis

Goal: kai charges ya kai ideas combine karo; symmetry aur signs subtle hote hain.

L3.1 Do charges $x$ -axis par rakhe hain: $q_{1} = + 3 μ C$ , $x = 0$ par aur $q_{2} = + 3 μ C$ , $x = 6 m$ par. (i) Midpoint $x = 3 m$ par potential $V$ nikalo, aur (ii) wahan field $E$ nikalo.

Figure — Potential of point charge, potential from field and vice versa

Recall Solution L3.1

Figure dekho: dono charges equal aur positive hain; midpoint (yellow dot) har ek se $3 m$ door hai. (i) Potential — scalars add karo: $V = k \frac{q _{1}}{r _{1}} + k \frac{q _{2}}{r _{2}} = 9 \times 1 0^{9} (\frac{3 \times 1 0 ^{- 6}}{3} + \frac{3 \times 1 0 ^{- 6}}{3})$ $= 9 \times 1 0^{9} \times 2 \times 1 0^{- 6} = 1.8 \times 1 0^{4} V .$ (ii) Field — vectors add karo: $q_{1}$ ka field $+ x$ direction mein point karta hai (red arrow, left charge se door push karta hai), $q_{2}$ ka field $- x$ direction mein (equal magnitude). Woh cancel ho jaate hain: $E = 0.$ Punchline yeh hai: usi point par $E = 0$ lekin $V \neq = 0$ . Field cancellation directions ke baare mein hai; potential sirf numbers ka ek plain sum hai jisme cancel karne ke liye koi direction nahi.

L3.2 Ab unhe opposite banao: $q_{1} = + 3 μ C$ , $x = 0$ par; $q_{2} = - 3 μ C$ , $x = 6 m$ par. Midpoint $x = 3 m$ par $V$ aur $∣ E ∣$ nikalo.

Recall Solution L3.2

Potential: $V = 9 \times 1 0^{9} (\frac{3 \times 1 0 ^{- 6}}{3} + \frac{- 3 \times 1 0 ^{- 6}}{3}) = 0 V .$ Dono signed contributions bilkul cancel ho jaati hain. Field: har charge ka field midpoint par $+ x$ direction mein point karta hai ( $+ q$ right push karta hai, $- q$ right pull karta hai), isliye woh add ho jaate hain: $E_{1} = E_{2} = \frac{k ∣ q ∣}{r ^{2}} = \frac{9 \times 1 0 ^{9} \times 3 \times 1 0 ^{- 6}}{3 ^{2}} = 3000 V/m,$ $∣ E ∣ = E_{1} + E_{2} = 6000 V/m .$ Punchline yeh hai: yahan $V = 0$ lekin $E \neq = 0$ — yeh L3.1 ka bilkul ulta hai. Yeh parent note ki headline mistake ko concrete banata hai.

Level 4 — Synthesis

Goal: field↔potential machinery chain karke koi naya result banao.

L4.1 Ek potential field hai $V (x, y) = 4 x^{2} - 3 y$ (volts, metres). Poora vector field $E$ nikalo, phir point $(1, 2)$ par evaluate karo. Magnitude bhi do.

Recall Solution L4.1

KYA: $E = - \nabla V = - (\frac{\partial V}{\partial x} \overset{x}{^} + \frac{\partial V}{\partial y} \overset{y}{^})$ . KYUN: $V$ do coordinates par depend karta hai, isliye humein gradient chahiye, single derivative nahi. Har partial derivative poochhti hai "agar main sirf is axis ke saath move karun, doosre ko fixed rakhte hue, toh $V$ kitni tezi se change hoti hai?"

$\frac{\partial V}{\partial x} = 8 x$ ( $y$ ko constant maano, isliye $- 3 y$ zero ho jaata hai).
$\frac{\partial V}{\partial y} = - 3$ ( $x$ ko constant maano, isliye $4 x^{2}$ zero ho jaata hai). $E = - (8 x) \overset{x}{^} - (- 3) \overset{y}{^} = - 8 x \overset{x}{^} + 3 \overset{y}{^} .$ $(1, 2)$ par: $E = - 8 \overset{x}{^} + 3 \overset{y}{^} V/m .$ Magnitude: $∣ E ∣ = (- 8)^{2} + 3^{2} = 64 + 9 = 73 \approx 8.54 V/m .$

L4.2 Consistency check. Point-charge potential $V (r) = \frac{k q}{r}$ se shuru karo, $E_{r} = - \frac{d V}{d r}$ compute karo, aur confirm karo ki Coulomb's field wapas milta hai. Phir, doosri direction mein jaate hue, us field ko $\infty$ se $r$ tak integrate karo aur confirm karo ki $V$ wapas milti hai.

Recall Solution L4.2

Forward (potential → field): $E_{r} = - \frac{d}{d r} (\frac{k q}{r}) = - k q \cdot (- \frac{1}{r ^{2}}) = \frac{k q}{r ^{2}} . ✓$ Yeh exactly Coulomb field hai. Back (field → potential): $V (r) = - \int_{\infty}^{r} \frac{k q}{r ^{'2}} d r^{'} = - k q [- \frac{1}{r ^{'}}]_{\infty}^{r} = - k q (- \frac{1}{r} + 0) = \frac{k q}{r} . ✓$ Baat yeh hai: derivative aur integral inverse operations hain, isliye $V \leftrightarrow E$ bridge bilkul do-taraf kaam karta hai (isliye field conservative hona chahiye — Conservative Fields and Curl dekho).

