Electrostatic Potential and Capacitance: Class 12 Physics Chapter 2 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

170 min
2 simulations
20 practice questions
55 MCQs

Is chapter mein aap kya seekhenge

Samjho ki electrostatic force conservative kyun hai, aur infinity par zero lekar electrostatic potential energy aur potential define karo
Point charge, electric dipole, charges ke system aur charged spherical shell ke potentials derive karo aur use karo
Equipotential surfaces describe karo aur electric field ko us rate se jodo jis par potential badalta hai
Charges ke assembly ki, external potential mein rakhe charge ki aur uniform field mein dipole ki potential energy nikaalo
Conductors ki electrostatic properties, electrostatic shielding samet, batao aur unhe justify karo
Samjho ki polar aur non-polar dielectrics kaise polarise hote hain aur isse unke andar field kaise kamzor hota hai
Dielectric ke saath aur uske bina parallel plate capacitor ki capacitance derive karo, aur capacitors ko series aur parallel mein jodo
Capacitor mein stored energy aur electric field ki energy density calculate karo

NCERT Physics Class 12, Chapter 2, pp. 45–80 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

2.1 Introduction

Paani ki balti seedhiyon se chhat ki tank tak le jao, aur tumhari mehnat bekaar nahi jaati: woh wahin intezaar karti hai, aur jab paani pipe se kitchen ke tap tak neeche aata hai to wapas mil jaati hai. Class 11 mein tumne is stored energy ko gravitational potential energy kaha tha. Electric charges bhi bilkul yahi sauda dete hain. Ek positive charge ko doosre ki taraf dhakelo to tumhe unke repulsion ke against work karna padta hai; chhod do, aur charges ek doosre se door bhaagte hain, aur woh work kinetic energy ke roop mein wapas de dete hain.

Stored work wapas mil hi kyun sakta hai? Gravity ki ek khaas property hai: balti par uska work sirf is par depend karta hai ki balti aakhir mein kitni oonchi pahunchi, is par nahi ki tum kaunsi seedhi chadhe ya kitni baar ruke. Is property wale forces ko conservative kehte hain; stretched spring ka force bhi aisa hi hai. Rest par rakhe do charges ke beech ka force bhi isi family ka hai. Do masses ke beech gravity ke pull ki tarah, yeh bhi dono ko jodne wali line ke along lagta hai. Iska size sirf is par depend karta hai ki woh kitni door hain, aur is tarah bana har force conservative hota hai, jaisa Section 2.2 ka staircase argument dikhayega. Is chapter ki baaki har cheez isi ek property par tiki hai.

Is neev par chapter do ideas banata hai. Pehla hai electrostatic potential, jo charges ke kisi bhi arrangement ki energy ka hisaab numbers ke ek map mein badal deta hai. Doosra hai capacitor, ek device jo do conductors ke beech ke field mein charge, aur uske saath energy, store karta hai.

Balti uthane ka electrical version yeh raha (Fig. 2.1). Ek positive charge $Q$ origin par insulating stand par rakha hai. Hum ek bahut chhote positive charge $q$ ko andar ki taraf le jaate hain, R se shuru karke P par khatam, jo $Q$ ke zyada paas hai. Hum use itne aaram se chalate hain ki uski speed kabhi nahi badhti: har pal hamara push $𝐅_{ext}$ electric repulsion $𝐅_{E}$ jitna hi strong hota hai aur ulti direction mein point karta hai. Stand $Q$ ko apni jagah rakhta hai, aur $q$ itna chhota hai ki woh $Q$ ko waise bhi mushkil se kheenchta hai, isliye jis arrangement ke against hum push karte hain woh poore samay same rehta hai.

2/3External force $𝐅_{ext} = - 𝐅_{E}$ , $q$ ko bina speed badhaye andar le jaata hai; work potential energy $U$ ke roop mein jama hota jaata hai.

Fig. 2.1Repulsion ke against dheere-dheere kiya gaya work potential energy ke roop mein store hota hai; chhodne par charge use wapas kinetic energy mein badal deta hai.NCERT Fig. 2.1 · p. 46

Left mein ek bada positive charge Q stand par fixed hai. Ek chhota positive test charge q right mein door se shuru hota hai; chhote arrows bahar ki taraf electric force aur utna hi andar ki taraf external force dikhate hain. Neeche do horizontal bars potential energy U aur kinetic energy K track karte hain. Doosre step mein q ko dheere-dheere Q ke kaafi paas dhakela gaya hai: dono force arrows lambe hain aur U bar bada ho gaya hai jabki K abhi bhi zero hai. Teesre step mein q ko chhod diya gaya hai aur woh velocity arrow ke saath right ki taraf tezi se bhaag raha hai; U bar chhota ho gaya hai aur K bar utna hi bada ho gaya hai.

DerivationPotential energy ke roop mein stored work

$q$ par koi net force nahi hai, isliye uski kinetic energy kabhi nahi badalti, aur external agent ka kiya har joule work store ho jaata hai. R se P tak path ke along $𝐅_{ext} \cdot d 𝐫$ jodne par work milta hai

W_{R P} = \int_{R}^{P} 𝐅_{ext} \cdot d 𝐫 = - \int_{R}^{P} 𝐅_{E} \cdot d 𝐫

(2.1)

Is beech electric force utna hi aur ulta work, $- W_{R P}$ , karta hai. Hum $q$ ki electrostatic potential energy mein badhat ko isi stored work ke roop mein define karte hain:

Δ U = U_{P} - U_{R} = W_{R P}

(2.2)

Dekho ki Eq. (2.2) kya fix karti hai aur kya khula chhodti hai. Yeh sirf $U$ mein change fix karti hai, aur woh change sirf end points R aur P par depend karta hai: dono ke beech har route par work same hai (kyun, yeh tum Fig. 2.2 mein dekhoge), aur isi wajah se hum har point ko ek number de paate hain. Jo khula rehta hai woh ek constant hai: har point par $U$ mein $5 J$ , ya koi bhi amount, jod do, har difference same rahega.

Isliye hum $U$ ka zero jahan sabse convenient ho wahan chun sakte hain. Kisi finite region mein bandhe charges ke liye sabse seedhi jagah bahut door hai, jahan woh ek doosre ko bilkul affect nahi karte. $U_{\infty} = 0$ lekar aur R ko infinity tak bahar le jaakar,

W_{\infty P} = U_{P} - U_{\infty} = U_{P}

(2.3)

Shabdon mein: P par $q$ ki potential energy woh work hai jo external agent ko $q$ ko infinity se dheere-dheere P tak laane mein karna padta hai, jabki baaki charges apni jagah rakhe rehte hain.

Definition

Electrostatic potential energy

Kisi charge ko bina accelerate kiye infinity se diye gaye point tak laane mein external agent dwara kiya gaya work, jabki baaki saare charges fixed rakhe hon. Sirf iske differences ka physical matlab hota hai.

ExampleLike charge ke paas charge ke liye yeh positive hoti hai aur unlike charge ke paas negative.

2.2 Electrostatic Potential

Eq. (2.2) mein work $q$ ke saath-saath badhta hai: test charge double karo to us par har force $q 𝐄$ double ho jaata hai, aur work bhi. Visitor ki jagah source charges ke arrangement ko describe karne ke liye $q$ se divide karo. Gravity ke liye yeh role height nibhaati hai: uthaye gaye object ki energy $m g h$ uske mass par depend karti hai, lekin height $h$ jagah ki property hai. Height ka electrical counterpart hai electrostatic potential $V$ :

Potential difference

V_{P} - V_{R} = \frac{U_{P} - U_{R}}{q} = \frac{W_{R P}}{q}

(2.4)

Eq. (2.4) do points ko compare karti hai; akele kisi ek ke baare mein kuch nahi kehti. Har point ko apna number dene ke liye hume ek reference level chahiye, jaise heights ke liye sea level hota hai. Finite region mein bandhe charges ke liye natural reference infinity hai, jahan unka asar khatam ho chuka hota hai. $V_{\infty} = 0$ lekar, kisi point par potential woh work per unit positive charge hai jo test charge ko infinity se wahan laane mein lagta hai.

Definition

Electrostatic potential

Test charge ko bina acceleration ke infinity se kisi point tak laane mein external agent dwara per unit positive test charge kiya gaya work. Iski SI unit volt hai: $1 V = 1 J / C$ .

ExampleTorch cell apne positive terminal ko negative terminal se lagbhag $1.5 V$ upar rakhta hai, isliye unke beech guzarne wala har coulomb $1.5 J$ ka len-den karta hai.

Ek technical baat: asli test charge source charges ko thoda sa apni jagah se kheench dega. Bahut hi chhota test charge $δ q$ use karke, aur uske liye zaroori work $δ W$ ko $δ q$ se divide karke, yeh chinta khatam ho jaati hai aur answer nahi badalta.

Kyunki force conservative hai, $V_{P} - V_{R}$ dono points ke beech har route ke liye same hai. Fig. 2.2 ek single point charge ke liye iski wajah dikhata hai. Kisi bhi route ki nakal, jitni paas chaho utni, charge par centred chhote radial steps aur chhote arcs ki staircase se ki ja sakti hai. Arc ke along force motion ke right angle par hota hai aur koi work nahi karta, isliye sirf radial steps count hote hain, aur woh hamesha andar ki taraf ek hi trip jitne jud jaate hain.

3/3Kisi bhi path ko arcs (koi work nahi: force motion ke right angle par) aur radial steps mein todo, jo hamesha same total tak jud jaate hain.

Fig. 2.2R se P tak route kuch bhi ho, work per unit charge,

V_{P} - V_{R}

, same rehta hai.NCERT Fig. 2.2 · p. 48

Ek positive charge Q neeche left mein hai, aur uske chaaron taraf usi par centred halke circular arcs hain. Point R right mein door hai aur point P upar left mein zyada paas. Pehle step mein ek seedha path R ko P se jodta hai, jis par work W likha hai. Doosre step mein ek tedha-medha orange path unhi points ke beech neeche jaakar wapas upar aata hai, jis par same work likha hai. Teesre step mein ek zigzag path seedhi line ke saath-saath chalta hai: yeh circles ke along chhote arcs se bana hai, jinmein koi work nahi lagta, aur Q ki taraf chhote radial steps se, jo saara work uthate hain.

Potential difference woh hai jo voltmeter padhta hai aur jo cell ya socket ke label par likha hota hai. Torch cell lagbhag $1.5 V$ deta hai, car battery $12 V$ aur India mein ghar ka socket $230 V$ (ek alternating value, jaisa tum Chapter 7 mein dekhoge). Thundercloud aur zameen ke beech potential difference dasiyon crore volts tak pahunch sakta hai.

History seNCERT p. 47

Alessandro Volta (1745–1827)

20 March 1800 ko London ki Royal Society ko likhe ek letter mein Alessandro Volta ne ek naye device ka description diya. Yeh zinc aur copper (ya silver) ki baari-baari rakhi discs ka ek column tha, jinke beech namak ke paani mein bhigoe kapde ke pads the. Yeh column apne across jude kisi bhi cheez mein lagaataar charge dhakelta tha: yeh pehli battery thi, voltaic pile. Isne ek lambi behes bhi khatam kar di. Unke hamwatan Luigi Galvani maante the ki phadakti frog legs ek 'animal electricity' dikhati hain jo living tissue banata hai. Volta ka kehna tha ki geele tissue ko chhoone wali do alag metals hi asli source hain, aur unke pile ne, bina kisi frog ke, yeh baat saabit kar di. Baad mein Napoleon ne unhe count banaya, aur potential difference ki unit unke naam par hai.

Quick check

3 mein se 1

+ 2.0 μ C

charge ko point R se point P tak dheere-dheere le jaane mein

6.0 m J

work lagta hai.

V_{P} - V_{R}

kitna hai, volts mein?

Answer dekho

Answer 3000 V

V_{P} - V_{R} = W / q = (6.0 \times 10^{- 3} J) / (2.0 \times 10^{- 6} C) = 3.0 \times 10^{3} V

2.3 Point Charge ka Potential

Ab ek asli calculation karte hain. Origin par ek point charge $Q$ rakho aur usse $r$ doori par point P par $V$ poochho. Route se fark nahi padta, isliye sabse aasaan chuno: infinity se P tak radius ke along seedhe andar.

DerivationPoint charge ka potential

$Q$ se $r^{'}$ doori par ek unit positive test charge Coulomb force mehsoos karta hai

𝐅 = \frac{1}{4 π ε_{0}} \frac{Q}{r^{' 2}} {\hat{𝐫}}^{'}

(2.5)

jahan ${\hat{𝐫}}^{'}$ , $Q$ se door point karta hai. Test charge ko ek chhote radial step $Δ r^{'}$ se hilao. Andar aate hue $r^{'}$ ghatta hai, isliye $Δ r^{'}$ negative hai. External agent $- 𝐅$ se push karta hai, aur step par uska work hai

Δ W = - \frac{Q}{4 π ε_{0} r^{' 2}} Δ r^{'}

(2.6)

jo $Q > 0$ ke liye positive hai, jaisa hona chahiye. $r^{'} = \infty$ se $r^{'} = r$ tak steps jodne par sum ek integral ban jaata hai:

W = - \int_{\infty}^{r} \frac{Q}{4 π ε_{0} r^{' 2}} d r^{'} = {[\frac{Q}{4 π ε_{0} r^{'}}]}_{\infty}^{r} = \frac{Q}{4 π ε_{0} r}

(2.7)

Yeh work per unit charge, definition se, P par potential hai:

Point charge ka potential

V (r) = \frac{1}{4 π ε_{0}} \frac{Q}{r}

(2.8)

Eq. (2.7) ka ek badhiya graphical matlab bhi hai: $V (r)$ , distance ke against $E$ ke graph ke neeche ka area hai, $r$ se lekar infinity tak.

Agar $Q$ negative ho, jaise sookhe baalon par ragde gaye balloon ka charge, to? Tab Eq. (2.8) har distance par $V < 0$ deti hai. Socho ki unit test charge balloon ki taraf kheencha ja raha hai. Use andar bhaagne se rokne ke liye tumhare haath ko poore raaste use peeche pakde rakhna padta hai, charge andar jaata hai jabki haath bahar ki taraf push karta hai, aur jo force motion ka virodh karta hai woh negative work karta hai. Point jitna paas, uska potential utna hi zyada negative.

Fig. 2.3 mein compare karo ki $V$ aur $E$ distance ke saath kaise fade hote hain. Field $1 / r^{2}$ ki tarah girta hai, potential sirf $1 / r$ ki tarah. Distance double karo to $V$ aadha ho jaata hai, jabki $E$ ek chauthai reh jaata hai.