Level 5 — Mastery

Goal: poora multi-step reasoning, energy, aur ek limiting case.

L5.1 Teen charges ek square ke corners par hain jiska side $a = 1 m$ hai: $+ 2 μ C$ , $- 2 μ C$ , aur $+ 2 μ C$ (teen corners par ghoomte hue). Khaali chauthe corner par potential nikalo.

Recall Solution L5.1

Figure dekho: khaali corner ko $P$ label karo. Do charges adjacent corners par hain ( $P$ se $a$ door); ek charge diagonal corner par hai ( $P$ se $a 2$ door). Figure se do adjacent charges $+ 2 μ C$ aur $+ 2 μ C$ hain (har ek $a = 1$ par), aur diagonal charge $- 2 μ C$ hai ( $a 2 = 2$ par). Scalars add karo, har ek apni distance ke saath: $V = k (\frac{+ 2 \times 1 0 ^{- 6}}{1} + \frac{+ 2 \times 1 0 ^{- 6}}{1} + \frac{- 2 \times 1 0 ^{- 6}}{2}) .$ $V = 9 \times 1 0^{9} \times 1 0^{- 6} (2 + 2 - \frac{2}{2}) = 9 \times 1 0^{3} (4 - 2) .$ $V = 9000 \times (4 - 1.4142) = 9000 \times 2.5858 \approx 2.327 \times 1 0^{4} V .$ Diagonal distance $a 2$ kyun hai: side $a$ wale square ka diagonal $a^{2} + a^{2} = a 2$ hota hai (Pythagoras).

L5.2 Ek external agent ko kitna work karna hoga ek charge $Q = + 1 μ C$ ko infinity se L5.1 ke corner $P$ tak laane ke liye (kinetic energy mein koi change nahi)?

Recall Solution L5.2

KYA: work $= Q \times V$ jahan $V$ destination par potential hai. KYUN: definition se (parent §1) $V$ work per unit charge hai, isliye actually move kiye gaye charge se multiply karo. Yeh Potential Energy of Charge System se $U = Q V$ ke zariye connect hota hai. $W_{ext} = Q V = (1 \times 1 0^{- 6}) (2.327 \times 1 0^{4}) \approx 2.33 \times 1 0^{- 2} J .$ Positive work: hume positive $Q$ ko positive potential ke region ki taraf push karna padega.

L5.3 (limiting case) L3.2 wala dipole lo ( $+ q$ aur $- q$ jo $d$ se separated hain) aur perpendicular bisector par ek point dekho — dono ke beech symmetry ki line. Dikhao ki us line par har jagah $V = 0$ hai, aur check karo ki yeh consistent hai jab point infinity tak jaata hai.

Recall Solution L5.3

Geometry: perpendicular bisector par har point dono charges se equidistant hai — us distance ko $s$ kaho. Dono charges same distance par hain. Potential: $V = k \frac{+ q}{s} + k \frac{- q}{s} = \frac{k q}{s} - \frac{k q}{s} = 0.$ Har $s$ ke liye do equal-distance contributions cancel ho jaati hain, isliye poori bisector ek zero-potential (equipotential) plane hai. Limiting check: jab point door jaata hai ( $s \to \infty$ ), har term individually $\to 0$ hoti hai, isliye $V \to 0$ — consistent hai. Bisector ka $V = 0$ isliye nahi hai ki point door hai, balki exact cancellation ki wajah se hai; limit sirf confirm karta hai ki door kuch bhi anokha nahi hota. Degenerate case: agar dono charges equal hote ( $+ q, + q$ ), toh bisector deta $V = \frac{2 k q}{s} \neq = 0$ — cancellation poori tarah opposite signs par depend karta hai.

Recall wrap-up

Recall One-line answers (inhe cover karo)

$+ 5 μ C$ se $2 m$ par $V$ ? ::: $2.25 \times 1 0^{4} V$ .
Do equal $+$ charges ka midpoint: $E$ ? $V$ ? ::: $E = 0$ , $V \neq = 0$ .
$+ q, - q$ ka midpoint: $V$ ? $E$ ? ::: $V = 0$ , $E \neq = 0$ .
$V = 4 x^{2} - 3 y$ se $E$ ? ::: $- 8 x \overset{x}{^} + 3 \overset{y}{^}$ .
$Q$ ko potential $V$ par laane ka work? ::: $W = Q V$ .

Connections

Coulomb's Law — $1/ r^{2}$ field jo yahan har $E$ ke peeche hai.
Electric Field of Point Charge — L4.2 mein recover kiya gaya.
Potential Energy of Charge System — L5.2 mein $U = Q V$ use hua.
Equipotential Surfaces — L5.3 ka $V = 0$ plane.
Conservative Fields and Curl — kyun L4.2 ka bridge do-taraf kaam karta hai.
Gradient Operator — L4.1 ki machinery.