Fig. 2.3Dono distance ke saath fade hote hain, lekin field tezi se fade hota hai: double distance par

V

aadha hota hai jabki

E

ek chauthai.NCERT Fig. 2.4 · p. 49

Ek positive point charge Q distance axis ke left end par hai, jis par r0, 2r0, 3r0 aur 4r0 mark hain. Har distance par bars ka ek pair potential, ek solid blue bar, ko field, ek hatched orange bar, se compare karta hai; dono r0 par 1 ke scale par hain, jahan ek dashed line unki common height mark karti hai. 2r0 par potential bar aadha lamba hai lekin field bar sirf ek chauthai. 3r0 par bars ek tihaai aur ek nauvaan hissa hain, aur 4r0 par ek chauthai aur ek solahvaan hissa, isliye field bars potential bars se kahin tezi se chhote hote hain.

Example 2.1NCERT Example 2.1 · p. 49

Charged sphere ke paas ek dust grain

Insulating stand par rakhe ek chhote metal sphere par $+ 6.0 n C$ hai; bahar se yeh apne centre par rakhe point charge ki tarah behave karta hai. $- 3.0 n C$ wala ek dust grain door se behta hua aata hai aur centre se $12 c m$ door point P par pahunchta hai. (a) P par potential kya hai? (b) Andar aate hue grain ki potential energy kitni badalti hai? Field grain ki speed badhata hai ya ghatata hai? (c) Phir grain ko sphere par centred $12 c m$ radius ke circle par ek baar dheere-dheere ghumaya jaata hai, aur aakhir mein $24 c m$ tak bahar le jaaya jaata hai. In dono stages mein agent kitna work karta hai? $1 / 4 π ε_{0} = 9.0 \times 10^{9} N m^{2} C^{- 2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.4 Electric Dipole ka Potential

Chapter 1 ka electric dipole yaad karo: $\pm q$ charges ka ek pair jo ek chhoti distance $2 a$ par rakha hai. Ise uske dipole moment $𝐩$ se describe karte hain, jiska size $2 a q$ hai aur jo negative charge se positive charge ki taraf jaata hai. Iska field nikaalne mein thodi mehnat lagi thi, kyunki field vectors ko head to tail jodna padta hai. Potentials sign wale simple numbers hain, isliye superposition bas addition hai, aur dipole ka potential sirf kuch lines mein nikal aata hai.

2/3Door se dekhne par $r_{2}$ , $r_{1}$ se OP par dipole ke projection jitna bada hai: $r_{2} - r_{1} \approx 2 a \cos θ$ .

Fig. 2.4Dipole se door, P tak ki dono distances mein

2 a \cos θ

ka fark hota hai; potential

+ q

wali side positive,

- q

wali side negative aur equatorial plane par zero hai.NCERT Fig. 2.5 · p. 50

Ek vertical dipole ke upar +q aur neeche -q hai, centre O hai aur dipole moment p upar point karta hai. Ek point P upar right mein hai. Pehle step mein lines r1 aur r2, P ko dono charges se jodti hain, ek dashed line r, O ko P se jodti hai, aur p aur OP ke beech angle theta mark hai. Doosre step mein dono charges se line OP par giraye gaye perpendiculars 2a cos theta lambai ka ek highlighted segment alag karte hain: +q ke comparison mein -q se P tak ki extra distance. Teesre step mein plane ka upar wala aadha hissa tinted hai aur us par V zero se bada likha hai, neeche wale par V zero se chhota, aur O se guzarti horizontal equatorial line par V barabar zero likha hai.

DerivationDoor se dipole ka potential

Positions ko dipole ke midpoint O se naapo. Point P, position $𝐫$ par, $+ q$ se $r_{1}$ aur $- q$ se $r_{2}$ door hai (Fig. 2.4). Dono point-charge potentials jodne par,

V = \frac{1}{4 π ε_{0}} (\frac{q}{r_{1}} - \frac{q}{r_{2}}) = \frac{q}{4 π ε_{0}} \frac{r_{2} - r_{1}}{r_{1} r_{2}}

(2.9)

Cosine rule dono distances exactly deta hai:

r_{1}^{2} = r^{2} + a^{2} - 2 a r \cos θ, r_{2}^{2} = r^{2} + a^{2} + 2 a r \cos θ

(2.10)

Dipole se door ( $r ≫ a$ ) dono charges se P tak ki lines lagbhag parallel hoti hain. $a / r$ mein sirf first order ke terms rakhne par, har distance $r$ se line OP par aadhe dipole ke projection $a \cos θ$ jitni alag hoti hai:

r_{1} \approx r - a \cos θ, r_{2} \approx r + a \cos θ

(2.11)

(Yahi result tum Eq. (2.10) se chhote $x$ ke liye binomial approximation $\sqrt{1 - x} \approx 1 - x / 2$ lekar bhi pa sakte ho.) Isliye hume jo difference aur product chahiye woh hain

r_{2} - r_{1} \approx 2 a \cos θ, r_{1} r_{2} \approx r^{2}

(2.12)

Eq. (2.9) mein substitute karke aur $p = 2 q a$ likhkar,

V = \frac{q}{4 π ε_{0}} \frac{2 a \cos θ}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{p \cos θ}{r^{2}}

(2.14)

Kyunki $p \cos θ = 𝐩 \cdot \hat{𝐫}$ , jahan $\hat{𝐫}$ , O se P ki taraf unit vector hai, result ek compact vector form le leta hai:

Dipole ka potential

V = \frac{1}{4 π ε_{0}} \frac{𝐩 \cdot \hat{𝐫}}{r^{2}} (r ≫ a)

(2.15)

Approximation kitna achha hai? Jo terms humne chhod diye woh rakhe gaye term se lagbhag $(a / r)^{2}$ factor se chhote hain. Jaise axis par exact result $V = k p / (r^{2} - a^{2})$ hai, isliye das dipole lengths ki distance ( $r = 20 a$ ) par approximate formula sirf chauthai percent galat hai. Ek ideal point dipole ke liye, jise $p$ fixed rakhte hue zero size tak chhota kar diya jaaye, yeh exact hai.

Do directions ko khaas dhyaan se dekhna chahiye. Axis par $θ = 0$ ya $π$ , isliye $\cos θ = \pm 1$ :

V = \pm \frac{1}{4 π ε_{0}} \frac{p}{r^{2}}

(2.16)

$+ q$ wali side plus sign ke saath aur $- q$ wali side minus sign ke saath. Equatorial plane par $θ = π / 2$ aur $V = 0$ : wahan har point $+ q$ aur $- q$ se barabar door hai. Single charge ke saath rakhkar dekho to dipole potential teen tarah se alag dikhta hai:

Yeh direction par depend karta hai. Dipole ke chaaron taraf fixed distance par ghoomo to $V$ badalta hai, $+ q$ end ke aage positive se, side mein zero se hote hue, $- q$ end ke aage negative tak. Sirf axis se angle $θ$ matter karta hai, isliye pattern axis ki har side same hai, jaise stand ke chaaron taraf lampshade ki shape.
Yeh tezi se fade hota hai. Dipole potential $1 / r^{2}$ ki tarah girta hai, jabki single charge ka $1 / r$ ki tarah, kyunki door se $+ q$ aur $- q$ lagbhag cancel ho jaate hain. (Field ek power aur tezi se, $1 / r^{3}$ ki tarah, girta hai.)
Zero potential ka matlab zero field nahi hai. Equatorial plane par $V = 0$ , phir bhi Chapter 1 ne wahan $p / 4 π ε_{0} r^{3}$ size ka field diya tha, jo $𝐩$ ke ulta point karta hai. Field is par depend karta hai ki $V$ point se point kaise badalta hai, uski value par nahi (Section 2.6.1).

Example

Water molecule ke paas potential

Water molecule ka permanent dipole moment lagbhag $6.2 \times 10^{- 30} C m$ hai. Ise point dipole maankar woh potential nikaalo jo yeh banata hai (a) apne axis ke along $1.0 n m$ door, positive side par; (b) utni hi distance par, lekin axis se $60^{\circ}$ par; (c) axis ke along $2.0 n m$ door. (a) aur (c) ko compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Electric dipole ke equatorial plane ke kisi point par kya sahi hai?

V = 0

aur

E = 0

V \neq 0

aur

E = 0

V \neq 0

aur

E \neq 0

V = 0

lekin

E \neq 0

Answer dekho

Answer (D) $V = 0$ lekin $E \neq 0$

Point

+ q

aur

- q

se barabar door hai, isliye unke potentials cancel ho jaate hain. Unke fields nahi hote: dono ke components

+ q

wali side se

- q

wali side ki taraf point karte hain, jo jud jaate hain.

2.5 Charges ke System ka Potential

Superposition general case ko aasaan bana deta hai. Collection ka har charge P par apna point-charge potential deta hai, aur kyunki potentials sign wale numbers hain, total unka algebraic sum hai. P se $r_{1 P}, r_{2 P}, \dots, r_{n P}$ distances par rakhe charges $q_{1}, q_{2}, \dots, q_{n}$ ke liye,

V = V_{1} + V_{2} + \dots + V_{n}

(2.17)

Charges ke system ka potential

V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1 P}} + \frac{q_{2}}{r_{2 P}} + \dots + \frac{q_{n}}{r_{n P}}) = \frac{1}{4 π ε_{0}} \sum_{i = 1}^{n} \frac{q_{i}}{r_{i P}}

(2.18)

Na angles, na components: positive charge $V$ mein judta hai aur negative ghatta hai, chahe woh kisi bhi direction mein ho. Unhi charges ke liye $𝐄$ nikaalne mein har contribution ko pehle components mein todna padta, isliye aksar pehle $V$ nikaalna faayde ka hota hai.

Jo charge phaila hua hai, kisi volume mein ya surface par, use bhi isi tarah handle karte hain. Use chhote-chhote tukdon mein baanto, har tukda itna chhota ki P se apni distance $r$ par point charge $d q$ maana ja sake. Unke contributions $d q / 4 π ε_{0} r$ jodo; bahut hi chhote tukdon ke liye sum ek integral ban jaata hai. Sirf distances aati hain, directions kabhi nahi, isliye phaile hue charge ka potential aksar uske field se kahin aasaani se nikal aata hai.

Chapter 1 ne radius $R$ aur charge $q$ wale patle spherical shell ke liye mushkil kaam pehle hi kar diya hai. Gauss's law ne dikhaya ki iska bahar ka field centre par rakhe point charge $q$ ke field se bilkul alag nahi kiya ja sakta. Jo do fields P se infinity tak har jagah match karte hain unhe same work per unit charge chahiye, isliye shell ke bahar

V = \frac{1}{4 π ε_{0}} \frac{q}{r} (r \geq R)

(2.19(a))

Shell ke andar field zero hai. Test charge wahan bina kisi work ke ghoom sakta hai, isliye andar ke har point par $V$ ki wahi value hai jo shell par hai:

V = \frac{1}{4 π ε_{0}} \frac{q}{R} (r \leq R)

(2.19(b))

Fig. 2.5Charged shell ki surface ke paar field jump karta hai, lekin potential continuous rehta hai.NCERT p. 52

Do graphs aamne-saamne hain jinmein ek shaded strip radius R tak shell ke andar ka hissa mark karti hai. Left graph mein field andar zero hai, R par jump karke kq over R squared ho jaata hai aur phir 1 over r squared ki tarah girta hai. Right graph mein potential andar kq over R par flat hai aur bahar 1 over r ki tarah smoothly girta hai, R par koi jump nahi.

Example 2.2NCERT Example 2.2 · p. 52

Potential kahan zero hai?

$+ 4.0 n C$ ka charge origin par hai aur $- 1.0 n C$ ka charge $x$ -axis par $x = 30 c m$ par hai (Fig. 2.6). Infinity par $V = 0$ lekar $x$ -axis par woh saari jagahein dhoondho jahan $V$ zero ho jaata hai. Kya un jagahon par electric field bhi zero hota hai?

Fig. 2.6Dono charges ki line ke along

V (x)

do baar zero cross karta hai: unke beech, aur chhote charge ke aage.NCERT Fig. 2.7 · p. 52

Centimetres mein marked ek horizontal x-axis par 0 par +4.0 nC charge aur 30 cm par -1.0 nC charge hai. Ek blue curve axis ke along potential V dikhata hai: positive charge ke paas bada aur positive, negative charge ke paas bade negative values tak gir jaata hai, aur uske aage phir badhkar right mein door ek chhota positive ubhaar banata hai. Curve do points par zero cross karta hai, jo 24 cm aur 40 cm par circles se mark hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractivePotential probe

Ek interactive stage jismein ek horizontal line par 30 cm door do point charges hain, jinki values tum 1 nC ke steps mein badal sakte ho. Ek draggable probe apni position par potential volts mein aur field strength newtons per coulomb mein dikhata hai, saath mein field ki direction ke liye ek arrow aur un saare points se guzarta ek highlighted curve jinka potential probe jitna hai. Stage ke neeche ek graph charges se guzarti line ke along potential plot karta hai aur mark karta hai ki yeh kahan zero hai. Preset buttons Example 2.2 ke charges, ek dipole aur do barabar positive charges load karte hain.

Fig. 2.7

V

aur

E

padhne ke liye probe ko drag karo. Highlighted curve probe se guzarta equipotential hai; graph charges ki line ke along

V

dikhata hai.

Try karke dekho

Potential probe ke saath explore karo

Example 2.2 load karo aur Probe to axis dabao. Probe ko charges ke beech axis ke along chalao aur $V$ ka sign badalte dekho: woh crossing pehla zero hai (precise reading ke liye arrow keys probe ko $0.5 c m$ ke steps mein chalati hain). Probe kisi charge ke aar-paar nahi ja sakta, isliye $- 1.0 n C$ charge ke aage axis par tap karo, ya probe ko uske upar se utha kar le jao, aur doosra zero dhoondho. Kya tumhari positions $24 c m$ aur $40 c m$ se match karti hain?
Probe ko axis se hata kar drag karo aur highlighted equipotential dekho: field arrow hamesha use right angle par cross karta hai.
Like charges load karo aur probe ko unke theek beech mein rakho. Wahan field readout zero ho jaata hai (precise position ke liye arrow keys use karo), lekin potential nahi. Ab Dipole load karo: uske midpoint par $V = 0$ lekin $E \neq 0$ .

Example 2.3NCERT Example 2.3 · p. 53

Charged sphere ke paas ek electron aur ek proton

$+ 5.0 n C$ wala ek chhota metal sphere evacuated chamber ke andar O par fixed hai. Point N, O se $3.0 c m$ door hai aur point F, O se $9.0 c m$ door. (a) Kaunsa point higher potential par hai, aur kitna? (b) Ek electron ko F par rest se chhoda jaata hai. Woh kis taraf chalta hai, field us par positive work karta hai ya negative, aur N se guzarte waqt uski kinetic energy kitni hai? (c) F aur N ke beech electron ki potential energy ka kya hota hai? (d) Ek proton ko F se N tak dheere-dheere le jaana hai. Agent ko use push karna padega ya peeche rokna padega, aur agent ka work positive hai ya negative? (e) Agar phir proton ko N par chhod diya jaaye, to uski kinetic energy badhti hai ya ghatti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

+ 2.0 n C

+ 2.0 n C

aur

- 3.0 n C

ke charges har ek point P se

10 c m

door, teen alag directions mein rakhe hain. P par potential kitna hai, volts mein? (

k = 9.0 \times 10^{9} N m^{2} C^{- 2}

lo.)

Answer dekho

Answer 90 V

V

ke liye directions matter nahi karti:

V = k (2.0 + 2.0 - 3.0) n C / 0.10 m = (9.0 \times 10^{9}) (1.0 \times 10^{- 9}) / 0.10 = 90 V

2.6 Equipotential Surfaces

Western Ghats mein trekkers contour lines wale maps rakhte hain, har line sea level se same height wale points ko jodti hai. Jahan lines paas-paas hoti hain wahan dhalaan steep hoti hai; jahan door-door hoti hain wahan zameen halki dhalaan wali hoti hai. Potential bhi bilkul aisa hi map banane ka mauka deta hai.

Definition

Equipotential surface

Aisi surface jiske har point par electrostatic potential same ho.

ExampleSingle point charge ke liye, charge par centred koi bhi sphere.

Akele point charge ke liye map banana aasaan hai. $V = Q / 4 π ε_{0} r$ sirf distance par depend karta hai, isliye har equipotential surface charge par centred ek sphere hai, aur yeh saare pyaaz ki parton ki tarah ek ke andar ek hote hain. Field lines radii hain, isliye har parat ko seedhe, right angle par chhedti hain. Yeh right angle wali crossing ek general law nikalti hai, is simple case ka sanyog nahi:

PropertyField equipotentials ke normal hota hai

Field lines equipotential surfaces ko hamesha right angle par cross karti hain, aur higher se lower potential ki taraf chalti hain.

Maan lo kisi equipotential surface ke point S par field ka surface ke along ek component $E_{∥}$ hota (Fig. 2.8). Ek chhote charge $q$ ko surface ke along, $E_{∥}$ ki direction mein, chhoti distance $δ l$ tak le jao. Tab field us par $q E_{∥} δ l$ work karta. Lekin conservative field ke liye yeh work potential energy ki girawat, $- q δ V$ , ke barabar hona chahiye, aur $δ V = 0$ kyunki step ke dono ends same equipotential par hain. Isliye $E_{∥} = 0$ : field ka surface ke along koi component nahi hai aur woh uske normal hai. Yeh us taraf point karta hai jidhar positive charge potential energy khota, yaani lower $V$ ki taraf.

3/3Isliye $𝐄$ har equipotential se right angle par milta hai, lower potential ki taraf point karte hue.

Fig. 2.8Equipotential ke along charge le jaane mein koi work nahi lagta, isliye field ka uske along koi component nahi ho sakta.NCERT Fig. 2.9 · p. 54

Teen dashed equipotential curves upar se neeche chalte hain, left se right 60 V, 40 V aur 20 V likhe hain, aur 40 V curve par point S par ek chhota positive test charge hai. Pehle step mein curve ke along ek arrow charge ko us par slide hote dikhata hai, jis par W barabar zero likha hai. Doosre step mein S par ek hypothetical tilted field ka dashed arrow do hisson mein toda gaya hai, ek curve ke across aur ek uske along; along wala hissa work karta. Teesre step mein S par field arrow curve ko right angle par cross karta hai, jo ek chhote square se mark hai, aur kai aur field arrows teeno curves ko right angle par cross karte hain, 60 V se 20 V ki taraf point karte hue.

Uniform field sabse simple pattern deta hai. Potential $𝐄$ ki direction mein lagaataar girta hai aur side mein nahi badalta, isliye iske equipotentials field ke across khadi flat, parallel sheets hain, jo $V$ ke barabar steps ke liye barabar door hoti hain. Yahi pattern tumhe capacitor ki plates ke beech milega (Section 2.12). Dipole ke, ya like charges ke pair ke, har charge ke paas equipotentials chhote lagbhag-spheres hote hain; aur door jaakar yeh bade closed surfaces mein mil jaate hain. Neeche ka simulation inmein se kai patterns unki field lines ke saath dikhata hai.

Scroll karke explore karo

Kuch arrangements ke equipotentials

Stage par point charges, unki field lines aur dashed equipotential curves ke saath, jo potential ki kuch fixed values par bane hain: positive potential ke liye green, negative ke liye orange aur zero ke liye grey. Scroll karte hue stage ek positive charge, ek negative charge, opposite charges ki do rows, ek dipole aur do barabar positive charges dikhata hai. Tum charges ko drag kar sakte ho aur field lines, field vectors aur equipotentials ko on ya off kar sakte ho.

Ek positive charge. Dashed equipotentials circles hain, three dimensions mein spheres, jo charge par centred hain; yeh green hain kyunki $V > 0$ . Field lines har circle ko right angle par cross karti hain. Charge ke paas circles bheed laga lete hain, jahan field sabse strong hai.

Ek negative charge. Wahi circles, ab orange kyunki potential har jagah negative hai. Field lines andar ki taraf point karti hain, higher (kam negative) se lower (zyada negative) potential ki taraf.

Opposite charges ki do rows. Rows ke theek beech equipotentials seedhe hokar lagbhag flat, parallel lines ban jaate hain, aur field lines unke across seedhi chalti hain. Yeh charged capacitor ka ek rough model hai, jiska plates ke beech ka field uniform hota hai (Section 2.12).

Ek dipole. $+ q$ aur $- q$ ke theek beech grey dashed line zero-potential surface hai, yaani equatorial plane. Positive equipotentials $+ q$ end ke chaaron taraf lipte hain aur negative wale $- q$ end ke chaaron taraf.

Do barabar positive charges. Har charge ke paas equipotentials chhote circles hain; door jaakar yeh mil kar ek peanut jaisa curve bana lete hain jo dono ko gherta hai. Charges ke theek beech dono fields cancel ho jaate hain, isliye us point se koi field line nahi guzarti, phir bhi wahan potential positive hai: do positive contributions jud jaate hain.

Chalo explore karein

Namak ke paani ki tray mein equipotentials ka map banao

Tumhe chahiye

graph paper ki sheet par rakhi ek uthli transparent plastic tray
namak wala paani
electrodes ke liye aluminium foil ki do strips aur ek coin
crocodile-clip leads ke saath ek 9 V battery
DC volts par set ek digital multimeter

Pehle guess karo

Tray ke aamne-saamne ends par do lambi strip electrodes ke saath, barabar voltage wali lines ki shape kaisi hogi?

har strip ke chaaron taraf circlesdono strips ko jodti seedhi linesstrips ke parallel seedhi lineskoi regular pattern nahi

Graph paper ko transparent tray ke neeche rakho aur usmein lagbhag $1 c m$ gehra namak ka paani daalo. Tray ke har chhote end ke along ek foil strip khadi karo aur strips ko battery se jodo.
Multimeter ki black lead ko negative strip se clip karo. Red lead ki tip ko paani mein kisi bhi point par dubao aur reading note karo.
Paanch aise points dhoondho jo sab $3.0 V$ padhte hon, unhe graph paper par mark karo aur ek smooth line se jodo. $4.5 V$ aur $6.0 V$ ke liye dohrao.
Ek strip ki jagah apne edge par khada ek coin rakho (ek chhota, lagbhag point jaisa electrode) aur wahi teen voltages phir se map karo.

Table 2.1Tumhara equipotential map

Equipotential	Do strips ke saath shape	Strip aur coin ke saath shape
3.0 V
4.5 V
6.0 V

Tumhari entries apne aap save hoti hain.

Do strips ke saath barabar voltage wali lines seedhi aati hain, electrodes ke parallel aur barabar door: unke beech field lagbhag uniform hai. Coin ke saath lines uske chaaron taraf mudti hain aur uske paas bheed laga leti hain, jahan field sabse strong hai. Har point par, jis direction mein reading sabse tezi se badalti hai woh tumhari lines ke right angle par hai, bilkul jaise $𝐄$ equipotentials ke right angle par hota hai. (Meter paani mein bahut chhota current chalata hai, lekin potentials ka pattern electrostatic wale se match karta hai.)

Abhi nahi kar sakte? Simulation try karo · Fig. 2.11

2.6.1 Field aur potential ka relation

Contour-map wali picture pehle hi $𝐄$ aur $V$ ka link batati hai: jahan contours bheed lagate hain wahan steep zameen, jahan equipotentials bheed lagate hain wahan strong field. Fig. 2.10 ise quantitative banata hai.

3/3A par $10 V$ ki girawat $1 c m$ mein hoti hai, isliye $E \approx 1000 V / m$ . B par isme $4 c m$ lagte hain, isliye $E \approx 250 V / m$ .

Fig. 2.10Field wahan sabse strong hai jahan equipotentials sabse paas hain:

E \approx ∣ Δ V ∣ / Δ l

, surfaces ke right angle par naapa hua.NCERT Fig. 2.12 · p. 55

50, 40, 30, 20 aur 10 volts likhi curved equipotential lines upar se neeche chalti hain, left mein paas-paas aur right ki taraf badhti doori par; ek scale bar 1 cm dikhata hai. Doosre step mein blue field arrows lines ko right angle par cross karte hain, 50 V se 10 V ki taraf point karte hue; yeh left mein sabse lambe hain jahan lines bheed lagati hain aur right mein sabse chhote. Teesre step mein do measurement bars point A ko, 1 cm door 50 V aur 40 V lines ke beech, aur point B ko, 4 cm door 20 V aur 10 V lines ke beech, mark karte hain; ek panel par likha hai A par E lagbhag 1000 V/m aur B par E lagbhag 250 V/m.

DerivationEquipotentials ki spacing se field

Ek unit positive charge ko kisi aisi direction mein chhoti distance $δ l$ tak le jao jo $𝐄$ se angle $θ$ banati hai. Field us par $E δ l \cos θ$ work karta hai, isliye uska potential $δ V = - E δ l \cos θ$ badalta hai. Do directions khaas hain.

Agar step equipotential ke along hai, to $δ V = 0$ , isliye $\cos θ = 0$ : field surface ke right angle par hai, jaisa humne upar paaya. Agar iske bajaye step seedhe field ke along ( $θ = 0$ ) liya jaaye, ek equipotential se agle tak, to potential jitni tezi se ho sake girta hai, aur

δ V = - ∣ 𝐄 ∣ δ l \Rightarrow ∣ 𝐄 ∣ = - \frac{δ V}{δ l}

(2.20)

Yahan $δ l$ do padosi equipotentials ke beech ki perpendicular distance hai aur $δ V$ , jo field ki direction mein negative hai, unke potentials ka difference hai. Isliye field ka size hai

Potential se field

E = \frac{∣ δ V ∣}{δ l}

(2.21)

Kis taraf: dhalaan ki taraf. $𝐄$ , $V$ ki sabse steep girawat ke along point karta hai, jo hamesha equipotentials ke seedhe across chalti hai.
Kitna strong: jitni steep dhalaan, utna strong. $E$ per metre $V$ ki girawat hai, equipotentials ke seedhe across naapi hui, isliye jahan $V$ ke barabar steps bheed lagate hain wahan field strong hai.

Kisi bhi seedhi line ke along, maan lo $x$ -axis, yahi reasoning us line ke along field ka component deti hai: $E_{x} = - d V / d x$ . Field component potential ka dhalaan wala slope hai. Isse unit $V / m$ bhi samajh aati hai, jo tumhe $N / C$ jitni hi baar dikhegi: dono same hain, kyunki $1 V / m = 1 J C^{- 1} m^{- 1} = 1 N / C$ .

SimulationField aur equipotentials

Ek dipole jiske dashed equipotential curves potential ki kuch fixed values par bane hain: positive ke liye green, negative ke liye orange aur zero potential ke liye grey. Ek chhote positive test charge ko kahin bhi drag kiya ja sakta hai; ek arrow us par force dikhata hai, aur readouts field strength aur force ki direction dete hain. Buttons single charge ya do like charges par switch karte hain, field lines ya field vectors dikhate hain, aur test charge ko release karte hain taaki woh chale.

Fig. 2.11Dashed curves equipotentials hain (green positive, orange negative, grey zero). Test charge ko drag karo aur force arrow dekho.NCERT p. 55

Try karo

Test charge ko ek dashed curve ke along dheere-dheere drag karo. Force arrow poore raaste curve ke right angle par rehta hai.
Test charge ko charges ke theek beech grey dashed line par rakho, jahan $V = 0$ . Kya wahan field readout zero ho jaata hai?
Test charge ko $+ q$ ke thoda right drag karo aur Release test charge dabao. Woh kis taraf bhaagta hai? Kya woh dashed curves ko green se grey hote hue orange tak, yaani lower potential ki taraf, cross karta hai, ya ulta?

Example

Equipotential spacing se field

Uniform field wale region mein $120 V$ aur $70 V$ ke equipotential planes $2.0 c m$ door hain. (a) Field ka size aur direction nikaalo. (b) $120 V$ plane se $70 V$ plane tak jaane wale $+ 5.0 μ C$ charge par field kitna work karta hai? (c) $70 V$ plane se $0.80 c m$ aage wale plane par potential kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.7 Charges ke System ki Potential Energy

Ab tak source charges ka ek fixed set potential banata tha, aur hum ek visitor ki energy ke baare mein poochhte the. Ab sawaal ulta karo: charges ke arrangement mein khud kitni energy band hai? Answer hai use banane mein laga total work, har charge ko infinity se laate hue jabki pehle se rakhe charges ko roke rakha jaaye (Fig. 2.12).

3/3 $q_{3}$ , $q_{1}$ aur $q_{2}$ dono ko mehsoos karta hai. Ledger ka total arrangement ki potential energy $U$ hai.

Fig. 2.12Charges ko ek-ek karke jodna: har naya charge pehle se rakhe charges ke potential ki keemat chukata hai.NCERT Fig. 2.14 · p. 56

Left mein charges ek-ek karke triangle ke corners par aate hain; right mein ek ledger har aagman ka work record karta hai. Pehle step mein sirf q1 hai aur ledger mein W1 barabar 0 likha hai. Doosre step mein q2, q1 se distance r12 par aa gaya hai aur ledger mein W2 barabar k q1 q2 over r12 judta hai. Teesre step mein q3 upar, distances r13 aur r23 par aa gaya hai, ledger mein W3 barabar k q3 guna q1 over r13 aur q2 over r23 ka sum judta hai, aur aakhri line U ko teeno ke sum ke roop mein deti hai.

DerivationPoint charges ke assembly ki energy

Do charges. Work ka ek ledger rakho, jaisa Fig. 2.12 mein. Pehla charge, $q_{1}$ , khaali space mein aata hai: koi peeche nahi dhakelta, isliye ledger mein uski entry zero hai. Lekin uske aane se landscape badal jaata hai, kyunki ab har point ka potential $q_{1} / 4 π ε_{0} r$ hai. Doosra charge us landscape mein aane ki keemat chukata hai: $q_{2}$ ko $q_{1}$ se $r_{12}$ door ke point tak laane mein wahan ke potential ka $q_{2}$ guna lagta hai. Ledger ka total pair ki energy hai:

Do point charges ki energy

U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.22)

Teen charges. Pehle do charges ke aane ki keemat utni hi hai jitni pehle thi. Teeno positions ke liye $𝐫_{1}, 𝐫_{2}, 𝐫_{3}$ likho:

W_{2} = q_{2} V_{1} (𝐫_{2}) = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.23)

Ab $q_{1}$ aur $q_{2}$ milkar kisi bhi point P par yeh potential banate hain

V_{1, 2} (P) = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1 P}} + \frac{q_{2}}{r_{2 P}})

(2.24)

aur teesre charge ko dono ke against andar dhakelna padta hai:

W_{3} = q_{3} V_{1, 2} (𝐫_{3}) = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}})

(2.25)

Teeno steps jodne par:

Teen point charges ki energy

U = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}})

(2.26)

Pattern saaf hai: charges ke har pair ke liye ek term. Chaar charges chhe pair terms dete hain, aur $n$ charges $n (n - 1) / 2$ .

Order ulta kar do, pehle $q_{3}$ aur aakhir mein $q_{1}$ lao, to bhi tumhe bilkul wahi teen pair terms milte hain. Conservative force ke liye aisa hona hi tha. $U$ poore bane arrangement ki property hai, kuch-kuch bank account ke balance ki tarah, jo yeh record nahi karta ki deposits kis order mein kiye gaye.

Har pair term mein $q_{i} q_{j}$ ka sign hota hai, aur us sign ka matlab hai. $5.3 \times 10^{- 11} m$ door ek proton aur ek electron lo, jo hydrogen atom mein unki lagbhag doori hai: $U = - e^{2} / 4 π ε_{0} r \approx - 4.3 \times 10^{- 18} J$ . Negative $U$ ek bound pair dikhata hai: dono ko infinity tak door kheenchne mein itna work lagega. (Asli atom mein electron ka motion bhi hisaab mein aata hai, jaisa tum Chapter 12 mein dekhoge.) Electron ki jagah doosra proton rakho to wahi number positive aata hai: ab energy store hai, jo pair ke door udne par release hone ko taiyaar hai.

Example 2.4NCERT Example 2.4 · p. 57

Charges ke triangle ki energy

$+ 3.0 n C$ ka charge A par, $+ 3.0 n C$ B par aur $- 3.0 n C$ C par, $10 c m$ side wale equilateral triangle ke corners par hain (Fig. 2.13). (a) Inhe jodne mein kitna work laga? (b) A aur B ko fixed rakhte hue C wale charge ko AB ke midpoint M par le jaaya jaata hai. Ismein kitna work lagta hai? Answer ko do tarah se check karo.

Fig. 2.13Example 2.4 ka arrangement: C wala charge M par le jaaya jaata hai.NCERT p. 57

10 cm side wale equilateral triangle ke neeche left corner A par +3.0 nC charge, neeche right corner B par +3.0 nC charge aur upar wale corner C par -3.0 nC charge hai. Base AB ka midpoint M mark hai, aur ek dashed arrow C se seedha neeche M tak jaata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.8 External Field mein Potential Energy

Ab tak jin charges ki energy hum gin rahe the wahi field bhi bana rahe the. Aksar yeh do roles alag hote hain. Electron microscope ke beam mein ek electron socho, ya chimney ke electrostatic precipitator ki plates ke beech behta dhuein ka ek particle. Har ek electrodes ke banaye field mein chalta hai jinhe detail mein describe karne ki hume kabhi zaroorat nahi padti: hume bas unka banaya potential $V (𝐫)$ diya jaata hai. Ek hi assumption hai ki jab hamara charge unke beech chalta hai to woh sources jaise hain waise hi rehte hain.

2.8.1 Single charge ki potential energy

Kyunki $V (𝐫)$ pehle se hi infinity se andar ke safar ka work per unit charge naapta hai, $𝐫$ par rakhe charge $q$ ko bas $q$ guna chahiye. External field mein uski potential energy hai

External potential mein charge ki energy

U = q V (𝐫)

(2.27)

Eq. (2.27) se atomic aur nuclear physics ke liye energy ki ek kaam ki unit milti hai. $1 V$ ke potential difference se guzarne wale electron ya proton ki energy $e \times 1 V = 1.6 \times 10^{- 19} J$ badalti hai. Yeh amount ek electron volt hai.

Definition

Electron volt (eV)

Charge $e$ wale particle ko ek volt ke potential difference se guzarne par mili ya khoyi energy: $1 e V = 1.6 \times 10^{- 19} J$ .

ExampleVisible light ka ek photon lagbhag 2 se 3 eV le kar chalta hai.

Table 2.2Electron volt ke multiples

Unit	eV mein	Joules mein	Kahan milta hai
keV	$10^{3}$	$1.6 \times 10^{- 16}$	dental aur medical X-rays (dasiyon keV)
MeV	$10^{6}$	$1.6 \times 10^{- 13}$	ek single nuclear reaction mein release hone wali energy
GeV	$10^{9}$	$1.6 \times 10^{- 10}$	proton ki rest energy (lagbhag 0.94 GeV)
TeV	$10^{12}$	$1.6 \times 10^{- 7}$	Large Hadron Collider mein proton beams

Example

Ek electron gun

Electron gun mein, jaise electron microscope wali, ek electron rest se shuru hokar $2.5 k V$ ke potential difference se accelerate hota hai. Uski kinetic energy electron volts aur joules mein nikaalo, aur uski speed bhi. ( $m_{e} = 9.1 \times 10^{- 31} k g$ , $e = 1.6 \times 10^{- 19} C$ )

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.8.2 External field mein do charges ke system ki potential energy

Ab do charges, $𝐫_{1}$ par $q_{1}$ aur $𝐫_{2}$ par $q_{2}$ , ko same external field mein rakho, aur arrangement ko ek-ek charge karke banao. $q_{1}$ rakhne mein $q_{1} V (𝐫_{1})$ lagta hai, kisi bhi single charge ki tarah. Jab tak $q_{2}$ aata hai, $q_{1}$ pehle se wahan hai, isliye $q_{2}$ ko external field ke saath-saath $q_{1}$ ke field ke against bhi dhakelna padta hai. Fields superpose hote hain, isliye work ke dono contributions bas jud jaate hain:

W_{2} = q_{2} V (𝐫_{2}) + \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.28)

Total field mein pair ki potential energy hai:

External field mein do charges

U = q_{1} V (𝐫_{1}) + q_{2} V (𝐫_{2}) + \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}}

(2.29)

Teeno terms ko do tarah ki energy samjho: external field mein har charge ki energy, plus Section 2.7 se pair ki mutual energy.

Example 2.5NCERT Example 2.5 · p. 59

Uniform field mein rod par do beads

$10 c m$ lambi insulating rod ke ends par do chhote beads fixed hain. Bead 1 par $q_{1} = + 2.0 μ C$ hai aur bead 2 par $q_{2} = - 3.0 μ C$ . (a) Beads ki mutual potential energy kya hai? (b) Rod ko $+ x$ ke along point karte $4.0 \times 10^{5} N / C$ ke uniform external field ke along rakha jaata hai, bead 1 origin par aur bead 2 $x = + 10 c m$ par. External potential ko origin par zero lekar, taaki $V (x) = - E_{0} x$ , pair ki total electrostatic energy nikaalo. (c) Phir rod ko uske midpoint ke around ulta ghuma diya jaata hai, taaki beads apni jagah badal lein. Ismein kitna work lagta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.8.3 External field mein dipole ki potential energy

Uniform field mein dipole par koi net force nahi lagta lekin ek torque $𝝉 = 𝐩 \times 𝐄$ lagta hai (Chapter 1), jo $𝐩$ ko ghuma kar $𝐄$ ki line mein laane ki koshish karta hai. Use doosri taraf ghumaane mein work lagta hai, aur woh work potential energy ke roop mein store hota hai jo $𝐩$ aur $𝐄$ ke beech angle $θ$ par depend karti hai (Fig. 2.14).

2/4 $60^{\circ}$ ghumaaya hua: couple ise wapas ghumaane ki koshish karta hai; $U = - p E \cos 60^{\circ} = - \frac{1}{2} p E$ .

Fig. 2.14

U (θ) = - p E \cos θ

: sabse kam jab

𝐩

𝐄

ke along ho, sabse zyada jab woh ulti taraf point kare.NCERT Fig. 2.16 · p. 60

Left mein horizontal field lines right ki taraf point karti hain aur beech mein ek dipole hai jiske har charge par ek force arrow hai. Right mein ek graph U(theta) ko 0 degrees par minus pE se, 90 degrees par zero se hote hue, 180 degrees par plus pE tak smoothly badhte dikhata hai, aur ek dot current state mark karta hai. Pehle step mein dipole field ke along point karta hai, forces ek hi line ke along lagte hain aur dot minimum par hai, jis par stable likha hai. Doosre step mein dipole 60 degrees tak ghuma hai aur dot minus one half pE tak chadh gaya hai. Teesre step mein dipole field ke right angle par hai aur dot zero par hai. Chauthe step mein dipole field ke against point karta hai aur dot maximum par hai, jis par unstable likha hai.

DerivationAngle

θ

par dipole ki energy

Dipole par electric torque hai

Dipole par torque

𝝉 = 𝐩 \times 𝐄

(2.30)

jiska size $p E \sin θ$ hai; yeh hamesha $θ$ ko chhota karne ki koshish karta hai. Maan lo tum dipole ko pakad kar angle ko $θ_{0}$ se $θ_{1}$ tak bahut dheere kholte ho, use kabhi ghoomne nahi dete. Har angle par tumhara twist electric twist ko cancel karta hai, aur ek chhote turn $d θ$ par tum $p E \sin θ d θ$ work karte ho. Chhote turns jodne par,

W = \int_{θ_{0}}^{θ_{1}} p E \sin θ d θ = p E (\cos θ_{0} - \cos θ_{1})

(2.31)

Woh work potential energy ban jaata hai, isliye $U (θ_{1}) - U (θ_{0}) = p E (\cos θ_{0} - \cos θ_{1})$ . Hamesha ki tarah formula sirf differences fix karta hai. Agar hum maan lein ki jab dipole field ke across khada ho ( $θ_{0} = 90^{\circ}$ ) tab $U = 0$ , to starting angle hat jaata hai aur result compact ho jaata hai:

Uniform field mein dipole ki energy

U (θ) = p E (\cos \frac{π}{2} - \cos θ) = - p E \cos θ = - 𝐩 \cdot 𝐄

(2.32)

Charges se bhi wahi result. Dipole ko do alag charges maan kar Eq. (2.29) lagao. Mutual term, $- q^{2} / (4 π ε_{0} \cdot 2 a)$ , rigid dipole ka hai aur ghoomne par kabhi nahi badalta:

U^{'} (θ) = q [V (𝐫_{1}) - V (𝐫_{2})] - \frac{q^{2}}{4 π ε_{0} \cdot 2 a}

(2.33)

Yahan $𝐫_{1}$ , $+ q$ ki position hai aur $𝐫_{2}$ , $- q$ ki. Uniform field apne along chale har metre ke liye potential ko $E$ volts kam karta hai, aur $+ q$ end, $- q$ end se $2 a \cos θ$ zyada 'downstream' hai, isliye $V (𝐫_{1}) - V (𝐫_{2}) = - E (2 a \cos θ)$ aur

U^{'} (θ) = - p E \cos θ - \frac{q^{2}}{4 π ε_{0} \cdot 2 a}

(2.34)

Fixed mutual term ko chhodkar, jise hum hata sakte hain, yeh phir se Eq. (2.32) hai. Isse yeh bhi samajh aata hai ki $90^{\circ}$ natural zero kyun hai. Field ke across pade dipole ke dono ends ek hi equipotential par hote hain, isliye field ek end ko utni hi energy deta hai jitni doosre se leta hai.

Energy sabse kam, $- p E$ , tab hai jab $𝐩$ , $𝐄$ ke along point kare: stable equilibrium. Sabse zyada, $+ p E$ , tab jab $𝐩$ , $𝐄$ ke against point kare: unstable equilibrium. Dipole ko aligned se reversed tak ghumaane mein $2 p E$ lagta hai.

Example 2.6NCERT Example 2.6 · p. 61

Ek dipole jo jhoolta hai aur ruk jaata hai

$2.0 c m$ lambi ek halki insulating rod ke dono ends par $+ 5.0 n C$ aur $- 5.0 n C$ ke charges hain aur yeh $3.0 \times 10^{4} N / C$ ke uniform field mein apne centre ke around free ghoom sakti hai. Ise field se $120^{\circ}$ par, kuch field ke against point karte hue, pakda jaata hai aur chhod diya jaata hai. (a) Iska dipole moment aur $120^{\circ}$ par iski potential energy nikaalo. (b) Field ki direction se guzarte waqt iske paas kitni kinetic energy hai? (c) Air resistance aakhir mein ise field ke along rok deta hai. Kitni heat bani? (d) Phir ise ghuma kar bilkul field ke against point karwane mein kitna work lagega?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek proton rest se shuru hokar

500 V

se accelerate hota hai. Uski kinetic energy kitni hai, joules mein? (

e = 1.6 \times 10^{- 19} C

)

Answer dekho

Answer 8e-17 J

K = e Δ V = (1.6 \times 10^{- 19}) (500) J = 8.0 \times 10^{- 17} J

, jo seedha

500 e V

hai.

2.9 Conductors ki Electrostatics

Ek charged metal sphere ko uncharged sphere se chhuao aur charge lagbhag turant dono par phail jaata hai. Wajah yeh hai ki metal mein har atom ke kuch electrons kisi ek atom se bandhe nahi hote: woh poore crystal mein ghoomte hain, jabki positive ions apni jagah locked rehte hain. Jab metal ko akela chhod diya jaata hai aur saara motion ruk jaata hai, tab yeh ghoomte electrons kya karte hain? Answer se metals ki kuch kamaal ki electrical properties milti hain.

Fig. 2.15 uniform field mein daale gaye ek metal sphere ko follow karta hai. Field free electrons ko ek side ki taraf kheenchta hai, aur doosri side mein unki kami ho jaati hai. Dono sides par jama hote charges apna khud ka field banate hain, jo metal ke andar ulti direction mein point karta hai. Yeh drift tab tak chalta hai jab tak dono fields cancel nahi ho jaate. Achhe conductor mein yeh poora adjustment ek microsecond se bhi bahut kam samay mein ho jaata hai, aur settled state se chhe results nikalte hain.

3/3Metal ke andar dono fields cancel ho jaate hain. Bahar field lines ab surface se right angle par milti hain.

Fig. 2.15Free electrons tab tak khiskte hain jab tak induced charges ka field andar applied field ko cancel na kar de; bahar lines surface se right angle par milti hain.NCERT Fig. 2.20 · p. 65

Ek metal sphere uniform field mein hai jiski lines left se right chalti hain. Pehle step mein seedhi field lines sphere ke aar-paar guzarti hain aur andar free electrons ke arrows left ki taraf, field ke against, point karte hain. Doosre step mein sphere ki left side par minus signs aur right side par plus signs jama ho gaye hain, aur andar dashed orange arrows in induced charges ka field left ki taraf, applied field ke against, dikhate hain. Teesre step mein bahar ki field lines mud jaati hain taaki woh minus charges par khatam hon aur plus charges se phir shuru hon, surface se right angle par milte hue, aur andar E barabar 0 likha hai.

PropertyResult 1: Conductor ke andar koi field nahi

Electrostatic equilibrium mein conductor ke material ke andar har point par electric field zero hota hai, chahe conductor par net charge ho ya na ho aur uske chaaron taraf koi bhi field ho.

PropertyResult 2: Field lines surface se right angle par milti hain

Equilibrium wale conductor se jahan bhi field lines nikalti ya ghusti hain, woh uski surface se right angle par milti hain.

Surface ke theek bahar ke field ko do hisson mein todo: ek surface ke along aur ek usse seedha bahar. Surface ke free electrons side mein utni hi aasaani se khisak sakte hain jitna andar chalte hain, isliye koi bhi sideways hissa unhe tab tak dhakelta jab tak woh khatam na ho jaaye. Perpendicular hissa surface charges ko sirf metal mein dabata hai ya bahar kheenchta hai, aur kyunki woh metal se nikal nahi sakte, yeh surface ke along kuch nahi hilata. Isliye equilibrium mein sirf perpendicular hissa bachta hai.

PropertyResult 3: Extra charge surface par rehta hai

Equilibrium wale conductor ke material ke andar koi net charge nahi hota; use diya gaya koi bhi extra charge aakhir mein uski surface par pahunch jaata hai.

Maan lo metal ke andar gehrai mein kisi chhote region mein net charge hota. Use poori tarah metal ke andar padi ek chhoti closed surface se ghero. Result 1 se us surface ke har point par field zero hai, isliye usse flux zero hai, aur phir Gauss's law kehta hai ki woh koi net charge enclose nahi karti, jo maani gayi baat ke khilaaf hai. Extra charge sirf metal ki surface par tik sakta hai, jahan koi Gaussian surface poori tarah material ke andar nahi banayi ja sakti.

PropertyResult 4: Poora conductor ek hi potential par hai

Equilibrium wale conductor ka ek hi potential hota hai: uske andar aur surface par har point par $V$ ki same value hoti hai.

Ek test charge ko metal ke kisi bhi point se kisi doosre point tak le jao, andar se ya surface ke along. Andar, Result 1 kehta hai ki us par koi force nahi hai; surface par, Result 2 kehta hai ki force path ke perpendicular hai. Dono tarah koi work nahi hota, isliye dono points ka potential ek hai. $V$ ke map par conductor ek flat plateau hai, uski shape kitni bhi tedhi ho, aur $V$ tabhi badalta hai jab tum usse utar kar bahar ke field mein jaate ho. Do alag conductors alag heights ke plateaus par ho sakte hain.

DerivationResult 5: Charged conductor ke theek bahar field

Theek bahar field kitna strong hai? Yeh sirf is par depend karta hai ki surface ke paas wale patch par kitna charge hai, jise surface charge density $σ$ se naapte hain:

Conductor ke theek bahar field

𝐄 = \frac{σ}{ε_{0}} \hat{𝐧}

(2.35)

Yahan $\hat{𝐧}$ outward unit normal hai. Proof mein surface ke across khade flat coin jaisi ek Gaussian surface use hoti hai, jise 'pillbox' kehte hain (Fig. 2.16). Ek face, area $δ S$ ka, metal ke theek bahar hai aur doosra theek andar, aur rim itna chhota hai ki mushkil se bahar nikalta hai.

Fig. 2.16Pillbox ke sirf bahar wale face se flux guzarta hai.NCERT Fig. 2.17 · p. 63

Conductor surface ka ek zoom kiya hua patch ek horizontal line ki tarah bana hai jis par plus signs hain, aur neeche metal shaded hai. Field arrows surface se seedhe upar point karte hain. Ek dashed cylinder surface ko cross karta hai: area delta S wala uska upar ka face theek bahar hai, jahan se ek field arrow guzarta hai, aur area delta S wala neeche ka face metal ke andar hai, jahan koi field nahi hai.

Flux ko face-by-face gino. Metal ke andar wale face ko kuch nahi milta, kyunki wahan $𝐄 = 0$ . Chhote rim ko bhi kuch nahi milta: uska andar wala hissa koi field nahi dekhta, aur bahar wala hissa apne parallel chalta field dekhta hai (Result 2). Bacha bahar wala face, jahan field normal hai aur chhote se area par lagbhag uniform hai, isliye positively charged patch ke liye flux $E δ S$ hai. Pillbox charge $σ δ S$ enclose karta hai, aur Gauss's law deta hai

E δ S = \frac{σ δ S}{ε_{0}} \Rightarrow E = \frac{σ}{ε_{0}}

(2.36)

Result 2 se direction jodne par Eq. (2.35) milti hai, aur yeh dono signs ke liye kaam karti hai: jahan $σ < 0$ wahan vector metal ke andar point karta hai. Factor par dhyaan do. Ek isolated charged sheet har side sirf $σ / 2 ε_{0}$ banati hai. Conductor par baaki charge khud ko aise arrange karta hai ki metal ke andar patch ka field cancel ho jaaye, aur aisa karte hue woh theek bahar ka field double kar deta hai.

PropertyResult 6: Electrostatic shielding

Conductor ke andar ki khaali cavity mein kabhi electric field nahi hota. Cavity ki shape, conductor par charge ya aas-paas ka field, kuch bhi ise nahi badal sakta, aur cavity ki deewar par khud koi charge nahi hota.

Result 4 se cavity ki deewar ek single equipotential hai. Ab maan lo khaali cavity mein koi field hota. Field line ko charges par shuru aur khatam hona padta hai, aur cavity mein koi charge nahi hai, isliye line ko deewar se nikal kar wapas deewar par hi aana padta. Lekin field line ke along potential lagaataar girta hai, isliye uske dono ends us deewar par alag potentials par hote jiska sirf ek potential hai. Yeh impossible hai, isliye cavity mein koi field nahi hai, aur phir deewar par ek pillbox dikhata hai ki deewar par koi charge nahi hai. (Spherical shell ke liye Chapter 1 symmetry se isi result tak pahuncha tha; is argument ko kisi symmetry ki zaroorat nahi.)

1/3Field lines bahar wali surface ke induced charges par khatam hoti hain; andar cavity field-free rehti hai.

Fig. 2.17Closed conductor apni cavity ko bahar ke fields se bachata hai, lekin cavity ke andar rakha charge bahar phir bhi mehsoos hota hai.NCERT Fig. 2.18 · p. 64

Beech mein ek square metal box hai jismein ek square cavity hai. Pehle step mein left se aati horizontal field lines left bahari deewar par induced minus charges par khatam hoti hain aur right bahari deewar ke plus charges se phir shuru hoti hain; khaali cavity par E barabar 0 likha hai. Doosre step mein upar aur neeche ke bahari faces par bhi extra plus charges jod diye gaye hain, aur cavity par abhi bhi E barabar 0 likha hai. Teesre step mein bahar ka field hat gaya hai aur cavity mein ek positive charge hai: field lines usse cavity ki deewar ke minus charges tak jaati hain, bahar wali surface par plus charges aate hain, aur field lines box se har direction mein bahar phailti hain.

Yeh hai electrostatic shielding: ek closed metal enclosure bahar ke electric fields ko apne andar ki cheezon se door rakhta hai. Sensitive instruments metal cases mein rakhe jaate hain, TV ya network cable ka signal wire metal braid mein lipta hota hai, aur car ya aircraft ki metal body bijli girne par andar ke logon ko bachati hai. Lekin yeh protection sirf ek taraf kaam karta hai, jaisa Fig. 2.17 ka aakhri step dikhata hai: cavity ke andar rakha charge bahar apni maujoodgi mehsoos karwata hai.

Try karke dekhoNCERT p. 64

Steel tiffin box se phone ko chup karao

Tumhe chahiye

ek mobile phone
use call karne ke liye doosra phone
tight metal lid wala steel tiffin box ya biscuit tin
milte-julte size ka cardboard box

Pehle guess karo

Band metal tin ke andar rakha phone call karne par bajega kya?

Sirf tab jab tin earth se juda hoHaan; patle tin se koi fark nahi padtaAam taur par nahi, ya bahut halka

Check karo ki phone mein achha signal hai. Ise cardboard box mein rakho, lid band karo aur doosre phone se call karo.
Ab ise metal tin mein rakho, lid ko mazbooti se dabakar band karo aur phir call karo.
Lid thodi si kholo aur ek baar aur call karo.

Cardboard ke through call normally lag jaati hai. Achhe se band metal tin mein call aam taur par fail ho jaati hai ya signal bars bahut gir jaate hain. Phone ka signal tezi se badalta electric (aur magnetic) field hai; metal ke free electrons andar use cancel karne ke liye rearrange ho jaate hain, bilkul jaise woh static field ko cancel karte hain. Lid mein gap ho to field andar leak ho jaata hai. Microwave oven yahi idea ulti taraf use karta hai: uski metal deewarein aur darwaze ki mesh microwaves ko andar rakhti hain.

Example 2.7NCERT Example 2.7 · p. 64

Rozmarra ki zindagi mein static electricity

Inmein se har ek ko samjhao. (a) Sookhe kapde se ragdi plastic ruler tap se nikalti paani ki patli dhaar ko apni taraf mod deti hai, ruler par chahe kisi bhi sign ka charge aaya ho. (b) Phone circuit boards assemble karne wale technicians ek wrist strap pehente hain jo wire se earth se juda hota hai, aur ek conducting mat par kaam karte hain jo bhi earthed hoti hai. (c) Car ke andar baitha insaan kaafi had tak safe hai agar car par bijli gire. (d) Sardiyon ke ek sookhe din tum nylon carpet par chalkar metal door handle ki taraf haath badhate ho: ek spark tumhari ungli tak koodta hai. Turant handle ko phir chhuo to kuch nahi hota.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

(a) Charged ruler ka field paas ke paani par lagta hai. Tap ka paani thoda conduct karta hai aur tap plumbing se juda hai, isliye opposite sign ka charge dhaar ke us hisse mein aa jaata hai jo ruler ke sabse paas hai; polar water molecules bhi kuch field ki taraf ghoom jaate hain. Opposite charge ruler ke us se zyada paas hota hai jitna same sign ka koi charge, aur paas mein field zyada strong hota hai, isliye attraction jeet jaata hai aur dhaar mud jaati hai. Ruler ka charge ulta karo to har induced charge bhi ulta ho jaata hai, isliye dhaar phir bhi uski taraf mudti hai.
(b) Chalne aur hilne-dulne se friction ke kaaran body hazaaron volts tak charge ho sakti hai. Agar charged ungli kisi chip ko chhooti, to charge ek chhote spark mein usse hokar bhaagta, aur modern chips ke andar ki insulating layers, jo sirf nanometres moti hoti hain, aasaani se toot jaati hain. Strap aur mat technician, bench aur board ko ek common potential (earth) par rakhte hain, isliye spark chalane ke liye koi potential difference nahi hota. (Strap mein ek bada resistor hota hai, isliye isse khatarnaak current nahi guzar sakta.)
(d) Carpet par ragadne se tumhari body charged ho jaati hai aur handle ke potential se bahut alag potential par hoti hai. Jab tumhari ungli paas aati hai, to hawa ke chhote gap mein field hawa ki dielectric strength se zyada ho jaata hai aur charge tab tak koodta hai jab tak tum aur handle same potential par na aa jao (jude conductors ke liye Result 4). Doosri baar chhoone par koi potential difference nahi bacha hota, isliye charge nahi behta aur spark nahi hota. Charge sirf alag potentials wale points ke beech behta hai.

2.10 Dielectrics aur Polarisation

Dielectric ek insulator hai: glass, mica, rubber, plastic, sookha kaagaz, shuddh paani. Iske charges apne atoms aur molecules se bandhe hote hain aur idhar-udhar nahi ghoom sakte. Phir bhi electric field dielectric se bina badle nahi guzarta. Har molecule ke andar positive aur negative charges thoda khisak sakte hain, ya poora molecule ghoom sakta hai, aur iska mila-jula asar andar field ko kamzor kar deta hai.

Compare

Same field mein ek conductor aur ek dielectric

Conductor

Free electrons poore piece mein drift karte hain
Iski surfaces par induced charges aate hain
Unka field andar applied field ko bilkul cancel kar deta hai
Andar: $E = 0$

Dielectric

Bound charges sirf har molecule ke andar khisakte hain
Iski surfaces par bound (polarisation) charges aate hain
Unka field applied field ko sirf kuch had tak cancel karta hai
Andar: $E$ kam, lekin zero nahi

Molecule kaise respond karta hai yeh uski shape par depend karta hai. Water ( $H_{2} O$ ), ammonia ( $N H_{3}$ ) ya hydrogen chloride ( $H C l$ ) mein ek atom shared electrons ko apne partners se zyada zor se kheenchta hai, aur shape ek taraf jhuki hoti hai. Isliye molecule ka ek end hamesha thoda negative aur doosra thoda positive hota hai. Yeh polar molecules hain, har ek ka apna built-in dipole moment. $N_{2}$ , $C O_{2}$ ya $C H_{4}$ jaise symmetric molecule mein pulls cancel ho jaate hain aur positive aur negative charge ke centres ek hi jagah hote hain. Aise molecule ka apna koi dipole moment nahi hota: yeh non-polar hai.

2/4Har electron cloud field ke against khisakta hai (induced dipole); polar molecules kuch field ki taraf ghoom jaate hain.

Fig. 2.18Non-polar molecules dipoles ban jaate hain aur polar molecules kuch line up ho jaate hain; slab mein isse do faces par bound charge bachta hai, jo andar field ko kamzor karta hai.NCERT Fig. 2.22 · p. 66

Left mein non-polar atoms ki ek row hai, har ek ek gol electron cloud ke andar chhota positive nucleus, aur polar molecules ki ek row jo ovals ki tarah bane hain aur jinke dipole moment ke liye ek arrow hai. Pehle step mein koi field nahi hai: har nucleus apne cloud ke centre mein hai aur polar molecules random directions mein point karte hain. Doosre step mein field right ki taraf point karta hai: har electron cloud apne nucleus ke left khisak gaya hai aur polar molecules zyaadatar right ki taraf ghoom gaye hain. Teesre step mein right mein ek slab aligned dipoles se bhara hai, har plus end apne padosi ke minus end ke paas. Chauthe step mein sirf slab ke do faces highlighted hain, left face par minus charges aur right face par plus charges, aur andar arrows applied field, face charges ka ulta field aur chhota net field dikhate hain.

Non-polar case se shuru karo, Fig. 2.18 ki upar wali row. Applied field har atom ke nucleus ko ek taraf aur uske electron cloud ko doosri taraf kheenchta hai. Cloud ud nahi jaata, kyunki uske apne nucleus ka attraction use pakde rakhta hai, isliye woh centre se thoda hatkar tik jaata hai. Ab har atom ke paas field ke along ek chhota induced dipole hai, aur zyaadatar materials mein do guna strong field do guna bada shift deta hai. Is chapter mein hum sirf aise linear isotropic dielectrics dekhte hain, jo field ke proportion mein respond karte hain aur field ki direction kuch bhi ho, ek hi tarah.

Polar molecules ke paas dipoles pehle se hote hain, lekin bina field wale sample mein thermal dhakka-mukki unki directions ko bikher deti hai, aur average mein woh cancel ho jaate hain. Socho ek table par compass needles ki tray hai jise hilaya ja raha hai aur paas mein ek magnet pakda hai. Magnet needles ko line up karne ki koshish karta hai, hilana uska kaam bigaadta rehta hai, aur result sirf ek adhoori, average alignment hota hai. Field polar molecules par isi tarah kaam karta hai, kyunki $𝐄$ ke along dipole ki energy kam hoti hai, $- 𝐩 \cdot 𝐄$ (Section 2.8.3). Sample jitna thanda aur field jitna strong, alignment utna behtar.

Induced hon ya aligned, molecular dipoles ab ek hi taraf jhuke hain, isliye material ke har chhote volume ka ek net dipole moment hai. Material polarised ho gaya hai, aur kitna strongly, yeh hum uske polarisation se naapte hain:

Definition

Polarisation (

𝐏

)

Dielectric ka net dipole moment per unit volume. Iski SI unit $C m^{- 2}$ hai.

ExampleLinear isotropic dielectric ke liye $𝐏$ field ke along point karta hai aur uske proportional hai.

Linear isotropic dielectric

𝐏 = ε_{0} χ_{e} 𝐄

(2.37)

jahan

$χ_{e}$: material ki electric susceptibility, ek dimensionless constant
$𝐄$: dielectric ke andar ka field

Bada $χ_{e}$ matlab aisa material jo aasaani se polarise hota hai. Hamare kaam ke liye yeh bas har material ki ek measured property hai, uski density ki tarah.

Is sab se field par kya asar hota hai? Dielectric ka ek slab uniform field $𝐄_{0}$ mein rakho jo do aamne-saamne ke faces se seedha milta hai (Fig. 2.18, aakhri do steps), aur molecules ki ek single row ko ek face se doosre tak follow karo. Yeh aisi padhi jaati hai $- + - + - + \dots$ : jahan bhi do padosi milte hain, ek $+$ end ek $-$ end se laga hota hai aur bahar se dekhne par pair neutral lagta hai. Sirf row ke do ends par charges bina jode ke bachte hain, jis face se field ghusta hai wahan $-$ aur jahan se nikalta hai wahan $+$ .

Har row isi tarah khatam hoti hai, isliye slab charge ki do patli sheets ki tarah behave karta hai, ek face par $- σ_{p}$ aur doosre par $+ σ_{p}$ , aur andar neutral. Un sheets ke beech unka apna field $+$ se $-$ ki taraf chalta hai, $𝐄_{0}$ ke comparison mein ulta, isliye material ke andar ka field aas-paas ke field se kamzor hai. Yeh surface charge bound charge hai: yeh molecules ka hai aur, metal ke free charge ke ulat, wire se bahar nahi le jaaya ja sakta.

2.11 Capacitors aur Capacitance

Phone ka circuit board ya puraane radio ka andar ka hissa dekho, tumhe darjanon chhote components milenge, jinmein se kai capacitors hain. Packaging hata do to har capacitor ek hi cheez hai: insulator se alag rakhe do conductors ka pair (Fig. 2.19). Dono conductors ko battery ke terminals se jodo to woh charge ko ek se doosre tak tab tak le jaati hai jab tak un par barabar aur opposite charges, $+ Q$ aur $- Q$ , potential difference $V$ par na aa jaayein. Jab hum 'capacitor $50 μ C$ tak charged hai' kehte hain to hamara matlab uske positive conductor ka charge hota hai; poore capacitor ka net charge phir bhi zero hai.

Fig. 2.19Beech mein insulator wala conductors ka koi bhi pair capacitor banata hai. Fixed aur variable capacitor ke symbols right mein hain.NCERT Fig. 2.24 · p. 67

Left mein do tedhe-medhe conductors ek shaded insulating gap ke aar-paar aamne-saamne hain. Left wale ki saamne wali side par plus charges hain aur us par plus Q likha hai; right wale par minus charges hain aur us par minus Q likha hai. Field lines gap ke across positive se negative conductor tak chalti hain. Right mein do circuit symbols hain: fixed capacitor, do chhoti parallel lines ki tarah bana, aur variable capacitor, wahi symbol jis par ek diagonal arrow hai.

Ek capacitor ko $3 V$ cell se charge karo, phir $6 V$ se aur $9 V$ se, aur har baar charge naapo: results $1 : 2 : 3$ ke ratio mein aate hain. Superposition iski wajah batata hai. Conductors par charge ka koi bhi pattern, kisi factor se bada kiya jaaye, to har point par field bhi usi factor se bada hota hai, aur isliye potential difference bhi. Isliye $Q$ aur $V$ hamesha saath-saath badalte hain, aur unka ratio sirf conductors par depend karta hai:

Capacitance

C = \frac{Q}{V}

(2.38)

Definition

Capacitance (

C

)

Capacitor ke kisi bhi conductor par charge aur unke beech potential difference ka ratio, $C = Q / V$ . Iski SI unit farad hai: $1 F = 1 C / V$ .

ExampleJo capacitor $10 V$ par $\pm 5 μ C$ rakhta hai uska $C = 0.5 μ F$ hai.

Balti ke volume ki tarah, capacitance device banate waqt hi fix ho jaati hai. Bade conductors paas-paas rakhe hon to $C$ bada hota hai, aur Section 2.13 dikhata hai ki unke beech kya bhara hai yeh bhi matter karta hai; tum kitna charge daalte ho, yeh nahi. Ek farad bahut bada hai: practical capacitors microfarads ( $1 μ F = 10^{- 6} F$ ), nanofarads ( $10^{- 9} F$ ) aur picofarads ( $10^{- 12} F$ ) mein rate kiye jaate hain. Akele metal sphere ki bhi capacitance hoti hai, jo uske partner conductor ko infinity par phaila hua maan kar nikalti hai; neeche ka example ise nikaalta hai.

Capacitor ko charge ki tank samjho, jismein $V$ pressure ka role nibhaata hai. Chaudi tank (bada $C$ ) kam pressure par bahut kuch rakhti hai; same charge ko chhote $C$ mein thoosne ke liye bada $V$ chahiye, aur bade $V$ ka matlab insulator mein strong field. Ek had ke baad insulator fail ho jaata hai. Hawa mein bhatakte ions itni tezi se uchhalte hain ki aur electrons ko free kar dete hain, charge carriers ka avalanche banta hai, aur ek spark gap ko paar karta hai, stored charge ko khaali kar deta hai. Insulator jo sabse bada field seh sakta hai woh uski dielectric strength hai.

Definition

Dielectric strength

Sabse bada electric field jo insulating material toot kar conduct karna shuru karne se pehle seh sakta hai. Sookhi hawa ke liye yeh lagbhag $3 \times 10^{6} V / m$ hai.

ExampleUngli se door handle tak $5 m m$ koodne wale static spark ko lagbhag $15 k V$ chahiye.

Isiliye capacitor ke case par capacitance ke saath maximum working voltage bhi likha hota hai, jaise ' $470 μ F$ , $25 V$ '. Agar design ko zyada stored charge chahiye, to safe tareeka bada $C$ hai, kyunki $V$ badhane se jaldi breakdown aa jaata hai. Capacitors ka storage ke alawa ek doosra kaam bhi hai: alternating-current circuits (Chapter 7) mein woh steady currents ko rokte hain aur badalte currents ko jaane dete hain, aur isi tarah woh signals ko filter aur shape karte hain.

Example

Hawa mein sphere kitna charge rakh sakta hai?

$15 c m$ radius ka ek metal sphere sookhi hawa mein insulating support par khada hai. (a) Uske paas ki hawa ke breakdown se pehle woh sabse zyada kitna charge rakh sakta hai? (b) Tab uska potential kya hai? (c) Doosre conductor ko infinity par maan kar uski capacitance kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

12 V

se jodne par ek capacitor

\pm 24 μ C

store karta hai. Uski capacitance kitni hai, µF mein?

µF

Answer dekho

Answer 2 µF

C = Q / V = 24 μ C / 12 V = 2.0 μ F

2.12 Parallel Plate Capacitor

Analyse karne ke liye sabse simple capacitor hai area $A$ ki do flat metal plates jo ek patle gap $d$ ke aar-paar aamne-saamne hain (Fig. 2.20). Gap ko khaali maan lo (hawa lagbhag same behave karti hai) aur ek plate par $+ Q$ aur doosri par $- Q$ rakho, taaki charge densities $\pm σ$ hon, jahan $σ = Q / A$ . Agar plates gap ke comparison mein chaudi hain ( $d^{2} ≪ A$ ), to unke beech ka point har plate ko ek kabhi na khatam hone wali sheet ki tarah dekhta hai. Chapter 1 ne aisi sheet ka field diya tha: dono taraf $σ / 2 ε_{0}$ , positive charge se door aur negative ki taraf point karta hua.

3/4Plates ke bahar dono fields cancel ho jaate hain; unke beech jud kar $σ / ε_{0}$ bante hain.

Fig. 2.20Plates ke bahar dono sheet fields cancel ho jaate hain; unke beech jud kar

σ / ε_{0}

bante hain, edges ke paas ko chhodkar ek uniform field.NCERT Fig. 2.25 · p. 68

Do vertical plates aamne-saamne hain: left wali positive, right wali negative. Pehle step mein negative plate halki bani hai aur blue arrows sirf positive plate ka field dikhate hain, dono taraf usse door point karte hue, har jagah sigma over 2 epsilon nought size ke. Doosre step mein dashed arrows negative plate ka field jodte hain, dono taraf uski taraf point karte hue. Teesre step mein sirf sum bana hai: plates ke beech mote arrows, sigma over epsilon nought size ke, aur bahar ke regions par E barabar 0 likha hai. Chauthe step mein plates ke upar aur neeche ke edges par curved lines bahar phool jaati hain, aur plates ke beech dashed vertical lines flat equipotential planes mark karti hain.

DerivationParallel plates ka field aur capacitance

Positive plate ke aage wale region ko I aur negative plate ke aage wale ko II kaho. Har bahar wale region mein dono sheet fields opposite directions mein point karte hain aur cancel ho jaate hain:

E_{I} = \frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} = 0

(2.39)

E_{II} = \frac{σ}{2 ε_{0}} - \frac{σ}{2 ε_{0}} = 0

(2.40)

Plates ke beech dono fields positive plate se negative plate ki taraf point karte hain, isliye jud jaate hain:

E = \frac{σ}{2 ε_{0}} + \frac{σ}{2 ε_{0}} = \frac{σ}{ε_{0}} = \frac{Q}{ε_{0} A}

(2.41)

(Section 2.9 ka Result 5 seedha yahi value deta hai: positive plate ke andar wale face ke theek bahar field $σ / ε_{0}$ hai.) Kyunki $E$ poore gap mein same hai, barabar steps ke liye potential barabar girta hai, aur poore gap ke across yeh girta hai

V = E d = \frac{Q d}{ε_{0} A}

(2.42)

aur capacitance hai

Parallel plate capacitor

C = \frac{Q}{V} = \frac{ε_{0} A}{d}

(2.43)

Dhyaan do ki answer mein na $Q$ bachta hai na $V$ : sirf plate area aur gap. Badi plates same voltage par zyada charge rakhti hain, aur patle gap ko same charge ke liye kam voltage chahiye.

Asli plates anant nahi hoti. Unke kinaaron ke paas kuch field lines bahar ki space mein mud jaati hain (Fig. 2.20 ka aakhri step), is effect ko fringing kehte hain, aur charge edges ke paas beech se thoda zyada ghana hota hai. Isse kitna fark padta hai? Disturbed region sirf lagbhag ek gap-width chauda hota hai, isliye gap se kai guna chaudi plates ke liye unke beech ki lagbhag saari space mein Eq. (2.41) wala uniform field hota hai.

Kuch numbers dikhate hain ki farad kitna bada hai. A4 page ke size ( $0.21 m \times 0.297 m$ ) ki do metal sheets, hawa mein $0.10 m m$ door rakhi hon, to deti hain

C = \frac{(8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}) (0.0624 m^{2})}{1.0 \times 10^{- 4} m} \approx 5.5 \times 10^{- 9} F = 5.5 n F

Sawaal ulta karo: cricket field jitni badi plates, lagbhag $150 m \times 150 m$ , se $1 F$ paane ke liye gap kitna patla hona chahiye? Answer, $d = ε_{0} A / C \approx 2 \times 10^{- 7} m$ , ek micrometre ka paanchvaan hissa hai, kisi bhi sheet se bahut patla jise tum alag pakad sako. (Units check karne ke liye dekho ki $1 F = 1 C / V = 1 C^{2} / J = 1 C^{2} N^{- 1} m^{- 1}$ , isliye $ε_{0} A / C$ metres mein aata hai.) Supercapacitors aisi values tak sirf bahut badi internal surface wale electrodes par kuch molecules moti insulating layers use karke pahunchte hain (Section 2.15 ka box dekho).

Example

Ghar par bana parallel plate capacitor

$20 c m$ side wali do square aluminium plates hawa mein $1.5 m m$ door rakhi hain aur $9.0 V$ battery se judi hain. Nikaalo (a) capacitance, (b) har plate par charge, (c) plates ke beech field aur (d) plates par surface charge density.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Chalo explore karein

Kitchen foil ka capacitor

Tumhe chahiye

kitchen aluminium foil ki do sheets, lagbhag 15 cm square
ek clear plastic file sleeve
kuch bhaari kitaabein
capacitance range wala digital multimeter

Pehle guess karo

Agar tum foils ke beech plastic ki motai double kar do, to measured capacitance…

double ho jaayegisame rahegilagbhag aadhi ho jaayegizero ho jaayegi

Ek foil sheet flat bichhao, use plastic sleeve se dhako, aur doosri foil upar aise rakho ki foils overlap karein lekin kahin bhi touch na karein.
Stack ko kitaabon ke neeche dabakar flat karo. Meter ki leads ko dono foils se chhuao aur capacitance note karo.
Upar wali foil ko aise khiskao ki uska sirf aadha area neeche wali se overlap kare, phir se dabakar flat karo aur naapo.
Poora overlap wapas lao, lekin foils ke beech plastic ki do layers rakho, aur ek baar aur naapo.

Table 2.3Tumhare measurements

Arrangement	Naapa gaya C (nF)	Pehle se ratio
Poora overlap, ek layer
Aadha overlap, ek layer
Poora overlap, do layers

Tumhari entries apne aap save hoti hain.

Quick check

4 mein se 1

Ek parallel plate capacitor ka

C = 40 p F

hai. Iska plate separation double aur plate area teen guna kar diya jaata hai. Nayi capacitance kitni hai, pF mein?

Answer dekho

Answer 60 pF

C \propto A / d

40 p F \times 3 / 2 = 60 p F

2.13 Capacitance par Dielectric ka Effect

Capacitor ke gap ko mica ya glass jaise dielectric se bhar do to uski capacitance badh jaati hai, aksar kai guna. Wajah Section 2.10 ka polarisation hai (Fig. 2.21).

3/3Zyaadatar field lines bound charges par khatam hoti hain, isliye kam lines dielectric ko cross karti hain: $E = E_{0} / K$ . Same $Q$ ke liye $V$ , $K$ guna girta hai aur $C$ , $K$ guna badhta hai.

Fig. 2.21Dielectric ke faces par bound charges free charges ke field ka kuch hissa cancel karte hain, isliye same

Q

ke liye voltage girta hai aur capacitance badhti hai.NCERT p. 70

Do vertical plates aamne-saamne hain; left plate ke andar wale face par plus charges aur right plate ke andar wale face par minus charges hain. Pehle step mein aath field lines khaali gap ko plate se plate tak cross karti hain. Doosre step mein ek slab khisak kar gap mein aata hai; positive plate ke paas uske left face par minus signs, right face par plus signs, andar chhote dipoles bhare hain, aur applied field ki aath lines abhi bhi uske aar-paar halki dikhti hain. Teesre step mein aathon lines plates ke paas ki patli hawa wali jagah ko cross karti hain, lekin sirf do slab ke aar-paar jaati hain; baaki minus face par khatam hokar plus face se phir shuru hoti hain, aur ek label par E barabar E0 over K, V barabar V0 over K aur C barabar K C0 likha hai.

DerivationDielectric ke saath capacitance

Plates par free charge ko poore samay $\pm Q$ par fixed rakho, taaki $σ = Q / A$ kabhi na badle. Gap mein kuch na ho to Section 2.12 deta hai

E_{0} = \frac{σ}{ε_{0}}, V_{0} = E_{0} d, C_{0} = \frac{Q}{V_{0}} = \frac{ε_{0} A}{d}

(2.46)

Ab ek slab andar khiskao jo gap ko bhar de. Positive plate ke paas slab ka $- σ_{p}$ face hai, aur negative plate ke paas uska $+ σ_{p}$ face (Fig. 2.21). Slab ke andar se dekhne par har plate aur uske paas wala face milkar density $σ - σ_{p}$ ki ek single sheet ki tarah kaam karte hain, isliye material mein

E = \frac{σ - σ_{p}}{ε_{0}}

(2.47)

V = E d = \frac{(σ - σ_{p}) d}{ε_{0}}

(2.48)

Free charge double karo to linear dielectric mein polarisation bhi double hota hai, isliye $σ_{p}$ bhi double hota hai. Isliye bina cancel hua hissa $σ - σ_{p}$ diye gaye material ke liye hamesha $σ$ ka same fraction hota hai. Us fraction ko $1 / K$ kehkar,

σ - σ_{p} = \frac{σ}{K}

(2.49)

Bound charge free charge ke asar ko kamzor kar sakta hai lekin kabhi palat nahi sakta, isliye $0 < σ_{p} < σ$ aur $K > 1$ . Potential difference aur capacitance ho jaate hain

V = \frac{σ d}{ε_{0} K} = \frac{Q d}{ε_{0} K A}

(2.50)

C = \frac{Q}{V} = \frac{K ε_{0} A}{d}

(2.51)

$K$ ko material ke ek constant mein shaamil kar lena convenient hai. Kisi medium ki permittivity $ε = ε_{0} K$ hai, aur pure number $K$ , dono permittivities ka ratio, uska dielectric constant ya relative permittivity hai:

ε = ε_{0} K

(2.52)

K = \frac{ε}{ε_{0}}

(2.53)

Khaali space reference hai, jiske liye exactly $K = 1$ aur $ε = ε_{0}$ . Eqs. (2.46) aur (2.51) ko saath rakhne par result sabse simple form mein milta hai:

Dielectric ka effect

C = K C_{0}

(2.54)

Definition

Dielectric constant (

K

)

Woh factor jisse capacitor ki capacitance badhti hai jab uske conductors ke beech ki space poori tarah material se bhar di jaaye: $K = C / C_{0} = ε / ε_{0}$ . Yeh 1 se bada ek pure number hai.

ExampleMica ka $K \approx 6$ hai, shuddh paani ka lagbhag 80.

Eq. (2.54) hamari flat-plate derivation se zyada general hai: kisi bhi conductors ke pair ke chaaron taraf ki space ko material se poori tarah bharne par unki capacitance usi $K$ se multiply ho jaati hai. Isse capacitor ek kaam ka measuring instrument ban jaata hai, aur Table 2.4 ki values material ke saath aur bina $C$ ke comparison se aati hain.

Table 2.4Room temperature par dielectric constants aur approximate dielectric strengths

Material	Dielectric constant K	Dielectric strength (MV/m)
Vacuum	1 (exactly)	—
Sookhi hawa	1.0006	3
Teflon (PTFE)	2.1	60
Polystyrene	2.6	24
Kaagaz	3.5	16
Pyrex glass	5.6	14
Mica	5.4–6	—
Paani (20 °C)	80	—
Strontium titanate	310	8

InteractiveCapacitor lab

Parallel plate capacitor ka side view. Sliders plate area, plates ke beech gap, battery voltage aur dielectric slab ki motai set karte hain, aur ek menu slab ka material chunta hai. Ek switch battery ko jodta ya hatata hai; hatane ke baad plates par charge fixed rehta hai. Drawing plates par plus aur minus charges, slab ke faces par bound charges aur field lines dikhati hai jinki spacing field strength batati hai. Readouts capacitance, charge, potential difference, hawa mein aur slab mein field, aur stored energy dete hain, aur agar hawa breakdown hone wali ho to warning bhi.

Fig. 2.22Battery jodkar ya hatakar, plate area, gap aur dielectric badlo, aur

C

Q

V

E

aur stored energy dekho.NCERT p. 70

Example 2.8NCERT Example 2.8 · p. 71

Sheet kitni moti honi chahiye?

Hawa se bhare ek parallel plate capacitor ki plates ka area $0.020 m^{2}$ hai aur fixed spacers unhe $5.0 m m$ door rakhte hain, jisse lagbhag $35 p F$ milta hai. Circuit ko iska $1.5$ guna, lagbhag $53 p F$ , chahiye, lekin plates hil nahi sakti, isliye designer gap mein plates jitne area ki mica ( $K = 6.0$ ) ki ek sheet khiska deta hai. (a) Gap $d$ mein motai $t$ ki sheet ke saath capacitance nikaalo. (b) Mica sheet kitni moti honi chahiye? (c) Kya isse fark padta hai ki sheet gap mein kahan hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2.14 Capacitors ka Combination

Capacitors standard values mein bikte hain, isliye circuit designers aksar zaroori capacitance paane ke liye kai capacitors jodte hain. Kisi bhi combination ko ek equivalent capacitor se badla ja sakta hai jo same potential difference $V$ par supply se same charge $Q$ leta hai, taaki $C = Q / V$ . Capacitors jodne ke do basic tareeke do simple rules dete hain (Fig. 2.23).

1/4Series mein andar wali plates zero net charge wala ek isolated island banati hain, isliye har capacitor par same $\pm Q$ hota hai.

Fig. 2.23Series mein har capacitor par same charge hota hai aur voltages judte hain; parallel mein har capacitor par same voltage hota hai aur charges judte hain.NCERT Fig. 2.26 · p. 71

Upar do capacitors C1 aur C2 do terminals ke beech ek ke baad ek jude hain: series connection. Neeche wahi do capacitors do terminals ke beech parallel branches par hain. Pehle step mein series row highlighted hai: har capacitor par plus Q aur minus Q likha hai, aur andar ki do plates aur unke beech ke wire ke chaaron taraf ek dashed box par net 0 likha hai. Doosre step mein series row ke neeche brackets V1 aur V2 ko jud kar V banate dikhate hain. Teesre step mein parallel circuit highlighted hai, aur ek bracket dono branches ke across same V dikhata hai. Chauthe step mein parallel capacitors par Q1 barabar C1 V aur Q2 barabar C2 V likha hai, aur total Q barabar Q1 plus Q2.

2.14.1 Series mein capacitors

DerivationSeries mein equivalent capacitance

$C_{1}$ aur $C_{2}$ ko supply ke across ek ke baad ek jodo, jisse $C_{1}$ ki bahar wali plate par $+ Q$ aur $C_{2}$ ki bahar wali plate par $- Q$ aata hai. Pehle $C_{1}$ ko dekho, aur ek Gaussian box banao jiske dono ends uski plates ke metal mein hon. Metal mein field zero hai aur gap mein box ki sides ke along chalta hai, isliye box se koi flux nahi guzarta. Isliye aamne-saamne ki surfaces par barabar aur opposite charge hona chahiye: $C_{1}$ ki andar wali plate par $- Q$ . Woh andar wali plate $C_{2}$ ki andar wali plate se wire se judi hai. Dono milkar metal ka ek island banati hain jo neutral shuru hua tha aur kisi aur cheez ko nahi chhoota, isliye ek taraf ka $- Q$ doosri taraf $+ Q$ chhod deta hai. Isliye series chain ke har capacitor par same charge hota hai.

Capacitors ke across potential differences jud kar supply voltage banate hain:

V = V_{1} + V_{2} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}}

(2.55)

\frac{V}{Q} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

(2.56)

Equivalent capacitor same $V$ par same $Q$ rakhega:

C = \frac{Q}{V}

(2.57)

Eq. (2.56) se compare karne par,

\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

(2.58)

Zyada capacitors ke saath kuch nahi badalta: har ek par $Q$ hai aur voltages judte hain,

V = V_{1} + V_{2} + \dots + V_{n} = \frac{Q}{C_{1}} + \frac{Q}{C_{2}} + \dots + \frac{Q}{C_{n}}

(2.59)

Series mein capacitors

\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + \dots + \frac{1}{C_{n}}

(2.60)

Series chain ki equivalent capacitance usmein ke sabse chhote capacitor se bhi chhoti hoti hai: asal mein gaps jud jaate hain. Voltage capacitance ke inverse proportion mein bantta hai, $V_{i} = Q / C_{i}$ , isliye sabse chhota capacitor voltage ka sabse bada hissa leta hai, jo tab matter karta hai jab capacitors ki voltage ratings hon.

2.14.2 Parallel mein capacitors

DerivationParallel mein equivalent capacitance

Ab $C_{1}$ aur $C_{2}$ ko saath-saath jodo, taaki dono junctions ke same pair ke beech hon. Same do points se jude hone se dono same potential difference $V$ par hain, aur har ek utna charge leta hai jitna uski apni capacitance allow karti hai:

Q_{1} = C_{1} V, Q_{2} = C_{2} V

(2.61)

Supply ne total charge diya hai

Q = Q_{1} + Q_{2}

(2.62)

aur equivalent capacitor ko same $V$ par yeh charge rakhna chahiye:

Q = C V = C_{1} V + C_{2} V

(2.63)

C = C_{1} + C_{2}

(2.64)

Parallel mein $n$ capacitors ke liye charges isi tarah judte hain:

Q = Q_{1} + Q_{2} + \dots + Q_{n}

(2.65)

C V = C_{1} V + C_{2} V + \dots + C_{n} V

(2.66)

Parallel mein capacitors

C = C_{1} + C_{2} + \dots + C_{n}

(2.67)

Parallel mein plate areas asal mein jud jaate hain, isliye equivalent capacitance sabse bade capacitor se bhi badi hoti hai. Dhyaan do ki dono rules Chapter 3 ke resistors wale rules ke ulte hain.

Compare

Series aur parallel ek nazar mein

Series

Har ek par same charge $Q$
Voltages judte hain: $V = V_{1} + V_{2} + \dots$
$1 / C = 1 / C_{1} + 1 / C_{2} + \dots$
$C$ sabse chhote $C_{i}$ se bhi chhota hai

Parallel

Har ek ke across same voltage $V$
Charges judte hain: $Q = Q_{1} + Q_{2} + \dots$
$C = C_{1} + C_{2} + \dots$
$C$ sabse bade $C_{i}$ se bhi bada hai

Example 2.9NCERT Example 2.9 · p. 73

Network ko chhota karna

Fig. 2.24 ke network mein $C_{1} = 2.0 μ F$ aur $C_{2} = 4.0 μ F$ , A aur B ke beech parallel mein hain, aur yeh pair B aur D ke beech $C_{3} = 3.0 μ F$ ke saath series mein hai. A aur D ke across $18 V$ supply judi hai. Nikaalo (a) woh single capacitance jo poore network ki jagah le sake, (b) har capacitor mein stored charge aur (c) har ek ke across voltage.

Fig. 2.24Example 2.9 ka network.NCERT p. 73

Junctions A aur B ke beech do capacitors parallel branches par hain: upar 2 microfarads ka C1 aur neeche 4 microfarads ka C2. B se 3 microfarads ka ek single capacitor C3, D tak jaata hai. 18 volts ka ek cell, jiska positive terminal A wali taraf hai, figure ke neeche se D ko wapas A se jodta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

3.0 μ F

aur

6.0 μ F

ke capacitors series mein jude hain. Equivalent capacitance kitni hai, µF mein?

µF

Answer dekho

Answer 2 µF

1 / C = 1 / 3.0 + 1 / 6.0 = 1 / 2.0

, isliye

C = 2.0 μ F

, dono se chhota.

2.15 Capacitor mein Stored Energy

Capacitor charge karne ka matlab hai charge ko ek conductor se doosre tak le jaana, ek aise potential difference ke against jo charge jama hone ke saath badhta jaata hai. Kiya gaya work electrostatic potential energy ke roop mein store hota hai, aur capacitor use bahut jaldi wapas de sakta hai: camera ka flash, defibrillator ka jhatka aur kuch electric buses mein power ka burst, sab stored capacitor energy se aate hain.

3/3Strips jodne se line ke neeche ka triangle bhar jaata hai: $W = \frac{1}{2} Q V$ .

Fig. 2.25Capacitor charge karne ka work uski

V

–

Q

line ke neeche ka area hai:

W = \frac{1}{2} Q V

.NCERT Fig. 2.30 · p. 74

Ek graph ke horizontal axis par charge aur vertical axis par voltage hai. Pehle step mein ek seedhi line origin se final point tak chadhti hai jahan charge Q aur voltage V hai, jis par V prime barabar Q prime over C likha hai. Doosre step mein beech ke kisi charge par line ke neeche delta Q prime width ki ek patli orange strip khadi hai, jis par delta W barabar V prime delta Q prime likha hai. Teesre step mein line ke neeche ka poora triangle shaded hai aur us par W barabar one half Q V likha hai.

DerivationCapacitor mein stored energy

Charging ka matlab hai charge ko thoda-thoda karke ek conductor se doosre tak uthaana. Beech mein conductors par $\pm Q^{'}$ hai aur woh potential difference $V^{'} = Q^{'} / C$ par hain, isliye agla chhota parcel $δ Q^{'}$ negative conductor se positive tak uthaane mein lagta hai

δ W = V^{'} δ Q^{'} = \frac{Q^{'}}{C} δ Q^{'}

(2.68)

jo Fig. 2.25 mein line $V^{'} = Q^{'} / C$ ke neeche ek patli strip ka area hai. $Q^{'} = 0$ se $Q^{'} = Q$ tak strips jodne par triangle ka area milta hai, ya isi ke barabar integral

W = \int_{0}^{Q} \frac{Q^{'}}{C} d Q^{'} = \frac{Q^{2}}{2 C}

$Q = C V$ use karke stored energy ko teen equivalent tareekon se likha ja sakta hai:

Capacitor mein stored energy

U = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V

(2.69)

Half kyun? Saara charge poore voltage $V$ par nahi le jaaya gaya: shuru mein voltage chhota tha, aur average mein har hissa $V / 2$ par paar hua.

Charged capacitor ko flash tube, chhoti motor ya simple resistor ke across jodo, aur stored energy light, motion ya heat ban kar nikalti hai. Kitni nikalti hai yeh sirf us charge par depend karta hai jo capacitor ne rakha tha, uski history par nahi. Ek millisecond mein charge karne wali battery aur ek minute lene wala hand-cranked generator, agar same $Q$ chhodte hain, to same $U$ chhodte hain.

Energy asal mein kahan hai? Charges plates par hain, lekin field gap ko bharta hai. Yeh kehna bahut faayde ka nikalta hai ki energy field mein store hai, us poori space mein phaili hui jo field gherta hai. Uniform field wala parallel plate capacitor hume ispar ek number lagaane deta hai.

DerivationElectric field ki energy density

Stored energy ko surface charge density $σ = Q / A$ aur capacitance $ε_{0} A / d$ ke terms mein likho:

U = \frac{Q^{2}}{2 C} = \frac{(σ A)^{2}}{2} \times \frac{d}{ε_{0} A}

(2.70)

Plates ke beech field hai

E = \frac{σ}{ε_{0}}

(2.71)

isliye $σ = ε_{0} E$ , aur

U = \frac{1}{2} ε_{0} E^{2} \times (A d)

(2.72)

Yahan $A d$ gap ka volume hai, wahi ek jagah jahan field hai. Isliye field ki energy per unit volume hai

Electric field ki energy density

u = \frac{1}{2} ε_{0} E^{2}

(2.73)

Capacitor Eq. (2.73) ke liye bas ek convenient laboratory tha. Yahi energy density har electric field ki hai, chahe woh thundercloud ke chaaron taraf ho, charged kanghi ke ya atom ke: jahan field hai wahan energy hai, har cubic metre mein $\frac{1}{2} ε_{0} E^{2}$ . (Dielectric ke andar yahi argument $\frac{1}{2} ε E^{2}$ deta hai.) Yeh energy density pehla ishaara hai ki fields apne aap mein physical cheezein hain, ek theme jo Chapter 8 mein electromagnetic waves ke saath wapas aati hai.

Definition

Energy density (

u

)

Electric field mein per unit volume stored energy, vacuum mein $u = \frac{1}{2} ε_{0} E^{2}$ . Iski SI unit $J / m^{3}$ hai.

ExampleBreakdown ke kareeb hawa mein, har cubic metre mein lagbhag $40 J$ .

Example

Spark hone wali hawa mein energy

(a) Breakdown se theek neeche, jahan $E = 3.0 \times 10^{6} V / m$ , hawa mein field ki energy density kya hai? (b) Ek thundercloud aur uske neeche ki zameen lagbhag ek parallel plate capacitor ki tarah behave karte hain jiski plates ka area $1.0 \times 10^{7} m^{2}$ hai aur jo $1.0 k m$ door hain. Hawa ke breakdown se pehle yeh 'capacitor' sabse zyada kitni energy store kar sakta hai, iska estimate lagao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2.10NCERT Example 2.10 · p. 75

Energy kahan jaati hai?

$4.0 μ F$ capacitor ko $50 V$ tak charge karke supply se hata diya jaata hai. Phir uske terminals ek uncharged $1.0 μ F$ capacitor ke terminals se jode jaate hain (Fig. 2.26). Nikaalo (a) shuru mein stored energy, (b) charge bant jaane ke baad common potential difference, aur (c) final stored energy. Difference ki wajah batao.

Fig. 2.26Charge share karne se

Q

conserve hota hai lekin stored energy nahi.NCERT p. 75

Do panels. Before: 50 volts tak charged 4 microfarad capacitor par plus aur minus 200 microcoulombs hain, paas mein ek khaali 1 microfarad capacitor; ek energy bar 5.0 millijoules padhta hai. After: dono capacitors common 40 volts par parallel mein jude hain, un par 160 aur 40 microcoulombs hain; energy bar chhota hai, 4.0 millijoules, aur ek dashed box khoye hue 1.0 millijoule ko mark karta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

10 μ F

capacitor

20 V

tak charge hai. Yeh kitni energy store karta hai, mJ mein?

Answer dekho

Answer 2 mJ

U = \frac{1}{2} C V^{2} = \frac{1}{2} (1.0 \times 10^{- 5}) (20)^{2} J = 2.0 \times 10^{- 3} J = 2.0 m J

Table 2.5Is chapter mein mili quantitiesNCERT p. 78

Quantity	Symbol	SI unit	Dimensions	Aam values
Electrostatic potential	$V$	volt, $V = J / C$	$[M L^{2} T^{- 3} A^{- 1}]$	$1.5 V$ (torch cell) se lagbhag $10^{8} V$ (bijli) tak
Capacitance	$C$	farad, $F = C / V$	$[M^{- 1} L^{- 2} T^{4} A^{2}]$	radio circuits mein picofarads se supercapacitors mein hazaaron farads tak
Polarisation	$𝐏$	$C m^{- 2}$	$[L^{- 2} T A]$	$10^{5} V / m$ ke field mein glass mein lagbhag $4 \times 10^{- 6} C m^{- 2}$
Dielectric constant	$K$	koi nahi	dimensionless	vacuum ke liye 1, paani ke liye lagbhag 80, kuch ceramics mein hazaaron
Energy density	$u$	$J m^{- 3}$	$[M L^{- 1} T^{- 2}]$	breakdown se theek neeche hawa mein lagbhag $40 J m^{- 3}$

Flash cards · 13

Formula flashcards

1 / 13

Recap

Summary aur key points

Electrostatic force conservative hai. Charge $q$ ko R se P tak dheere-dheere le jaane mein $q (V_{P} - V_{R})$ work lagta hai, jo potential energy ke roop mein store hota hai (Section 2.1, Section 2.2).
Potential infinity se laaye gaye per unit positive charge ka work hai; sirf differences matter karte hain. Point charge $V = Q / 4 π ε_{0} r$ deta hai (Section 2.3).
Dipole ka potential, $𝐩 \cdot \hat{𝐫} / 4 π ε_{0} r^{2}$ , direction par depend karta hai, $1 / r^{2}$ ki tarah girta hai aur equatorial plane par zero hai (Section 2.4).
Kai charges ke potentials sign wale numbers ki tarah judte hain. Charged shell ka bahar $V = q / 4 π ε_{0} r$ hai aur andar constant $q / 4 π ε_{0} R$ (Section 2.5).
Equipotential surfaces barabar $V$ wale points ko jodti hain. Field unhe right angle par cross karta hai, lower potential ki taraf point karta hai aur uska size $∣ δ V ∣ / δ l$ hai (Section 2.6, Section 2.6.1).
Charges ke set ki potential energy use jodne ka work hai: har pair ke liye ek term $q_{i} q_{j} / 4 π ε_{0} r_{i j}$ (Section 2.7).
External potential mein charge ki energy $q V (𝐫)$ hai, aur uniform field mein dipole ki $U = - 𝐩 \cdot 𝐄$ ; $1 e V = 1.6 \times 10^{- 19} J$ (Section 2.8).
Equilibrium wale conductor mein: andar koi field nahi, extra charge sirf surface par, poore mein ek potential, aur theek bahar $𝐄 = (σ / ε_{0}) \hat{𝐧}$ . Khaali cavity bahar ke fields se shielded rehti hai (Section 2.9).
Field mein dielectric polarise hota hai, induced ya kuch aligned molecular dipoles se; uske faces par bound charge andar field ko kamzor karta hai (Section 2.10).
Capacitor beech mein insulator wale do conductors ka pair hai. Iski capacitance $C = Q / V$ , farads mein, sirf geometry aur dielectric par depend karti hai (Section 2.11).
Parallel plate capacitor mein uniform field $σ / ε_{0}$ hota hai aur $C = ε_{0} A / d$ ; ise dielectric se bharne par $C$ , $K$ se multiply ho jaata hai (Section 2.12, Section 2.13).
Series mein $1 / C = \sum 1 / C_{i}$ (same charge); parallel mein $C = \sum C_{i}$ (same voltage) (Section 2.14).
Capacitor $U = Q^{2} / 2 C = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$ store karta hai, jise density $\frac{1}{2} ε_{0} E^{2}$ wali field energy ki tarah socha ja sakta hai (Section 2.15).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 2.1	Practice Q1
NCERT Exercises 2.2	Practice Q2
NCERT Exercises 2.3	Practice Q3
NCERT Exercises 2.4	Practice Q4
NCERT Exercises 2.5	Practice Q5
NCERT Exercises 2.6	Practice Q6
NCERT Exercises 2.7	Practice Q7
NCERT Exercises 2.8	Practice Q8
NCERT Exercises 2.9	Practice Q9
NCERT Exercises 2.10	Practice Q10
NCERT Exercises 2.11	Practice Q11

Quick check: MCQs

2.1 Introduction3 questions
2.2 Electrostatic Potential3 questions
2.3 Point Charge ka Potential4 questions
2.4 Electric Dipole ka Potential3 questions
2.5 Charges ke System ka Potential3 questions
2.6.1 Field aur potential ka relation4 questions
2.7 Charges ke System ki Potential Energy4 questions
2.8.3 External field mein dipole ki potential energy4 questions
2.9 Conductors ki Electrostatics4 questions
2.10 Dielectrics aur Polarisation4 questions
2.11 Capacitors aur Capacitance3 questions
2.12 Parallel Plate Capacitor4 questions
2.13 Capacitance par Dielectric ka Effect4 questions
2.14.2 Parallel mein capacitors4 questions
2.15 Capacitor mein Stored Energy4 questions

Glossary · 14

Key terms (English + Hindi)

capacitanceधारिताdhaarita: Capacitor ke charge aur uske potential difference ka ratio $C = Q / V$ ; farads mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
capacitorसंधारित्रsandhaaritra: Insulator se alag kiye gaye do conductors, jo barabar aur opposite charges aur electrical energy store karne ke kaam aate hain.Chapter mein pehli baar yahan aaya →
conservative forceसंरक्षी बलsanrakshi bal: Aisa force jiska do points ke beech work liye gaye path par depend nahi karta, isliye uske liye potential energy define ki ja sakti hai.Chapter mein pehli baar yahan aaya →

dielectricपरावैद्युतparaavaidyut: Ek insulating material jiske bound charges field mein khisakte ya ghoomte hain, aur uske andar field ko kuch cancel karte hain.Chapter mein pehli baar yahan aaya →
dielectric constantपरावैद्युतांकparaavaidyutaank: Woh factor $K = ε / ε_{0}$ jisse capacitor ko kisi material se bharne par uski capacitance badhti hai.Chapter mein pehli baar yahan aaya →
dielectric strengthपरावैद्युत सामर्थ्यparaavaidyut saamarthya: Sabse bada field jo insulator tootne se pehle seh sakta hai; sookhi hawa ke liye lagbhag $3 \times 10^{6} V / m$ .Chapter mein pehli baar yahan aaya →

electron voltइलेक्ट्रॉन वोल्टelectron volt: $1 V$ se guzarne wale charge $e$ ki energy ka change: $1 e V = 1.6 \times 10^{- 19} J$ .Chapter mein pehli baar yahan aaya →
electrostatic potentialस्थिरवैद्युत विभवsthirvaidyut vibhav: Charge ko infinity se kisi point tak laane mein per unit positive test charge work; volts mein naapa jaata hai ( $1 V = 1 J / C$ ).Chapter mein pehli baar yahan aaya →
electrostatic potential energyस्थिरवैद्युत स्थितिज ऊर्जाsthirvaidyut sthitij oorja: Kisi charge ko laane, ya charges ke set ko jodne, mein infinity se bina acceleration ke laga work.Chapter mein pehli baar yahan aaya →
electrostatic shieldingस्थिरवैद्युत परिरक्षणsthirvaidyut parirakshan: Conductor ki khaali cavity ke andar electric field ka na hona, bahar charges aur fields kuch bhi hon.Chapter mein pehli baar yahan aaya →
energy densityऊर्जा घनत्वoorja ghanatva: Electric field ke per unit volume mein stored energy, $u = \frac{1}{2} ε_{0} E^{2}$ .Chapter mein pehli baar yahan aaya →
equipotential surfaceसमविभव पृष्ठsamvibhav prishth: Aisi surface jis par potential ki value har jagah same ho; field use right angle par cross karta hai.Chapter mein pehli baar yahan aaya →

polar moleculeध्रुवीय अणुdhruviya anu: Aisa molecule jiske positive aur negative charge ke centres hamesha alag hote hain, jisse uska apna dipole moment hota hai.Chapter mein pehli baar yahan aaya →
polarisationध्रुवणdhruvan: Woh dipole moment per unit volume jo electric field mein dielectric mein ban jaata hai.Chapter mein pehli baar yahan aaya →

Electrostatic Potential and Capacitance: Class 12 Physics Chapter 2 Hinglish mein

Is chapter mein aap kya seekhenge

2.1 Introduction

2.2 Electrostatic Potential

2.3 Point Charge ka Potential

2.4 Electric Dipole ka Potential

2.5 Charges ke System ka Potential

2.6 Equipotential Surfaces

2.6.1 Field aur potential ka relation

2.7 Charges ke System ki Potential Energy

2.8 External Field mein Potential Energy

2.8.1 Single charge ki potential energy

2.8.2 External field mein do charges ke system ki potential energy

2.8.3 External field mein dipole ki potential energy

2.9 Conductors ki Electrostatics

2.10 Dielectrics aur Polarisation

2.11 Capacitors aur Capacitance

2.12 Parallel Plate Capacitor

2.13 Capacitance par Dielectric ka Effect

2.14 Capacitors ka Combination

2.14.1 Series mein capacitors

2.14.2 Parallel mein capacitors

2.15 Capacitor mein Stored Energy

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye