Electric Charges and Fields: Class 12 Physics Chapter 1 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

155 min
3 simulations
29 practice questions
45 MCQs

Is chapter mein aap kya seekhenge

Rub karke charge karne ko electrons ka transfer samjhao, aur charge ki additivity, conservation aur quantisation use karo
Conductors aur insulators mein fark karo aur samjhao ki earthing conductor se charge kyun hata deti hai
Point charge par force nikaalne ke liye Coulomb's law ka vector form aur superposition principle lagao
Point charges aur charges ke systems ka electric field calculate karo, aur field-line diagrams padhna seekho
Electric flux define karo aur field mein rakhi flat surfaces ke liye use compute karo
Electric dipole ka field uske axis aur equatorial plane par derive karo, aur uniform field mein dipole par torque nikaalo
Gauss's law batao aur usse long wire, infinite plane sheet aur thin spherical shell ke fields nikaalo

NCERT Physics Class 12, Chapter 1, pp. 1–44 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

1.1 Introduction

December ki sookhi raat mein woollen sweater sir ke upar se utaaro, to tumhe halki chat-chat sunai de sakti hai, ya andhere kamre mein chhoti neeli sparks bhi dikh sakti hain. Nylon carpet par chal kar steel ke door handle ko chhuo, to ungli mein ek chhota sa jhatka lagta hai. Monsoon ka thunderstorm yahi trick bade scale par karta hai: lightning ka ek stroke kai kilometre lambi spark hota hai.

Fig. 1.1Storm cloud ke andar charge alag ho jaata hai; jab kaafi charge jama ho jaata hai, to ek badi spark ground tak koodti hai.NCERT p. 1

Samtal ground ke upar ek cloud tair raha hai. Cloud ke upar wale hisse mein plus signs bhare hain aur uske base par minus signs ki line hai. Neeche ground par plus signs ki ek line hai, jo cloud ke negative base se attract hui hai. Ek tedha-medha lightning bolt cloud base ko ground se jodta hai, aur ek taraf ek chhota ghar khada hai.

In sab mein charge kisi surface par jama hua, aksar rub karne se, aur phir achanak use nikalne ka raasta mil gaya. Ruke hue charges ki physics ko electrostatics kehte hain: aise charges ek doosre ko kaise push aur pull karte hain, aur apne aas-paas ke space ko kaise badalte hain. Is chapter mein hum electric charge aur uski properties, Coulomb's law, electric field aur field lines, flux, electric dipole aur aakhir mein Gauss's law dekhenge, jo ek shortcut hai aur kuch mushkil field calculations ko bas kuch lines mein badal deta hai.

1.2 Electric Charge

Electron aur electricity dono words elektron se aaye hain, jo amber ka Greek naam hai; amber ek sunehra fossil resin hai. Amber ke tukde ko fur se rub karo to woh sookhe patte aur bhoose ke tinke kheench leta hai; aam taur par maana jaata hai ki lagbhag 600 BCE mein Thales of Miletus ne yeh notice kiya tha. Baad ke experimenters ne yahi behaviour materials ke kai aur pairs mein bhi paaya: silk se rub kiya glass, fur se rub kiya plastic, wool se rub kiya ebonite.

Jab aisi rub ki hui cheezon ko ek doosre ke paas laate hain, to ek saaf pattern dikhta hai. Silk se rub ki hui do glass rods ek doosre ko door dhakelti hain. Fur se rub ki hui do plastic rods bhi ek doosre ko door dhakelti hain. Lekin rub ki hui glass rod aur rub ki hui plastic rod ek doosre ki taraf khinchti hain. Duniya ki har rub ki hui cheez in do families mein se kisi ek mein aati hai, isliye hum kehte hain ki electric charge ki bilkul do kinds hoti hain (विद्युत आवेश). Same family ki cheezein repel karti hain; alag families ki cheezein attract karti hain, jaisa Fig. 1.2 dikhata hai.

Fig. 1.2Like charges repel karte hain; unlike charges attract karte hain.NCERT Fig. 1.1 · p. 2

Teen panels mein ek common bar se chhoti charged balls ke pairs latke hain. Pehle panel mein dono balls positive hain aur dhaage door phail jaate hain; doosre mein dono negative hain aur woh bhi door phailte hain; teesre mein ek ball positive aur doosri negative hai, aur dhaage ek doosre ki taraf jhukte hain. Har ball ke neeche chhote arrows us par lagne wale electric force ki direction dikhate hain.

Rub ki hui glass rod ko wapas usi silk se chhuao, to dono ki paper ko attract karne ki power khatam ho jaati hai. Charge ki dono kinds ek doosre ko cancel kar deti hain, jaise withdrawal deposit ko cancel karta hai. Benjamin Franklin ne ise positive aur negative labels se pakda, jo is tarah chune gaye ki barabar amounts jud kar zero ho jaayein. Kis kind ko plus sign milta hai, yeh sirf convention hai: silk se rub kiye glass ko positive kehte hain, jisse silk, aur fur se rub ki hui plastic rod, negative ho jaati hai. Jis body ke charges bilkul cancel ho jaayein, woh electrically neutral hoti hai.

Definition

Electric charge

Matter ki ek basic property jiski wajah se woh electric forces lagata aur mehsoos karta hai. Iski do kinds hain, positive aur negative: like charges repel karte hain aur unlike charges attract karte hain.

ExampleSilk se rub ki hui glass rod positively charged ho jaati hai.

Charge ka ek simple detector hai electroscope (Fig. 1.3). Iska working part metal ka ek hi continuous piece hai: glass jar ke bahar ek knob, insulating stopper se guzarti ek rod, aur andar latki bahut halki foil leaves ka ek pair, jahan hawa ke jhonke unhe disturb nahi kar sakte. Kyunki metal ek hi conductor hai, knob ko diya gaya charge neeche leaves tak share ho jaata hai. Phir dono leaves par same kind ka charge hota hai aur woh ek doosre ko door dhakelti hain, aur zyada charge unhe aur zyada khol deta hai.

Fig. 1.3Rod ka charge metal par phail kar dono leaves tak pahunchta hai, jo phir ek doosre ko repel karti hain; zyada charge unhe aur zyada kholta hai.NCERT Fig. 1.2 · p. 3

Do identical electroscopes aamne-saamne khade hain. Har ek glass jar hai jismein insulating stopper se ek metal rod guzarti hai, upar ek flat metal knob hai aur rod ke neeche se do patli foil leaves latki hain. Left wale par uncharged likha hai; uski leaves seedhi neeche, saath-saath latki hain. Right wale par charged likha hai; ek negatively charged rod knob ko chhoo rahi hai, knob par aur dono leaves ke paas minus signs dikhte hain, aur leaves phail kar ulta V ban gayi hain.

Rub karne se body charge hoti hi kyun hai? Har atom ke nucleus mein positive protons hote hain aur uske around utne hi negative electrons, isliye aam matter charge se bhara hota hai jo bas balance mein hota hai. Kuch atoms ke sabse bahari electrons dheele bandhe hote hain. Jab do alag materials ko daba kar ek doosre par ghaseeta jaata hai, to inmein se kuch electrons boundary ke paar kheench liye jaate hain. Ek surface par electrons zyada ho jaate hain aur woh negative ban jaati hai; doosri mein kami reh jaati hai aur woh positive ban jaati hai.

3/3Ab rod positive hai aur silk negative, bilkul utne hi amount se.

Fig. 1.4Rub karna sirf electrons ko move karta hai; yeh kabhi charge create nahi karta.NCERT p. 3

Silk cloth ki ek patti ke upar ek glass rod bani hai. Dono mein paanch plus signs aur paanch electrons hain aur dono par neutral likha hai. Rubbing step mein cloth rod se dabta hai aur rod ke teen electrons cloth mein gir jaate hain, rod mein peeche dashed khaali circles chhod kar. Aakhri step mein cloth ko door kheencha jaata hai: rod, jismein teen electrons kam hain, par net plus likha hai, aur silk, jismein teen extra electrons hain, par net minus likha hai.

Do baatein khaas dhyaan dene layak hain. Pehli, rub karna charge ko transfer karta hai; use kabhi banata nahi. Glass utna hi positive hota hai jitna silk negative hota hai. Doosri, electrons ka bahut hi chhota hissa kabhi move karta hai. Ek charged comb ko arabon extra electrons mil sakte hain, phir bhi comb mein khud lagbhag $10^{24}$ electrons hote hain.

Electric forces sirf rub ki hui rods ka tamasha nahi hain. Steel beam ki stiffness, rubber band ka khinchaav, deewar par gecko ke toes ki pakad, woh friction jo tumhari chappal ko phisalne se rokti hai aur tumhare body ka har chemical reaction, sab charged particles ke ek doosre ko pull aur push karne par tika hai. Gravity ko chhod kar, roz ki life mein milne wala lagbhag har force asal mein electrical hai.

Ghar par try karoNCERT p. 2

Sticky tape se charge ki do kinds

Tumhe chahiye

transparent sticky tape ka ek roll
ek saaf, sookha table top
ek pen

Pehle guess karo

Do B strips table se ek hi tarah ukhaadi jaati hain. Jab tum unhe paas laoge to woh…

ek doosre ko attract karengiek doosre ko repel karengiek doosre par koi asar nahi daalengi

Tape ki do strips, har ek lagbhag 10 cm lambi, saaf table par saath-saath chipkao. Har ek ke ek end par chhota tab mod do aur dono tabs par B (bottom) likho.
Dono strips ko tezi se ukhaado. Unhe tabs se pakad kar paas lao. Dekho kya hota hai.
Table par do nayi strips chipkao aur har ek par ek teesri strip dabao, jis par T (top) likha ho. Har pair ko saath mein ukhaado, phir T strip ko uski B strip se tezi se kheench kar alag karo.
Har pairing test karo: T ke saath T, T ke saath B, aur har strip ke saath apni ungli. Apne observations table mein likho.

Table 1.1Tape test ke observations

Paas laaya gaya pair	Prediction	Tumne kya dekha
B aur B
T aur T
T aur B
Koi bhi strip aur tumhari ungli

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 1.9

Timeline

Electric charge ki khoj

c. 600 BCE
Rub kiya amber
Thales of Miletus record karte hain ki rub kiya amber halki cheezon ko attract karta hai.
1600
Ek naya word
William Gilbert amber jaisa behave karne wale kai materials ki list banate hain aur Latin word electricus gadhte hain.
1733
Do kinds
Charles du Fay dikhate hain ki electrification ki do kinds hain jo apni kind ko repel aur doosri ko attract karti hain.
1747
Plus aur minus
Benjamin Franklin dono kinds ko positive aur negative naam dete hain.
1785
Ek inverse-square law
Charles-Augustin de Coulomb charged spheres ke beech ka force measure karte hain.
1897
Electron
J. J. Thomson electron ko pehchaante hain, jo metals mein negative charge ka carrier hai.
1909–1913
Charge daanedaar hai
Robert Millikan ke oil-drop experiments dikhate hain ki charge $e$ ke whole-number multiples mein aata hai.

1.3 Conductors and Insulators

Plastic ruler ko sookhe baalon par rub karo to woh kaagaz ke tukde jhapat leta hai. Steel ruler ko haath mein pakad kar usi tarah rub karo, to kuch hota hi nahi lagta. Kya steel ko charge karna namumkin hai? Jawab is baat mein hai ki har material ke andar charge kitni aazaadi se move kar sakta hai.

Conductor (चालक) mein kuch charges ghoomne ke liye free hote hain. Metals dheele bandhe electrons se bhare hote hain jo ek atom se doosre atom tak drift karte hain. Nal ka paani, insaan ki body aur earth bhi conduct karte hain, kyunki unmein ions hote hain jo move kar sakte hain. Insulator (विद्युतरोधी) jaise rubber, PVC, mica, sookha kaagaz ya glass mein har charge apne atom ya molecule se bandha hota hai aur door nahi ja sakta. Ek teesra group, semiconductors jaise silicon, in dono ke beech aata hai; inse tum Chapter 14 mein miloge.

Extra electrons ka same chhota patch ek metal sphere par aur ek plastic rod par rakho (Fig. 1.5). Metal par electrons ek doosre ko repel karte hain aur, move karne ke liye free hone ki wajah se, tezi se door bhaagte hain jab tak poori surface par na chha jaayein, aur yeh sab second ke bahut chhote hisse mein hota hai. Plastic par woh move nahi kar sakte, isliye patch bilkul wahin rehta hai jahan rakha gaya tha. Chapter 2 dikhata hai ki conductor par charge aakhir mein poori tarah outer surface par hi tik jaata hai.

2/3Metal par electrons ek doosre ko repel karte hain aur poori surface par phail jaate hain; plastic par woh move nahi kar sakte.

Fig. 1.5Conductor par charge aazaadi se move karta hai lekin insulator par wahin tika rehta hai.NCERT p. 3

Left mein ek metal sphere insulating stand par khada hai; right mein ek plastic rod doosre stand par khadi hai. Pehle step mein dono par ek jagah electrons ka cluster rakha jaata hai. Doosre step mein metal ke electrons uski surface par barabar phail gaye hain jabki plastic ke electrons ek end par jhund mein hi hain. Teesre step mein ek wire metal sphere ko earth se jodti hai aur uske electrons beh kar nikal jaate hain, jabki plastic rod ka cluster bana rehta hai.

Ab steel ruler wali baat samajh aati hai. Rub karne par woh charge to leta hai, lekin steel, tumhara haath, tumhari body aur farsh, sab conduct karte hain, isliye charge lagbhag turant earth mein nikal jaata hai. Ruler ko apne haath se insulate karo, jaise use sookhe plastic bag ke through pakdo, to woh plastic wale ruler ki tarah apna charge bachaa ke rakhta hai. Conductor ko earth tak ka raasta dena taaki uska charge beh kar nikal jaaye, earthing ya grounding kehlata hai. Ghar ke plug ki teesri pin appliance ki metal body ko earth karti hai: agar kabhi live wire body ko chhoo le, to current seedha earth mein jaata hai aur fuse ya circuit breaker trip kar deta hai, na ki use chhoone wale insaan ke through behta hai.

Ghar par try karoNCERT p. 4

Kya steel ruler ko charge kiya ja sakta hai?

Tumhe chahiye

ek steel ruler
ek plastic ruler
ek sookha plastic bag
sookhe baal ya woollen kapda
kaagaz ke chhote tukde

Pehle guess karo

Kaagaz ke tukdon ko kaun uthayega?

nange haath mein pakda steel rulerplastic bag ke through pakda steel rulerkoi nahi, kyunki metals kabhi charge nahi hote

Kaagaz ki ek sheet ko phaad kar tukde karo, har tukda chaawal ke daane se chhota ho, aur unhe table par phaila do.
Steel ruler ko nange haath mein pakdo, use sookhe baalon ya wool par tezi se rub karo aur tukdon ke theek upar rakho.
Haath par ek sookha plastic bag chadhao, ruler ko sirf bag ke through pakdo, use phir se rub karo aur tukdon ke paas lao.
Yahi plastic ruler ko nange haath mein pakad kar dohrao.

Compare

Conductors aur insulators

Conductors

Inmein aise charges hote hain jo aazaadi se move karte hain (metals mein electrons, solutions mein ions)
Diya gaya charge poori surface par phail jaata hai
Rub karke tabhi charge ho sakte hain jab insulator se pakde jaayein
Examples: copper, aluminium, insaan ki body, earth

Insulators

Charges apne atoms ya molecules se bandhe hote hain
Diya gaya charge wahin rehta hai jahan rakha gaya
Rub karne se aasaani se charge ho jaate hain
Examples: rubber, PVC, glass, mica, sookha kaagaz

Definition

Conductor

Aisa material jismein electric charges aazaadi se move kar sakte hain, isliye us par rakha charge uski surface par phail jaata hai.

ExampleCopper, aluminium, insaan ki body.

1.4 Basic Properties of Electric Charge

Ab hum jaante hain ki charge ki do kinds hain aur woh ek doosre ko cancel karne ki koshish karti hain. Teen aur properties is book ki har calculation ki neenv hain. Pehle ek kaam ka idealisation: charged body ko point charge maana jaata hai jab uska size problem ki zaroori distances ke comparison mein negligible ho, aur tab hum uska poora charge ek hi point par rakh dete hain. Doosre se ek metre door rakha charged pollen grain ek badhiya point charge hai; ek millimetre ki doori par rakhi do charged plates nahi.

1.4.1 Additivity of charges

Charge bank account ke paise ki tarah judta hai: deposits positive gine jaate hain, withdrawals negative, aur balance unka sum hota hai. Usi tarah, system ke har charge ko uske sign ke saath jodo, to tumhe system ka total charge mil jaata hai. Charge ek scalar hai, iski koi direction nahi hoti, isi wajah se itna simple addition possible hai. $+ 3$ , $- 5$ , $+ 2$ , $- 1$ aur $+ 4 μ C$ wale system ka total charge $+ 3 μ C$ hai. Jis system mein dono signs ke barabar amounts hon, uska net charge zero hota hai, chahe usmein kitna bhi charge ho.

q_{total} = q_{1} + q_{2} + q_{3} + \dots + q_{n}

1.4.2 Charge conserved hota hai

Rub karne se humein ek gehre rule ki pehli jhalak mili: silk ko jitna charge mila, glass ne bilkul utna hi khoya, aur kuch bhi kahin se nahi aaya. Har tarah ke experiments, electrolysis se lekar particle collisions tak, is general law ko confirm karte hain. Ek isolated system ka total charge kabhi nahi badalta; uske andar charge ek body se doosri body mein ja sakta hai, lekin algebraic total fixed rehta hai. Yeh conservation of charge ka law hai.

Conservation ka matlab yeh nahi ki charged particles hamesha rehte hain. Ek free neutron ek proton, ek electron aur ek neutral antineutrino mein decay hota hai: pehle total charge $0$ , baad mein $(+ e) + (- e) + 0 = 0$ . Nucleus ke paas se guzarta high-energy gamma ray ek electron aur ek positron mein badal sakta hai, jinke charges $- e$ aur $+ e$ hain: phir se pehle aur baad mein zero. Particles aate-jaate rehte hain; net charge kabhi nahi badalta.

1.4.3 Quantisation of charge

Jab Robert Millikan ne 1909 aur 1913 ke beech sainkdon chhoti oil drops par charges measure kiye, to unhe kabhi koi random value nahi mili. Har charge ek basic amount ka whole-number multiple tha, jise ab $e$ kehte hain: proton par $+ e$ hota hai aur electron par $- e$ . Tab se measure hua har free charge isi rule ko maanta hai. Yeh quantisation of charge hai:

Quantisation of charge

q = n e, n = 0, \pm 1, \pm 2, \pm 3, \dots

Aam matter ke liye yeh rule ginti par aa jaata hai. $n_{p}$ protons aur $n_{e}$ electrons wali body ka charge $(n_{p} - n_{e}) e$ hota hai, aur do whole numbers ka difference whole number hi hota hai. Millikan se kaafi pehle, Faraday ke electrolysis ke laws ne bhi yahi ishaara kiya tha: kisi bhi element ka ek mole release karne ke liye zaroori charge hamesha ek hi amount, lagbhag $96 500 C$ , ka chhota whole-number multiple hota hai.

Charge coulombs (C) mein measure hota hai. SI pehle ampere ko define karta hai aur coulomb usi se nikalta hai: $1 A$ ka steady current har second kisi bhi point se $1 C$ le jaata hai, isliye $1 C = 1 A s$ . Coulombs mein elementary charge hai

e = 1.602 \times 10^{- 19} C

jahan

$e$: elementary charge, yaani electron ya proton ke charge ka magnitude

Ulta dekho to ek coulomb lagbhag $6.24 \times 10^{18}$ elementary charges hai. Rub karke charge kiye comb ya balloon par iska bahut chhota hissa hi hota hai, aam taur par kuch nanocoulombs se ek microcoulomb ke beech: $1 μ C = 10^{- 6} C$ aur $1 n C = 10^{- 9} C$ .

Agar charge daanedaar hai, to woh smoothly behta aur badalta kyun lagta hai? Kyunki daane behad chhote hain. $1 μ C$ bhi chhah million million se zyada elementary charges hai, aur unmein se ek jodne ya hatane se total $10^{12}$ mein ek part se bhi kam badalta hai. Yeh chaawal ki bori jaisa hai: bori alag-alag daanon se bani hai, phir bhi ek kilogram kam ya zyada ek smooth change lagta hai (Fig. 1.6). Bade scale ke charges ke liye hum charge ko continuous maan sakte hain. Darjan bhar electrons wali oil drop, ya do charges wale ion ke liye quantisation ko ignore nahi kar sakte.

2/2Trillions electrons par steps dikhne ke liye bahut chhote hain, isliye charge continuous lagta hai.

Fig. 1.6Charge

e

ke steps mein aata hai, jo roz ki cheezon mein notice karne ke liye bahut chhote hain.NCERT p. 5

Do graphs body par charge ko jode gaye electrons ki sankhya ke against plot karte hain. Pehle step mein sirf left graph saaf hai: paas se dekho to woh staircase hai, aur ek riser par e size ka bracket laga hai, kyunki har jode gaye electron se charge theek ek step badhta hai. Doosre step mein right graph aata hai: trillions electrons par wahi relation ek smooth seedhi line lagta hai, aur line ke ek chhote spot ke upar rakha gol magnifier andar chhupe chhote steps dikhata hai.

Example 1.1NCERT Example 1.1 · p. 6

Ek coulomb kitna bada hai?

Laboratory ke ek ion source mein ek steady beam har second $2.0 \times 10^{11}$ singly charged positive ions ek insulated metal cup tak pahunchata hai. Cup ko (a) $1.0 C$ aur (b) $1.0 μ C$ collect karne ke liye beam ko kitni der chalna hoga? $e = 1.6 \times 10^{- 19} C$ lo.

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

Example 1.2NCERT Example 1.2 · p. 6

Iron nail mein chhupa charge

Ek iron nail ka mass $10 g$ hai. Iron ka atomic number $26$ aur molar mass $55.85 g / m o l$ hai. Nail mein total positive charge aur total negative charge nikaalo. ( $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ , $e = 1.6 \times 10^{- 19} C$ )

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

Quick check

4 mein se 1

Neutral object ko

+ 8.0 n C

charge dene ke liye usse kitne electrons hatane honge? (Number likho, jaise 3e10.)

Answer dekho

Answer 50000000000

n = q / e = (8.0 \times 10^{- 9} C) / (1.6 \times 10^{- 19} C) = 5.0 \times 10^{10}

electrons.

1.5 Coulomb's Law

Ab tak humne bas itna kaha hai ki charges attract ya repel karte hain. Yeh push ya pull kitna strong hai, aur distance ke saath kaise badalta hai? 1785 mein Charles-Augustin de Coulomb ne chhote charged spheres par dhyaan se measurements karke iska jawab diya. Unka result, Coulomb's law (कूलॉम का नियम), aage aane wali har cheez ki neenv hai.

LawCoulomb's law

Ruke hue do point charges ek doosre par unhe jodne wali seedhi line ke along force lagate hain. Har force charges ke magnitudes ke product ke proportional aur unke beech ki doori ke square ke inversely proportional hota hai. Like charges repel karte hain; unlike charges attract karte hain.

Coulomb's law (magnitude)

F = k \frac{∣ q_{1} q_{2} ∣}{r^{2}}

(1.1)

jahan

$q_{1}, q_{2}$: dono point charges
$r$: unke beech ki doori
$k$: ek positive constant jo units ke choice par depend karta hai

Coulomb ka instrument, torsion balance, ek bahut chhote force ko dikhne wale twist mein badal deta tha. Ek halki horizontal rod ek patle wire se latki thi, jiske ek end par chhoti charged ball thi. Paas laayi gayi doosri charged ball rod ko ghuma deti thi, aur wire tab tak twist hota jab tak wire ki springiness electric force ko balance na kar de; scale se padha gaya twist ka angle us force ko measure karta tha. Charge mushkil tha, kyunki koi instrument use measure nahi kar sakta tha. Lekin Coulomb charge ko bina measure kiye aadha kar sakte the (Fig. 1.7). Jab charged metal ball ek identical uncharged ball ko chhooti hai, to symmetry kehti hai ki dono ko barabar hissa milega. Yeh trick dohraane se ek unknown $q$ se $q / 2$ , $q / 4$ , $q / 8$ , … bante hain.

3/4Har sphere ke paas $q / 2$ rehta hai.

Fig. 1.7Identical neutral sphere ko chhoone se charge bina measure kiye aadha ho jaata hai.NCERT p. 7

Do identical metal spheres, A aur B, insulating stands par khadi hain. Pehle step mein A par aath plus signs hain aur q likha hai, jabki B neutral hai. Doosre step mein B khisak kar A ko chhooti hai; ab har sphere par chaar plus signs hain, aur q over 2 likha hai. Teesre step mein B wapas apni jagah jaati hai, ab bhi q over 2 ke saath. Chauthe step mein B ko discharge karke phir se A se chhuaya gaya hai, isliye ab har sphere par q over 4 hai.

Charges ko waisa hi chhodkar doori double karne par force ek-chauthai reh gaya. Doori fixed rakh kar ek charge aadha karne par force aadha hua, aur dono ko aadha karne par ek-chauthai. Ye observations milkar Eq. (1.1) dete hain.

History seNCERT p. 7

Charles-Augustin de Coulomb

Charles-Augustin de Coulomb (1736–1806) ki training military engineer ke roop mein hui thi aur unhone lagbhag aath saal Martinique mein ek fort ke construction ki nigraani mein bitaaye. Unki shuruaati research poori tarah practical thi: friction, beams ki strength aur mitti ki deewaron ki stability, jo kaam civil engineers aaj bhi use karte hain. Unka torsion balance patle dhaagon se latki compass needles par ek prize jeetne wali study se nikla. Isse 1785 mein unhone chhote charged spheres ke beech ke forces measure kiye, aur baad mein dikhaya ki magnetic poles bhi inverse-square law maante hain. Joseph Priestley ne 1767 mein ek hi clue se electric law ka andaaza lagaya tha: charged khokhla metal can apne andar ke charges par koi force nahi lagata, jo Section 1.14.3 ka shell result hai. Henry Cavendish ne ise experiment se confirm kiya, lekin unke results ek sadi tak unpublished rahe.

Coulomb ke paas charge ki koi unit nahi thi; woh sirf charges ko compare karte the. Lekin jab law pata ho, to usse ek unit define ki ja sakti hai. $k$ ke liye koi bhi positive number chuno, to charge ki unit woh amount ban jaati hai jisse $1 m$ door rakhe do aise charges $k$ newtons ka force mehsoos karein. SI is tarah kaam nahi karta. Itihaas mein usne ampere ko current-carrying wires ke beech ke magnetic forces se fix kiya (aaj woh $e$ ki value fix karta hai), aur coulomb bas $1 A s$ hai. Isliye $k$ hum nahi chun sakte; use measure kiya jaata hai:

k = \frac{1}{4 π ε_{0}} \approx 9 \times 10^{9} N m^{2} C^{- 2}, ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}

jahan

$ε_{0}$: permittivity of free space (vacuum)

Factor $4 π$ jaan-boojh kar likha jaata hai; isse baad ke kai formulas, Gauss's law samet, simple ho jaate hain. Iske saath Coulomb's law aisa dikhta hai

F = \frac{1}{4 π ε_{0}} \frac{∣ q_{1} q_{2} ∣}{r^{2}}

(1.2)

Vacuum mein $1 m$ door rakhe $1 C$ ke do charges lagbhag $9 \times 10^{9} N$ ke force se repel karenge, jo karib ek million tonne ke weight jitna hai. Coulomb itna bada kyun hai? Iska size currents ke magnetic effects ke hisaab se rakha gaya tha. Ek ampere ek maamooli current hai, lekin wires ke beech magnetic forces charges ke beech electric forces se kahin kamzor hote hain, isliye ek ampere ek second mein jo charge le jaata hai woh electrostatics ke standard se bahut bada hai. Isi wajah se electrostatics $μ$ C aur nC mein kaam karti hai.

Force ek vector hai, isliye humein uska size aur woh kis taraf point karta hai, dono batana chahiye. Fig. 1.8 geometry set up karta hai.

DerivationVector form mein Coulomb's law

Maan lo charges $q_{1}$ aur $q_{2}$ origin O se position vectors $𝐫_{1}$ aur $𝐫_{2}$ par hain. Charge 1 se charge 2 ki taraf point karne wala vector hai

𝐫_{21} = 𝐫_{2} - 𝐫_{1}, r_{21} = ∣ 𝐫_{21} ∣, {\hat{𝐫}}_{21} = \frac{𝐫_{21}}{r_{21}}

Isi tarah $𝐫_{12} = 𝐫_{1} - 𝐫_{2} = - 𝐫_{21}$ 2 se 1 ki taraf point karta hai, aur ${\hat{𝐫}}_{12} = - {\hat{𝐫}}_{21}$ . $q_{1}$ ki wajah se $q_{2}$ par lagne wale force ko $𝐅_{21}$ likho. Iska size Eq. (1.2) deta hai aur yeh charges ko jodne wali line ke along hota hai, isliye

𝐅_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{𝐫}}_{21}

(1.3)

Yahan charges apne signs ke saath hain, aur ek hi formula har case cover karta hai. Agar $q_{1}$ aur $q_{2}$ ka sign same hai, to $q_{1} q_{2} > 0$ aur $𝐅_{21}$ , ${\hat{𝐫}}_{21}$ ke along, $q_{1}$ se door point karta hai: repulsion. Agar signs alag hain, to $q_{1} q_{2} < 0$ aur force wapas $q_{1}$ ki taraf point karta hai: attraction.

Labels 1 aur 2 ko aapas mein badalne se $𝐅_{12}$ milta hai, yaani woh force jo $q_{2}$ , $q_{1}$ par lagata hai:

𝐅_{12} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{𝐫}}_{12} = - 𝐅_{21}

Dono forces size mein barabar aur direction mein opposite hain, bilkul jaisa Newton's third law kehta hai.

2/3 $q_{1} q_{2} > 0$ : har force doosre charge se door point karta hai.

Fig. 1.8Force charges ko jodne wali line ke along lagta hai, aur har case mein

𝐅_{12} = - 𝐅_{21}

.NCERT Fig. 1.3 · p. 8

Origin O se position vectors r1 aur r2 do charges q1 aur q2 ki taraf point karte hain, aur ek teesra arrow r21 q1 se q2 tak jaata hai. Doosre step mein dono charges positive hain aur forces unhe jodne wali line ke along ek doosre se door point karte hain: q2 par F21, q1 se door, aur q1 par F12, q2 se door. Teesre step mein q2 negative hai aur dono forces ek doosre ki taraf point karte hain, size mein barabar aur direction mein opposite.

Ek savdhaani: Eq. (1.3) vacuum mein rakhe charges ko describe karta hai, aur hawa ise bahut kam badalti hai. Charges ke beech paani, oil ya glass rakho to us material ke molecules response mein khud ko rearrange karte hain aur force kamzor ho jaata hai; Chapter 2 isse deal karta hai.

SimulationCoulomb's law lab

Ek metre chaude stage par 2 microcoulombs ka fixed positive source charge hai aur ek chhota positive test charge jise tum kahin bhi drag kar sakte ho. Arrows source ke around field dikhate hain, aur jaise-jaise tum test charge ko move karte ho, ek readout us par force newtons mein aur uski direction dikhata hai.

Fig. 1.9Ek metre chaude stage par ek fixed

+ 2 μ C

source charge aur ek movable

+ 1 μ C

test charge. Readout force newtons mein deta hai.

Try karo

Scale bar ki madad se test charge ko source se $0.20 m$ par rakho aur force note karo. Use $0.40 m$ par le jao. Kya force ek-chauthai ho jaata hai?
Test charge ko fixed distance par source ke chaaron taraf circle mein ghumao. Readout mein kya badalta hai, aur kya same rehta hai?
$F = k q_{1} q_{2} / r^{2}$ se $0.10 m$ par force predict karo, phir stage par check karo.

Example 1.3NCERT Example 1.3 · p. 9

Hydrogen atom mein electricity versus gravity

Hydrogen atom mein electron aur proton average mein $5.3 \times 10^{- 11} m$ door hote hain. (a) Electron aur proton ke beech, aur do electrons ke beech, electric force aur gravitational force ka ratio nikaalo. (b) Atom mein electric force nikaalo aur woh har particle ko kitna acceleration dega. $k = 8.99 \times 10^{9} N m^{2} C^{- 2}$ , $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ , $e = 1.60 \times 10^{- 19} C$ , $m_{e} = 9.11 \times 10^{- 31} k g$ , $m_{p} = 1.67 \times 10^{- 27} k g$ use karo.

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

Inverse-square law se bina sab kuch dobara calculate kiye changes ke baare mein jaldi soch sakte hain, jaisa agla example dikhata hai.

Example 1.4NCERT Example 1.4 · p. 10

Chhuao, baanto, phir repel

Insulating stands par do identical chhoti metal spheres A aur B par $+ 8.0 n C$ aur $- 2.0 n C$ ke charges hain. Unke centres $10 c m$ door hain aur woh ek doosre ko attract karti hain. (a) Yeh force nikaalo. (b) Spheres ko aapas mein chhuakar wapas unhi positions par rakha jaata hai. Ab force kitna hai? (c) Naya force original jitna bada ho, iske liye centres ko kitna paas laana hoga? Jab spheres alag hon to unhe point charges maano.

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

1.6 Forces between Multiple Charges

Asli device mein charge shayad hi kabhi akela hota hai: uske darjanon padosi ho sakte hain, jo sab use pull ya push karte hain. Coulomb's law ek baar mein ek pair se deal karta hai. Total force nikaalne ke liye humein ek aur fact chahiye, jo sirf experiment de sakta hai: superposition principle (अध्यारोपण सिद्धांत).

PrinciplePrinciple of superposition

Ek charge doosre par jo force lagata hai, woh bilkul wahi hai jo Coulomb's law us pair ke liye deta hai, chahe aas-paas aur koi bhi charges hon. Kisi bhi charge par total force un sab forces ka vector sum hai jo har doosra charge us par alag-alag lagata hai.

Teen charges ke liye (Fig. 1.10), $q_{1}$ par force $q_{2}$ se Coulomb force plus $q_{3}$ se Coulomb force hai, jo parallelogram (ya triangle) rule se jode jaate hain:

𝐅_{1} = 𝐅_{12} + 𝐅_{13} = \frac{1}{4 π ε_{0}} [\frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{𝐫}}_{12} + \frac{q_{1} q_{3}}{r_{13}^{2}} {\hat{𝐫}}_{13}]

(1.4)

3/3Parallelogram rule total force $𝐅_{1}$ deta hai.

Fig. 1.10Har pair aise interact karta hai jaise baaki maujood hi na hon; phir forces vectors ki tarah judte hain.NCERT Fig. 1.5 · p. 11

Teen charges dikhaye gaye hain: q1 aur q2 positive, q3 negative. Doosre step mein q1 par do force arrows aate hain: F12 q2 se door point karta hai aur ek chhota F13 q3 ki taraf point karta hai. Teesre step mein dashed lines in do arrows par parallelogram poora karti hain aur uske diagonal ke along ek bold arrow total force F1 dikhata hai.

$n$ charges ke saath hum bas jodte jaate hain. Yahan ${\hat{𝐫}}_{1 i}$ woh unit vector hai jo $q_{i}$ se $q_{1}$ ki taraf point karta hai:

𝐅_{1} = 𝐅_{12} + 𝐅_{13} + \dots + 𝐅_{1 n} = \frac{q_{1}}{4 π ε_{0}} \sum_{i = 2}^{n} \frac{q_{i}}{r_{1 i}^{2}} {\hat{𝐫}}_{1 i}

(1.5)

Ye do ideas milkar, har pair ke liye Coulomb's law aur bheed ke liye superposition, poori electrostatics ko chalate hain. Is chapter aur agle chapter ka har result inhi tak trace kiya ja sakta hai.

Example 1.5NCERT Example 1.5 · p. 12

Symmetry, aur gayab charge

Side $a = 20 c m$ wale square ke chaaron corners par barabar charges $q = + 2.0 μ C$ hain. Centre par ek charge $Q = + 1.0 μ C$ rakha hai. (a) $Q$ par net force kitna hai? (b) Corner A ka charge hata diya jaata hai. Ab $Q$ par net force kitna hai?

Central charge

Q

par forces.NCERT p. 12

Left: side a ka square jiske har corner par barabar positive charge hai aur centre par ek chhota positive charge Q; Q par chaar barabar force arrows chaaron corners se door point karte hain. Right: wahi square jiska corner A khaali hai; Q par teen force arrows bache hain aur ek bold arrow khaali corner A ki taraf point karta hai.

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

Example 1.6NCERT Example 1.6 · p. 13

Teen charges: forces jinka sum zero hai

Charges $q_{A} = + 2.0 μ C$ , $q_{B} = - 2.0 μ C$ aur $q_{C} = - 2.0 μ C$ ek right-angled isosceles triangle ke corners par hain, right angle A par hai aur $A B = A C = a = 30 c m$ . Har charge par net force nikaalo, aur check karo ki teeno net forces ka sum zero hai.

Right-angled isosceles triangle ke corners par charges.NCERT p. 13

Dashed sides wala ek right-angled isosceles triangle bana hai. Neeche left ke corner A par positive charge aur right-angle ka mark hai. Neeche right ke corner B aur upar left ke corner C, dono par negative charge hai. Sides AB aur AC par a likha hai aur tirchi side BC par a root 2. Neeche right mein chhote axes dikhate hain ki x AB ke along aur y AC ke along hai.

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

Quick check

3 mein se 1

+ 1.0 μ C

charge

x = 0

par hai aur ek

+ 4.0 μ C

charge

x = 0.30 m

par. Inke beech kis position

x

(metres mein) par teesre charge ko koi net force nahi lagta?

Answer dekho

Answer 0.1 m

Dono repulsions tab balance hote hain jab

\frac{1}{x^{2}} = \frac{4}{(0.30 - x)^{2}}

, isliye

0.30 - x = 2 x

aur

x = 0.10 m

. Yeh point chhote charge ke zyada paas hai.

1.7 Electric Field

Ek field tum pehle se jaante ho, bhale use yeh naam na dete ho. Earth ki surface ke paas har kilogram lagbhag $9.8 N$ se neeche kheencha jaata hai, chahe wahan kheenchne ke liye kuch ho ya na ho: $g = 9.8 N / k g$ jagah ki property hai, object ki nahi. Electric forces ko bhi isi tarah describe kiya ja sakta hai.

'Charge $Q$ , charge $q$ ko push karta hai' kehne ki jagah hum interaction ko do hisson mein todte hain. Pehle, $Q$ ek electric field (विद्युत क्षेत्र) set up karta hai: ek vector $𝐄$ jiski uske aas-paas har point par ek pakki value hoti hai. Doosre, kisi bhi point par rakha charge us force ko mehsoos karta hai jo us point ke field se tay hota hai, chahe field kisi ne bhi banaya ho. Field tab bhi wahan hota hai jab use mehsoos karne ke liye koi charge maujood na ho.

Point charge $Q$ ka field seedha Coulomb's law se nikalta hai. $Q$ se measure kiye position vector $𝐫$ wale point par,

Point charge ka field

𝐄 (𝐫) = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \hat{𝐫}

(1.6)

jahan

$\hat{𝐫}$: $Q$ se point ki taraf point karne wala unit vector
$r$: $Q$ se distance

Us point par rakha charge $q$ phir Coulomb force mehsoos karta hai

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

(1.7)

jo charge guna uski location par field hai:

𝐅 (𝐫) = q 𝐄 (𝐫)

(1.8)

Eq. (1.8) yeh bhi batata hai ki field ko measure kaise karein: point par ek known charge rakho, us par force measure karo aur divide karo. Is probe ko test charge $q_{0}$ kehte hain; field banane wale charges source charges hain. Thermometer ke saath bhi yahi problem hai. Garam chai ke chhote cup mein ek bada, thanda thermometer daalo to tum usi chai ko thanda kar doge jise measure karna tha. Isi tarah bada test charge source charges ko unki jagah se hila dega, isliye probe itna chhota hona chahiye ki unhe disturb na kare. Field limit mein define hota hai

Electric field ki definition

𝐄 = \lim_{q_{0} \to 0} \frac{𝐅}{q_{0}}

(1.9)

Kyunki force $q_{0}$ ke proportional hai, ratio $𝐅 / q_{0}$ same aata hai chahe hum koi bhi chhota test charge use karein: field sources aur point ka hai, probe ka nahi. Doosre shabdon mein, $𝐄$ us point par rakhe har coulomb positive charge par force hai.

Definition

Electric field

Kisi point par per unit positive test charge electric force, $𝐄 = 𝐅 / q_{0}$ . Yeh ek vector hai jiski SI unit N/C hai.

ExampleSaaf mausam wale din ground ke paas lagbhag $100 N / C$ ka neeche ki taraf field hota hai.

Eq. (1.8) field ki SI unit, N/C, bhi tay karta hai. (Agla chapter ek equivalent unit, V/m, deta hai.) Point-charge field ki teen features yaad rakhne layak hain, aur Fig. 1.11 unhe dikhata hai:

Direction. Positive $Q$ ke liye field radially bahar ki taraf point karta hai; negative $Q$ ke liye radially andar ki taraf. Positive test charge $𝐄$ ke along dhakela jaata hai; negative ulti taraf.
Size. Yeh $1 / r^{2}$ ki tarah girta hai: do guna door, field ek-chauthai strong.
Symmetry. Size sirf $r$ par depend karta hai, isliye charge par centred sphere par har jagah same hota hai. Field spherically symmetric hai.

1/3Bahar ki taraf arrows; do guna distance par field ek-chauthai strong hai.

Fig. 1.11Point charge ka field radial hota hai aur

1 / r^{2}

ki tarah girta hai.NCERT Fig. 1.8 · p. 14

Centre mein ek point charge hai aur do circles par aath directions mein field arrows hain, bahari circle do guna door hai. Pehle step mein charge positive hai aur har arrow radially bahar point karta hai; bahari circle ke arrows ek-chauthai lambe hain, jin par E aur E over 4 likha hai. Doosre step mein dashed circles radius r aur 2r ke spheres mark karte hain jin par field strength har jagah same hai. Teesre step mein charge negative hai aur har arrow radially andar point karta hai.

1.7.1 Electric field due to a system of charges

Kai source charges $q_{1}, q_{2}, \dots, q_{n}$ ke liye hum phir superposition ka sahara lete hain. Point P par field, P par rakhe test charge par per unit charge force hai, aur woh force alag-alag Coulomb forces ka vector sum hai. Isliye fields bas jud jaate hain:

𝐄 (𝐫) = 𝐄_{1} + 𝐄_{2} + \dots + 𝐄_{n} = \frac{1}{4 π ε_{0}} \sum_{i = 1}^{n} \frac{q_{i}}{r_{i P}^{2}} {\hat{𝐫}}_{i P}

(1.10)

jahan

$r_{i P}$: $q_{i}$ se P tak ki distance
${\hat{𝐫}}_{i P}$: $q_{i}$ se P ki taraf point karne wala unit vector

Tarika hamesha same hai: P par har charge ka field nikaalo, use arrow ki tarah banao, phir arrows ko vectors ki tarah jodo, aam taur par unhe components mein resolve karke. Example Example 1.8 ise step by step karta hai.

1.7.2 Physical significance of electric field

Jab tum mobile phone mein bolte ho, to uske antenna mein electrons aage-peeche daudte hain. $3 k m$ door tower ke antenna ke electrons response karte hain, lekin turant nahi: lagbhag das microseconds baad. Un microseconds mein asar kisne pahunchaya? Electrons ne nahi, woh to apni jagah par hi rahe. Woh tha electromagnetic field, ek disturbance jo light ki speed $c$ se bahar ki taraf chalta hai.

Koi bhi asar light se tez nahi chalta. Isliye jab charge $q_{1}$ ko achanak hilaya jaata hai, to door ka charge $q_{2}$ tab tak purana force hi mehsoos karta rehta hai jab tak khabar na pahunche (Fig. 1.12). Coulomb's law, jo turant action maanta hai, is gap ko describe nahi kar sakta; field kar sakta hai. Fields energy store aur carry karte hain (sunlight woh field energy hai jo lagbhag aath minute pehle Sun se chali thi), momentum carry karte hain aur apne khud ke laws ke hisaab se badalte hain, isliye physicists unhe asli cheezein maante hain. Faraday sabse pehle is tarah sochne wale the. Ruke hue charges ke liye bhi field ek kaam ka summary hai: ek baar $𝐄$ nikaal lo to kahin bhi rakhe kisi bhi charge par force pata chal jaata hai. General mein, field koi bhi quantity hai jiski har point par value ho; kamre mein temperature ek scalar field hai, aur $𝐄$ ek vector field hai.

3/3Time $d / c$ ke baad lehar $q_{2}$ tak pahunchti hai aur uske force ki direction badal jaati hai.

Fig. 1.12Field mein changes light ki speed se phailte hain, isliye forces delay ke saath respond karte hain.NCERT p. 17

Dashed line par do positive charges door-door hain; right wale ko right ki taraf steady push lagta hai. Doosre step mein left wala charge upar uchhalta hai aur usse mudi hui lehrein phailne lagti hain, jabki right wale charge par force waisa hi rehta hai. Teesre step mein lehrein right wale charge tak pahunch gayi hain aur tabhi uska force arrow jhuk kar left charge ki nayi position se door point karta hai; ek label par delay equals d over c likha hai.

Example 1.7NCERT Example 1.7 · p. 17

Electric field mein girna

Ek evacuated tube mein do horizontal metal plates $2.0 c m$ door hain, aur unke beech $5.0 \times 10^{3} N / C$ ka uniform field hai. (a) Field upar point kar raha hai, aur ek electron ko top plate ke theek neeche rest se chhoda jaata hai. Bottom plate tak pahunchne mein use kitna time lagta hai? (b) Plates ke connections badal diye jaate hain, jisse field ulta ho jaata hai, aur usi jagah ek proton ko rest se chhoda jaata hai. Ise kitna time lagta hai? (c) Dono times ko $2.0 c m$ girte chhote patthar ke time se compare karo, aur check karo ki particles ke liye gravity ignore ki ja sakti hai. ( $e = 1.6 \times 10^{- 19} C$ , $m_{e} = 9.11 \times 10^{- 31} k g$ , $m_{p} = 1.67 \times 10^{- 27} k g$ )

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

Example 1.8NCERT Example 1.8 · p. 18

Do unequal charges ke fields jodna

Point charges $q_{1} = + 4.0 \times 10^{- 8} C$ aur $q_{2} = - 1.0 \times 10^{- 8} C$ $20 c m$ door fixed hain. Nikaalo (a) unke theek beech P par electric field, aur (b) perpendicular bisector par P se $10 c m$ door R par field. (c) Dono charges se guzarti line par field kahan zero hai? $k = 9 \times 10^{9} N m^{2} C^{- 2}$ use karo.

Example 1.8 ke charges aur points.NCERT p. 18

Ek positive charge q1 aur ek negative charge q2 ek dashed horizontal line par 20 cm door hain; line dono charges se aage tak jaati hai aur us par line through the charges likha hai. Point P unke theek beech hai, har ek se 10 cm. Point R, P ke theek 10 cm upar hai, aur dashed lines R ko dono charges se jodti hain.

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

1.8 Electric Field Lines

Field space ke har point ko ek arrow deta hai, jo banane ke liye bahut zyada hain. Fig. 1.13 dikhata hai ki positive point charge ke liye inhe kaise kaabu karein. Kuch directions mein kuch points par arrows banao: sab charge se door point karte hain, aur distance ke saath chhote hote jaate hain. Phir arrows ki har chain se ek smooth curve guzaaro. Ye curves electric field lines (विद्युत क्षेत्र रेखाएँ) hain. Har point par, us se guzarti line ka tangent $𝐄$ ki direction deta hai, aur ek arrowhead batata hai ki tangent ke along kis taraf.

Arrows ko jodne se unki lengths chali jaati hain, phir bhi picture strength ko dikhati hai. Jahan field strong hai wahan lines ghani hoti hain aur jahan weak hai wahan phaili hoti hain. Sirf relative spacing ka matlab hai: har jagah do guna lines banane se physically kuch nahi badalta. Teen dimensions mein 'ghanapan' ka matlab hai lines ke perpendicular rakhe chhote area se guzarne wali lines ki sankhya.

3/3Do guna distance par wahi lines chaar guna bade patch ko cross karti hain.

Fig. 1.13Jahan field strong hai wahan lines paas-paas hoti hain; jahan weak hai wahan door-door.NCERT Fig. 1.12 · p. 19

Centre mein ek positive charge hai. Pehle step mein chhote arrows aath directions mein bahar point karte hain, aur charge se door jaakar chhote hote jaate hain. Doosre step mein arrows ki jagah solah continuous field lines aa jaati hain jo arrowheads ke saath bahar phailti hain. Teesre step mein ek patla cone shade kiya gaya hai; do arcs, ek paas aur ek do guna door, dono ko same teen lines cross karti hain, aur labels dikhate hain ki door wale patch ka area paas wale se chaar guna hai.

DerivationLine density inverse-square law kyun follow karti hai

Pehle humein solid angle chahiye, jo aam angle ka three-dimensional cousin hai. Kisi point par tip wala patla cone us point par centred radius $r$ ke sphere se ek chhota area $Δ S$ kaatta hai. Cone ka solid angle yeh area divided by $r^{2}$ hai, jo steradians mein measure hota hai (poora sphere $4 π$ sr hai). Kyunki kata hua area $r^{2}$ ki tarah badhta hai, same cone har sphere par same jawab deta hai:

Δ Ω = \frac{Δ S}{r^{2}}

Ek patla cone lo jiska solid angle $Δ Ω$ hai aur tip ek point charge par hai. Point charge ki lines seedhi aur radial hoti hain, isliye cone ke andar ki har line uske andar hi rehti hai. Maan lo cone mein $n$ lines hain. Distances $r_{1}$ aur $r_{2}$ par cone ke cross-section ke areas $r_{1}^{2} Δ Ω$ aur $r_{2}^{2} Δ Ω$ hain, aur same $n$ lines dono ko cross karti hain. Isliye per unit area lines ki sankhya hai

\frac{n}{r_{1}^{2} Δ Ω} at r_{1}, \frac{n}{r_{2}^{2} Δ Ω} at r_{2}

Kyunki $n$ aur $Δ Ω$ fixed hain, lines ki density $1 / r^{2}$ ki tarah girti hai, bilkul point charge ke field ki tarah. Aisi lines banana jo charge par shuru hon aur khaali space mein kabhi khatam na hon, apne aap strength ko sahi dikhata hai.

Michael Faraday, jinhe formal mathematics kam aati thi, fields ko isi tarah imagine karte the aur lines of force ki baat karte the. Hum field lines naam pasand karte hain, kyunki magnetic field line moving charge par magnetic force ki direction nahi dikhati, jo uske perpendicular hoti hai. Charges ki har arrangement ka apna pattern hota hai. Do barabar positive charges aisi lines banate hain jo unke beech ke gap se door bulge karti hain, jaise charges ko door dhakela ja raha ho. Ek positive aur ek negative charge aisi lines banate hain jo ek se doosre tak arc banati hain, jaise unhe aapas mein baandh rahi hon. In patterns ko compare karne ke liye simulation mein scroll karo.

Scroll karke explore karo

Field-line patterns

Stage par point charges aur unki field lines bani hain. Scroll karne par stage ek akele positive charge, ek akele negative charge, do barabar positive charges aur barabar-opposite charges ke pair ke beech badalta hai; tum charges ko drag karke pattern ki shape badal sakte ho.

Ek akela positive charge. Lines radially bahar point karti hain aur infinity tak chali jaati hain. Distance ke saath unki spacing badhti hai, jo kamzor hota field dikhata hai.

Ek akela negative charge. Wahi pattern, bas har arrow ulta: lines infinity se aati hain aur charge par khatam hoti hain.

Do barabar positive charges. Lines charges ke beech ke region se door mud jaati hain. Unke theek beech mein field zero hai, aur lines charges ko door dhakelti hui lagti hain.

Barabar aur opposite charges. Lines $+ q$ se nikalti hain aur $- q$ par khatam hoti hain, gap mein ghani ho jaati hain. Yeh pair, ek electric dipole, Section 1.10 ka topic hai. Ek charge ko drag karo aur dekho pattern kaise peeche-peeche chalta hai.

In sab pictures mein kuch general properties common hain, jinhe tum kisi bhi field-line diagram par check kar sakte ho:

Har electrostatic field line positive charge par shuru hoti hai aur negative par khatam hoti hai. Akela charge apni lines infinity tak bhejta hai (agar positive hai) ya infinity se unhe ikattha karta hai (agar negative hai).
Jis region mein koi charge nahi, wahan line kabhi tootti nahi aur kabhi kuch nahi se shuru nahi hoti.
Do lines kabhi milti ya cross nahi karti: crossing point par field ko ek saath do directions chahiye hongi.
Electrostatic field line kabhi khud par band nahi hoti. Chapter 2 iski wajah ruke hue charges ke fields ki ek khaas property mein dikhata hai: woh conservative hote hain.

1.9 Electric Flux

Jaipur mein kisi chhat par rakhe solar panel ke baare mein socho. Sun ke bilkul saamne hone par woh sabse zyada light pakadta hai. Tircha karne par kam pakadta hai, kyunki beam ke across uska area chhota dikhta hai. Edge-on ghuma do to woh lagbhag kuch nahi pakadta. Kitni light pakdi jaati hai, yeh panel ke size, light ki strength aur beam aur panel ke beech ke angle par depend karta hai.

Electric flux (विद्युत फ्लक्स) isi tarah measure karta hai ki kisi surface se kitna electric field guzarta hai. Asal mein kuch behta nahi, lekin yeh idea field-line picture se kaafi milta hai: flux ginta hai ki surface ko kitni field lines cross karti hain.

Surface ke ek patch ka size bhi hota hai aur woh kisi taraf face bhi karta hai, isliye hum use ek vector se describe karte hain, $Δ 𝐒 = Δ S \hat{𝐧}$ . Iski length area $Δ S$ hai, aur direction unit normal $\hat{𝐧}$ hai, jo patch se seedha bahar nikalta hai. Do normals possible hain, har side par ek. Ek akele open patch ke liye hum bas ek chun lete hain aur us par tike rehte hain; closed surface ke hisse wale patch ke liye rule hamesha outward normal hai. Curved surface ko itne chhote patches mein kaata jaata hai ki woh flat maane ja sakein, har ek ka apna normal.

Ab ek chhote patch ko field $𝐄$ mein rakho, uska normal $𝐄$ se angle $θ$ par ho (Fig. 1.15). Use cross karne wali lines woh hain jo field ke across uske projection se guzarti hain, jiska area $Δ S \cos θ$ hai. Isliye hum patch se flux ko aise define karte hain

Chhote patch se electric flux

Δ ϕ = 𝐄 \cdot Δ 𝐒 = E Δ S \cos θ

(1.11)

2/3 $θ = 60^{\circ}$ : flux aadha ho jaata hai, kyunki $\cos 60^{\circ} = 0.5$ .

Fig. 1.15Flux field ke across dikhne wale area,

Δ S \cos θ

, par depend karta hai.NCERT Fig. 1.15 · p. 22

Gyaarah horizontal field lines left se right chalti hain, beech mein ek patla flat patch edge-on dikhta hai aur uska normal vector n. Pehle step mein patch field ke saamne hai, theta 0 degrees, aur aath lines use cross karti hain. Doosre step mein use tircha kiya gaya hai taaki uska normal field se 60 degrees par ho; sirf chaar lines cross karti hain, aur ek dashed segment projected area delta S cos theta dikhata hai. Teesre step mein patch field ke along pada hai, theta 90 degrees, aur koi line use cross nahi karti.

Factor $\cos θ$ ko kisi bhi quantity ke saath joda ja sakta hai. Area ke saath jodo to $Δ S \cos θ$ field ke perpendicular plane par patch ki parchhai hai. Field ke saath jodo to $E \cos θ$ field ka woh hissa hai jo patch ke seedha aar-paar jaata hai. Dono same flux dete hain. Flux ek scalar hai. Yeh positive hota hai jab field patch ko $\hat{𝐧}$ ki direction mein cross karta hai ( $θ < 90^{\circ}$ ) aur negative jab ulti taraf cross karta hai ( $θ > 90^{\circ}$ ), aur iski SI unit $N m^{2} / C$ hai.

Badi ya curved surface ke liye hum use chhote patches mein baantte hain, har ek se flux nikaalte hain aur jodte hain:

ϕ \approx \sum 𝐄 \cdot Δ 𝐒

(1.12)

Sum sirf approximate hai kyunki har patch par $𝐄$ ki ek hi value maani gayi hai. Patches ko chhote se chhota karne par error khatam ho jaati hai; us limit mein sum ek surface integral ban jaata hai.

Definition

Electric flux

Chhote flat patch ke liye $Δ ϕ = 𝐄 \cdot Δ 𝐒$ ; badi surface ke liye uske saare patches par iska sum. Yeh measure karta hai ki surface se kitna field guzarta hai.

Example $200 N / C$ field ke saamne rakhe $0.10 m^{2}$ patch ka flux $20 N m^{2} / C$ hai.

ExampleNCERT p. 22

Field mein frame ghumana

Ek flat rectangular frame, $30 c m \times 20 c m$ , $2.0 \times 10^{3} N / C$ ke uniform electric field mein rakha hai. Isse flux nikaalo jab iska normal field se (a) $0^{\circ}$ , (b) $60^{\circ}$ , (c) $90^{\circ}$ aur (d) $180^{\circ}$ ka angle banata hai.

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

Quick check

3 mein se 1

400 N / C

ka uniform field

0.050 m^{2}

area wali flat surface ke normal se

30^{\circ}

ka angle banata hai. Isse flux

N m^{2} / C

mein nikaalo.

N m²/C

Answer dekho

Answer 17.3 N m²/C

ϕ = E A \cos θ = 400 \times 0.050 \times \cos 30^{\circ} \approx 17.3 N m^{2} / C

1.10 Electric Dipole

Tumhare aas-paas ki zyadatar cheezein electrically neutral hain, phir bhi neutral object field bana sakta hai agar uske positive aur negative charges alag jagahon par hon. Sabse simple example hai electric dipole (विद्युत द्विध्रुव): charges $+ q$ aur $- q$ jo fixed distance $2 a$ par rakhe hain. Unke beech ka midpoint dipole ka centre hai aur unse guzarti line uska axis hai. Hum dipole ko axis ke along ek direction dete hain, jo $- q$ se $+ q$ ki taraf point karti hai.

Kaisa field expect karna chahiye? Dipole ke paas tum ek charge ke doosre se kaafi zyada paas ho, aur wahi charge haavi rehta hai. Door jaakar dono charges lagbhag same distance par hote hain, isliye unke fields lagbhag cancel ho jaate hain, lekin poori tarah nahi. To ek sensible andaaza: dipole ka field distance ke saath akele charge ke $1 / r^{2}$ field se tez girna chahiye. Neeche ki calculation ise confirm karti hai.

1.10.1 The field of an electric dipole

General point se jujhne ki jagah hum field wahan nikaalte hain jahan geometry sabse simple hai: axis ke along, aur equatorial plane mein, yaani centre se guzarta plane jo axis ke right angle par hai. Fig. 1.16 dono cases dikhata hai. Poore time $\hat{𝐩}$ axis ke along $- q$ se $+ q$ ki taraf unit vector hai.

3/3Barabar fields; sideways hisse cancel ho jaate hain aur Q par net field $𝐩$ ke opposite point karta hai.

Fig. 1.16Axis par net field

𝐩

ke along point karta hai; equatorial line par

𝐩

ke against.NCERT Fig. 1.17 · p. 24

Ek negative aur ek positive charge horizontal axis par 2a distance par hain, aur dipole moment p negative se positive ki taraf point karta hai. Doosre step mein axis par point P par ek lamba arrow E plus q positive charge se door point karta hai aur ek chhota ulta arrow E minus q hai; unka sum, ek bold arrow E, p ke along point karta hai. Teesre step mein perpendicular bisector par point Q do barabar field arrows dikhata hai, ek positive charge se door aur ek negative charge ki taraf; inka parallelogram ek bold resultant deta hai jo p ke opposite point karta hai.

DerivationAxis par kisi point par field

Point P ko axis par centre O se distance $r$ par, $+ q$ ke aage rakho. Tab P, $+ q$ se $r - a$ aur $- q$ se $r + a$ par hai. $- q$ ka field $- q$ ki taraf, yaani $- \hat{𝐩}$ ke along, point karta hai; $+ q$ ka field $+ q$ se door, $+ \hat{𝐩}$ ke along:

𝐄_{- q} = - \frac{q}{4 π ε_{0} (r + a)^{2}} \hat{𝐩}

(1.13(a))

𝐄_{+ q} = \frac{q}{4 π ε_{0} (r - a)^{2}} \hat{𝐩}

(1.13(b))

Inhe jodo aur bracket ko common denominator par lao, $(r + a)^{2} - (r - a)^{2} = 4 a r$ use karke:

𝐄 = \frac{q}{4 π ε_{0}} [\frac{1}{(r - a)^{2}} - \frac{1}{(r + a)^{2}}] \hat{𝐩} = \frac{q}{4 π ε_{0}} \frac{4 a r}{(r^{2} - a^{2})^{2}} \hat{𝐩}

(1.14)

Door ( $r ≫ a$ ) denominator mein $r^{2}$ ke saath $a^{2}$ ko chhod sakte hain:

𝐄 = \frac{4 q a}{4 π ε_{0} r^{3}} \hat{𝐩} (r ≫ a)

(1.15)

DerivationEquatorial plane par kisi point par field

Ab point ko equatorial plane par centre se distance $r$ par lo. Dono charges same distance $\sqrt{r^{2} + a^{2}}$ par hain, isliye unke fields size mein barabar hain:

E_{+ q} = \frac{q}{4 π ε_{0}} \frac{1}{r^{2} + a^{2}}

(1.16(a))

E_{- q} = \frac{q}{4 π ε_{0}} \frac{1}{r^{2} + a^{2}}

(1.16(b))

$𝐄_{+ q}$ , $+ q$ se door point karta hai aur $𝐄_{- q}$ , $- q$ ki taraf. Axis ke perpendicular unke components barabar aur opposite hain, isliye cancel ho jaate hain. Axis ke parallel unke components dono $+ q$ se $- q$ ki taraf, $\hat{𝐩}$ ke opposite, point karte hain aur jud jaate hain. Agar $θ$ woh angle hai jo har field axis se banata hai, to $\cos θ = a / \sqrt{r^{2} + a^{2}}$ , aur

𝐄 = - (E_{+ q} + E_{- q}) \cos θ \hat{𝐩} = - \frac{2 q a}{4 π ε_{0} (r^{2} + a^{2})^{3 / 2}} \hat{𝐩}

(1.17)

$r ≫ a$ ke liye:

𝐄 = - \frac{2 q a}{4 π ε_{0} r^{3}} \hat{𝐩} (r ≫ a)

(1.18)

Eqs. (1.15) aur (1.18) ko compare karo. Dono mein charge aur separation sirf combination $2 q a$ ke roop mein aate hain: aadhe charge aur do guna separation wale dipole ka far field bilkul same hota hai. Yeh combination ek naam aur direction ka haqdaar hai. Yeh hai electric dipole moment (विद्युत द्विध्रुव आघूर्ण):

Dipole moment

𝐩 = q \times 2 a \hat{𝐩}

(1.19)

jahan

$p = 2 q a$: iska magnitude, $C m$ mein
$\hat{𝐩}$: direction: axis ke along $- q$ se $+ q$ ki taraf

$𝐩$ ke terms mein far fields ka compact form aisa hai:

𝐄_{axis} = \frac{2 𝐩}{4 π ε_{0} r^{3}} (r ≫ a)

(1.20)

𝐄_{equator} = - \frac{𝐩}{4 π ε_{0} r^{3}} (r ≫ a)

(1.21)

Definition

Electric dipole moment

$2 a$ separation wale charges $\pm q$ ke liye vector $𝐩 = 2 q a \hat{𝐩}$ , jo $- q$ se $+ q$ ki taraf point karta hai. SI unit: $C m$ .

Teen features saaf dikhte hain. Pehla, field $1 / r^{3}$ ki tarah girta hai, point charge se ek power tez, jo hamara andaaza confirm karta hai. Doosra, barabar distances par axial field equatorial field ka do guna hai, aur dono $𝐩$ ke relative opposite senses mein point karte hain. Teesra, point charge ke field ke ulat, dipole ka field har direction mein same nahi hota: yeh $r$ ke saath-saath $𝐫$ aur $𝐩$ ke beech ke angle par bhi depend karta hai.

Exact expressions Eqs. (1.20) aur (1.21) se tabhi kaafi alag hote hain jab $r$ , $a$ se bahut bada na ho. Isliye physicists aksar ek idealised point dipole use karte hain: separation ko zero ki taraf chhota karo aur charge badhao, $p = 2 q a$ fixed rakhte hue. Point dipole ke liye dono equations har distance par sahi hain, aur kuch molecular diameters door se dekha gaya chhota molecule lagbhag point dipole jaisa hi behave karta hai.

Compare

Point charge aur dipole

Point charge $q$

Total charge $q \neq 0$
Field $1 / r^{2}$ ki tarah girta hai
Diye gaye $r$ par har direction mein same strength
Field lines infinity tak jaati hain

Dipole $𝐩$

Total charge zero
Field $1 / r^{3}$ ki tarah girta hai
Strength direction par depend karti hai: axis par equatorial value ka do guna
Field lines $+ q$ se $- q$ tak jaati hain

Example 1.9NCERT Example 1.9 · p. 25

Exact aur approximate dipole fields

Ek chhota dipole $\pm 2.0 μ C$ ke charges se bana hai jo $4.0 m m$ door rakhe hain. (a) Uske centre se $20 c m$ door do points par field nikaalo: A axis par, positive charge ke aage, aur B equatorial plane par. Har ek ko pehle exactly aur phir far-field formula se karo. (b) Axis ke along, centre ke kitne paas tak ja sakte ho jab tak far-field formula $1 %$ chhota na aane lage? ( $k = 9 \times 10^{9} N m^{2} C^{- 2}$ )

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

1.10.2 Physical significance of dipoles

Dipoles charged beads ke pairs se kahin zyada important hain, kyunki kai molecules dipoles hote hain. Water molecule mein oxygen atom shared electrons ko apni taraf kheenchta hai, aur molecule ki mudi hui shape dono hydrogen atoms ko ek hi side rakhti hai. Oxygen wala end thoda negative aur hydrogen wala end thoda positive hota hai, isliye bina koi field lagaaye bhi molecule ka dipole moment hota hai (Fig. 1.17). Aise molecules ko polar molecules kehte hain. Paani ki polarity use namak gholne mein madad karti hai, aur isi wajah se microwave oven, jo water dipoles ko ek second mein arabon baar aage-peeche ghumata hai, khana garam karta hai.

Carbon dioxide ( $C O_{2}$ ) aur methane ( $C H_{4}$ ) ulti kahani batate hain. Inke har bond mein thodi polarity hai, lekin molecules itne symmetric hain ki bond dipoles cancel ho jaate hain: positive charges ka centre negative charges ke centre se mil jaata hai, aur koi permanent dipole moment nahi hota. External field phir bhi aise molecule ko thoda kheench sakta hai, dono centres ko alag karke use ek induced dipole moment de sakta hai. (Positive charges ka centre centre of mass ki tarah nikalta hai, $𝐫_{c} = \sum q_{i} 𝐫_{i} / \sum q_{i}$ , sirf positive charges par sum karke; negative charges ka centre bhi isi tarah nikalta hai.)

Fig. 1.17Shape polarity tay karti hai: symmetric molecule apne bond dipoles cancel kar deta hai, muda hua nahi.NCERT p. 25

Left mein carbon dioxide molecule seedhi line mein hai, oxygen, carbon, oxygen, oxygens par partial charges delta minus aur carbon par delta plus; neeche do chhote arrows opposite directions mein point karte hain aur cancel ho jaate hain, jisse p equals zero aata hai. Right mein water molecule muda hua hai, upar oxygen aur neeche do hydrogens, oxygen par delta minus aur har hydrogen par delta plus; ek bold arrow p oxygen se hydrogens ki taraf point karta hai.

Definition

Polar molecule

Aisa molecule jiske positive aur negative charge ke centres mel nahi khaate, isliye uska permanent dipole moment hota hai.

ExamplePaani, $H_{2} O$ .

1.11 Dipole in a Uniform External Field

Ek rigid dipole lo, jaise water molecule, jiska moment $𝐩$ kisi lagaaye gaye field par depend nahi karta. Use uniform field $𝐄$ mein field se angle $θ$ par rakho (Fig. 1.18). Kya woh khiskega, ghoomega, ya dono?

DerivationTorque on a dipole

Charge $+ q$ par force $q 𝐄$ lagta hai aur charge $- q$ par $- q 𝐄$ . Dono charges par field same hai, isliye ye forces bilkul cancel ho jaate hain: net force zero hai, aur dipole ka centre accelerate nahi karta.

Lekin dono forces ek hi line ke along nahi lagte, isliye milkar ek couple banate hain jo dipole ko ghumata hai. Centre O ke about torques lo. Charge $+ q$ , O se displacement $𝐚$ par hai aur $- q$ , $- 𝐚$ par, jahan $𝐚$ ki length $a$ hai aur yeh $- q$ se $+ q$ ki taraf point karta hai. Dono charges ke liye $𝐫 \times 𝐅$ jodne se milta hai

𝝉 = 𝐚 \times (q 𝐄) + (- 𝐚) \times (- q 𝐄) = 2 q 𝐚 \times 𝐄

Vector $2 q 𝐚$ bilkul dipole moment $𝐩$ hai, isliye

Torque on a dipole

𝝉 = 𝐩 \times 𝐄

(1.22)

Iska size $τ = p E \sin θ$ hai, jo ek force, $q E$ , guna dono lines of action ke beech ki perpendicular distance $2 a \sin θ$ bhi hai. Kyunki net force zero hai, torque har point ke about same hai, sirf O ke about nahi. Fig. 1.18 wale dipole ke liye $𝐩 \times 𝐄$ page ke andar point karta hai, isliye dipole clockwise, $𝐄$ ki taraf, ghoomta hai.

1/3Net force zero, lekin forces alag lines par lagte hain: $2 a \sin θ$ arm ka couple.

Fig. 1.18Barabar aur opposite forces ek couple banate hain:

τ = p E \sin θ

90^{\circ}

par sabse bada aur aligned hone par zero.NCERT Fig. 1.19 · p. 27

Horizontal field lines right ki taraf point karti hain. Ek dipole jiska positive charge upar right mein hai, field se 50 degrees ka angle theta banata hai. Pehle step mein positive charge par right ki taraf force qE aur negative charge par left ki taraf force minus qE lagta hai; dashed lines of action ke beech 2a sin theta ki doori hai, aur ek muda hua arrow dipole ko clockwise ghoomta dikhata hai. Doosre step mein dipole 25 degrees tak ghoom gaya hai aur forces ke beech ka arm chhota hai. Teesre step mein dipole field ke along hai, dono forces ek hi line par lagte hain aur torque zero hai.

Torque hamesha $𝐩$ ko $𝐄$ ki line mein laane ki koshish karta hai. Yeh sabse bada tab hota hai jab dipole field ke right angle par ho ( $\sin 90^{\circ} = 1$ ) aur zero ho jaata hai jab $𝐩$ , $𝐄$ ke parallel ho, jo stable position hai. Yeh tab bhi zero hota hai jab $𝐩$ bilkul $𝐄$ ke against point kare, lekin woh balance unstable hai: zara sa dhakka aur dipole ghoom jaata hai.

ExampleNCERT p. 27

Chhote dipole par torque

Ek dipole $\pm 2.0 n C$ ke charges se bana hai jo $3.0 m m$ door rakhe hain. Use $4.0 \times 10^{5} N / C$ ke uniform field mein rakha gaya hai, uska axis field se $30^{\circ}$ par hai. Nikaalo (a) net force, (b) torque, aur (c) sabse bada torque jo yeh field us par laga sakta hai.

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

Agar field uniform nahi hai, to $+ q$ aur $- q$ par forces ka size alag hota hai, aur dipole ko net force bhi lagta hai, aur aam taur par torque bhi. Sabse simple case lo: $𝐩$ , $𝐄$ ke along, field $x$ ke along point karta hai aur $x$ ke saath strong hota jaata hai. Tab $+ q$ wahan hai jahan field $- q$ wali jagah se zyada strong hai, aur net force $q [E (x + a) - E (x - a)]$ strong field ki taraf point karta hai; chhote dipole ke liye yeh lagbhag $p d E / d x$ hai. Dipole ko ulta ghuma do, taaki $𝐩$ , $𝐄$ ke against point kare, to net force ulta ho jaata hai aur use weak field ki taraf dhakelta hai.

Isi wajah se charged body neutral cheezon ko attract karti hai: baalon par rub kiya balloon deewar se chipak jaata hai, aur charged comb kaagaz ke tukde utha leti hai. Charged body ka field neutral object ke charges ko thoda alag kar deta hai, aur ek induced dipole banata hai jo field ke along line up hota hai. Field charged body ke paas sabse strong hota hai, isliye aligned dipole uski taraf kheencha jaata hai. Charged body ka sign matter nahi karta, kyunki induced dipole hamesha field ke saath line up hota hai.

Ghar par try karoNCERT p. 28

Bina chhuye paani ko modo

Tumhe chahiye

ek plastic comb ya balloon
sookhe baal ya woollen kapda
patli steady dhaar wala ek nal
silk scarf ya polythene bag

Pehle guess karo

Jab charged comb dhaar ke paas aati hai, to paani…

comb ki taraf mud jaayegacomb se door mud jaayeganahi hilega, kyunki paani neutral hai

Nal ko itna kholo ki paani ki ek patli, smooth dhaar gire (lagbhag pencil lead jitni moti).
Plastic comb ya phoole hue balloon ko sookhe baalon ya wool par tezi se rub karke charge karo.
Use nal se kuch centimetre neeche dhaar ke paas lao, paani ko chhuye bina. Dhaar ko dekho.
Comb par ungliyan pher kar use discharge karo, phir use kisi alag material (silk scarf ya polythene bag) par charge karke dohrao.

Dhaar comb ki taraf mudti hai, aur jab comb par doosre sign ka charge ho tab bhi uski taraf hi mudti hai. Comb ka field paani ko polarise karta hai: dhaar ki comb wali side par opposite sign ka charge jama hota hai (water molecules polar hain, aur nal ka paani thoda conduct bhi karta hai). Field comb ke paas sabse strong hai, isliye paas wali side jitni zor se kheenchi jaati hai, door wali side utni zor se dhakeli nahi jaati, aur dhaar comb ki taraf jhool jaati hai.

Abhi nahi kar sakte? Simulation try karo · Fig. 1.19

SimulationNon-uniform field mein dipole

Stage ke left edge ke paas ek bada positive charge fixed hai. Uske right mein ek chhota dipole hai, negative charge bade charge ke zyada paas aur positive charge thoda door. Field lines bade charge se bahar phailti hain aur arrows har charge par force dikhate hain. Tum dono chhote charges ko drag karke dipole ko ghuma sakte ho, ya use paas ya door le ja sakte ho, aur uske dono ends par badalte forces dekh sakte ho.

Fig. 1.19Ek bade fixed charge

+ 4 q

ke paas ek chhota dipole (charges

\pm 0.25 q

); bade charge ka field uske paas strong aur door weak hai. Dipole ke kisi bhi end ko drag karo.

Try karo

Agar Forces off hai to on karo. Dipole ka kaunsa end zyada zor se kheencha jaata hai, aur dipole par net force kis taraf hai?
Dono chhote charges ki jagah badlo taaki positive end bade charge ki taraf ho. Ab dipole kis taraf dhakela jaata hai?
Dono chhote charges ko ek-ek karke right mein kaafi door le jao, unke beech ki doori same rakhte hue. Dono ends par forces ke difference ka kya hota hai, aur kyun?

Quick check

3 mein se 1

3.0 \times 10^{- 9} C m

moment wala dipole

2.0 \times 10^{4} N / C

ke uniform field se

90^{\circ}

par hai. Us par kitna torque lagta hai, N m mein? (jaise 6e-5)

N m

Answer dekho

Answer 0.00006 N m

τ = p E \sin 90^{\circ} = (3.0 \times 10^{- 9}) (2.0 \times 10^{4}) = 6.0 \times 10^{- 5} N m

1.12 Continuous Charge Distribution

Census batata hai ki kisi district ke har square kilometre mein kitne log rehte hain. Yeh figure koi bhi har insaan ki jagah list karke nahi nikaalta; bade area ki population gin kar area se divide ki jaati hai. Badi cheezon par charge ko bhi isi tarah describe kiya jaata hai. Ek charged metal sheet par $10^{12}$ extra electrons ho sakte hain, jo ek-ek karke track karne ke liye bahut zyada hain, isliye hum uske charge ko ek density se describe karte hain.

Charged surface ka ek chhota patch lo, jiska area $Δ S$ hai aur us par charge $Δ Q$ hai. Yahan 'chhota' ka khaas matlab hai: object ke comparison mein bahut chhota, taaki density jagah-jagah badal sake, lekin atoms ke comparison mein bahut bada, taaki usmein millions charges hon. Us jagah surface charge density hai

Surface charge density

σ = \frac{Δ Q}{Δ S}

(1.23)

jahan

$σ$: surface charge density, $C / m^{2}$ mein

Poori surface par yahi dohraane se $σ$ position ka ek smooth function ban jaata hai. Yeh ek average hai: yeh jaan-boojh kar alag-alag electrons aur unke beech ke khaali space ko dhundhla kar deta hai, jaise population density gharon ke beech ke gaps ko ignore karti hai. Patle wire ke along, ya kisi volume mein phaile charge ko bhi isi tarah describe karte hain (Fig. 1.20):

Linear charge density

λ = \frac{Δ Q}{Δ l}

(1.24)

jahan

$λ$: linear charge density, $C / m$ mein

Volume charge density

ρ = \frac{Δ Q}{Δ V}

(1.25)

jahan

$ρ$: volume charge density, $C / m^{3}$ mein

Fig. 1.20Har element hamare scale par bahut chhota hai lekin phir bhi usmein bahut saare elementary charges hain.NCERT Fig. 1.21 · p. 28

Teen panels. Pehla ek mudi hui charged wire dikhata hai jiska ek chhota highlighted piece delta l hai, aur formula lambda equals delta Q over delta l. Doosra ek charged flat sheet dikhata hai jis par chhota patch delta S hai, aur sigma equals delta Q over delta S. Teesra ek charged solid blob dikhata hai jiske andar ek chhota cube delta V hai, aur rho equals delta Q over delta V.

Definition

Charge density

Continuous distribution ka per unit length ( $λ$ , C/m mein), per unit area ( $σ$ , $C / m^{2}$ mein) ya per unit volume ( $ρ$ , $C / m^{3}$ mein) charge.

Aisi distribution ka field nikaalne ke liye hum wapas point charges par jaate hain. Charged region ko chhote volume elements $Δ V$ mein kaato, har ek par charge $ρ Δ V$ ho aur woh itna chhota ho ki point charge maana ja sake. P se distance $r^{'}$ par wala element contribute karta hai

Δ 𝐄 = \frac{1}{4 π ε_{0}} \frac{ρ Δ V}{r^{' 2}} {\hat{𝐫}}^{'}

(1.26)

jahan

${\hat{𝐫}}^{'}$: element se P ki taraf unit vector

aur superposition saare contributions ko jod deta hai:

𝐄 \approx \frac{1}{4 π ε_{0}} \sum_{all Δ V} \frac{ρ Δ V}{r^{' 2}} {\hat{𝐫}}^{'}

(1.27)

Yahan $ρ$ , $r^{'}$ aur ${\hat{𝐫}}^{'}$ har element ke liye alag hain, isliye aam taur par yeh ek lamba sum hai; calculus se yeh integral ban jaata hai, jo hum yahan nahi karenge. Message yeh hai ki Coulomb's law aur superposition kisi bhi charge distribution ka field dete hain, discrete, continuous ya dono ka mix. Symmetric distributions ke liye Gauss's law lagbhag bina kisi sum ke jawab de dega.

Quick check

3 mein se 1

1.5 m

lambi ek patli plastic rod par

6.0 μ C

charge uski length par uniformly phaila hai. Iski linear charge density

μ

C/m mein kitni hai?

μC/m

Answer dekho

Answer 4 μC/m

λ = Q / l = 6.0 μ C / 1.5 m = 4.0 μ C / m

1.13 Gauss's Law

Ab hum flux aur Coulomb's law ko milate hain. Sabse simple case se shuru karo: radius $r$ ke ek imaginary sphere ke centre par point charge $q$ . Sphere se bahar total flux kitna hai?

DerivationSphere se point charge ka flux

Sphere ko chhote patches mein baanto. Patch $Δ 𝐒$ par field Eq. (1.6) deta hai, isliye

Δ ϕ = 𝐄 \cdot Δ 𝐒 = \frac{q}{4 π ε_{0} r^{2}} \hat{𝐫} \cdot Δ 𝐒

(1.28)

Sphere par har point par outward normal radius ke along hota hai, isliye $Δ 𝐒$ , $\hat{𝐫}$ ke parallel hai aur $\hat{𝐫} \cdot Δ 𝐒 = Δ S$ :

Δ ϕ = \frac{q}{4 π ε_{0} r^{2}} Δ S

(1.29)

Har patch charge se same distance $r$ par hai, isliye aage ka factor saare patches ke liye same hai. Jodne par saare patches milkar poora area $4 π r^{2}$ banate hain:

ϕ = \frac{q}{4 π ε_{0} r^{2}} \sum Δ S = \frac{q}{4 π ε_{0} r^{2}} \times 4 π r^{2} = \frac{q}{ε_{0}}

(1.30)

Radius cancel ho gaya. Bade sphere ka area zyada hai, lekin us par field bilkul usi proportion mein kamzor hai, isliye har sphere ke liye flux same hai. Yahin par $k = 1 / 4 π ε_{0}$ ka $4 π$ kaam aata hai.

Yeh result single charge ke around sphere se kahin zyada general hai. Yeh kisi bhi charges ke collection ke around kisi bhi closed surface ke liye sahi hai, aur ise Gauss's law kehte hain.

LawGauss's law

Kisi bhi closed surface se bahar net electric flux us surface ke andar ke total charge divided by $ε_{0}$ ke barabar hota hai.

Humne ise ek charge par centred sphere ke liye dikhaya hai. Doosri shapes ke liye dhyaan do ki flux field lines ginta hai: andar ke charge se nikli har line ko uske around ki kisi bhi closed surface ko cross karna padta hai (net ek baar bahar), jabki bahar ke charge ki line andar aati hai aur nikal jaati hai, kuch nahi jodti. Phir superposition result ko kitne bhi charges tak le jaata hai. Poora proof integral calculus use karta hai.

Gauss's law

ϕ = \oint 𝐄 \cdot d 𝐒 = \frac{q_{enc}}{ε_{0}}

(1.31)

jahan

$q_{enc}$: closed surface ke andar ke saare charges ka algebraic sum
$\oint$: poori closed surface par $𝐄 \cdot Δ 𝐒$ ka sum

3/3Andar aane wali lines nikal bhi jaati hain; net flux zero hai.

Fig. 1.21Closed surface se net flux sirf andar ka charge tay karta hai.NCERT Fig. 1.22 · p. 29

Ek positive point charge se chaubees field lines bahar phailti hain. Pehle step mein ek dashed sphere charge ko gherta hai aur har line use ek baar cross karti hai; ek label flux ko q over epsilon zero batata hai. Doosre step mein sphere ki jagah ek oobad-khaabad closed surface aati hai jo ab bhi charge ko gherti hai aur har line ise ek baar cross karti hai, isliye flux nahi badalta. Teesre step mein ek chhota dashed sphere right mein hai, jo koi charge nahi gherta: dots dikhate hain ki lines kahan andar aati aur kahan nikalti hain, dono taraf barabar, isliye flux zero hai.

Jo surface koi charge nahi gherti, uska net flux zero hona chahiye. Ise ek band katori par try karo: radius $R$ ka hemisphere jis par flat lid hai, aur woh uniform field $𝐄$ mein khadi hai jo lid ke through seedha upar point karta hai. Mudi hui neeche ki side se andar aane wali har line lid se nikal jaati hai. Lid field ke bilkul saamne hai, isliye uska flux $+ E π R^{2}$ hai. Mudi hui part ki parchhai bhi same hai, area $π R^{2}$ ki disc, lekin field usmein andar aata hai, isliye uska flux $- E π R^{2}$ hai. Total zero hai, jaisa Gauss's law kehta hai.

Law use karte waqt ye points hamesha yaad rakho:

Shape matter nahi karti. Result har closed surface ke liye sahi hai, uski shape ya size chahe jo ho.
Sirf andar ka charge ginta hai, signs ke saath. $q_{enc}$ andar ke charges ka algebraic sum hai; woh andar kahan baithe hain, isse koi fark nahi padta.
Lekin field sabka hai. Flux mein $𝐄$ total field hai. Surface ke theek bahar ka charge us par har jagah $𝐄$ badal deta hai; bas woh jitna flux andar bhejta hai utna hi bahar.
Surface tum chunte ho. Is kaam ke liye banayi gayi closed surface Gaussian surface (गाउसीय पृष्ठ) hai. Yeh imaginary hai, isliye ise wahan banao jahan symmetry $E$ ko simple bana de. Ise point charges se door rakho, jahan $E$ bina limit ke badhta hai; phaile hue charge ke beech se kaatna theek hai.
Symmetry hi ise useful banati hai. Law tumhe $E$ seedha tabhi deta hai jab tum aisi surface dhoondh sako jis par jahan bhi flux cross kare, wahan $E$ constant aur normal ke along ho.
Yeh Coulomb's law ka test hai. Agar force $1 / r^{2.01}$ ki tarah chalta, to charge ke around sphere se flux sphere ke radius ke saath thoda badalta. Isliye Gauss's law ke precise checks inverse-square law ke bhi precise checks hain.

Definition

Gaussian surface

Gauss's law lagaane ke liye chuni gayi ek imaginary closed surface, jiski shape aam taur par charges ki symmetry se match ki jaati hai.

Example 1.10NCERT Example 1.10 · p. 31

Badhte field mein cube se flux

Ek region mein electric field $𝐄 = β x^{2} \hat{𝐢}$ hai, jahan $β = 500 N C^{- 1} m^{- 2}$ . Side $0.20 m$ wale cube ke faces coordinate planes ke parallel hain, uska left face $x = 0.10 m$ par aur right face $x = 0.30 m$ par hai. Nikaalo (a) cube se bahar net flux aur (b) uske andar ka charge.

Field

x

ke along point karta hai aur

x^{2}

ki tarah badhta hai.NCERT p. 31

0.2 m lambe edges wala ek cube x-axis par rakha hai, uska left face x equals 0.1 m par aur right face x equals 0.3 m par hai. Ek chhota field arrow E L left face mein andar point karta hai aur ek kaafi lamba arrow E R right face se bahar point karta hai, jo dikhata hai ki field x ke saath badhta hai.

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

Example 1.11NCERT Example 1.11 · p. 32

Charge kahan hai?

Ek region mein electric field $x > 0$ ke liye $𝐄 = + 350 \hat{𝐢} N / C$ aur $x < 0$ ke liye $𝐄 = - 350 \hat{𝐢} N / C$ hai. $8.0 c m \times 6.0 c m$ cross-section wala ek band box $x$ -axis ke along rakha hai, uske end faces $x = - 10 c m$ aur $x = + 20 c m$ par hain. Har face se flux nikaal kar box ke andar ka charge nikaalo. Agar box ko $x$ -axis ke along itna khiskaya jaaye ki uske dono ends positive $x$ par hon, to kya jawab badlega?

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

Dhyaan do ki ends ka $x = 0$ ke about symmetric hona zaroori nahi: field ka size har jagah same hai, isliye sirf yeh matter karta hai ki box plane $x = 0$ ke dono taraf phaila hai. Agar box poora $x > 0$ par hota, to field left face mein andar aata ( $- 1.68$ ) aur right face se nikalta ( $+ 1.68$ ), jisse flux zero aur charge zero hota. Isliye charge khud plane $x = 0$ mein hi hona chahiye, $σ = q / A$ wali ek sheet ki tarah.
$σ = \frac{3.0 \times 10^{- 11} C}{4.8 \times 10^{- 3} m^{2}} \approx 6.2 \times 10^{- 9} C / m^{2}$

Quick check

3 mein se 1

Ek closed surface

+ 3.0 μ C

aur

- 1.0 μ C

ke charges gherti hai. Ek

+ 5.0 μ C

charge theek bahar rakha hai. Surface se bahar net flux

N m^{2} / C

mein kitna hai? (jaise 1.5e5)

N m²/C

Answer dekho

Answer 226000 N m²/C

Sirf ghera hua charge ginta hai:

q_{enc} = 2.0 μ C

, isliye

ϕ = q_{enc} / ε_{0} = (2.0 \times 10^{- 6}) / (8.854 \times 10^{- 12}) \approx 2.26 \times 10^{5} N m^{2} / C

. Bahar ka charge surface par

𝐄

badalta hai lekin net flux nahi.

1.14 Applications of Gauss's Law

General charge distribution ke liye Eq. (1.27) ka sum bahut mehnat ka kaam hai. Jab distribution bahut symmetric ho, to Gauss's law kuch lines mein kaam kar deta hai. Tarika hamesha same hai: (1) symmetry se tay karo ki $𝐄$ ki direction kya hai aur uska size kis cheez par depend kar sakta hai; (2) aisi Gaussian surface chuno jis par jahan bhi flux zero nahi, wahan $E$ constant ho; (3) flux ko $q_{enc} / ε_{0}$ ke barabar rakho aur $E$ ke liye solve karo.

1.14.1 Field due to an infinitely long straight uniformly charged wire

Ek bahut lambi, patli seedhi wire lo jis par per unit length charge $λ$ hai. Pehle symmetry. Poore arrangement ko wire ke around ghumao, ya wire ke along khiskao: yeh bilkul same dikhta hai, isliye field sirf wire se perpendicular distance $r$ par depend kar sakta hai. Direction ke liye, wire ke har chhote piece ko P se perpendicular ke foot ki doosri taraf same distance par ek barabar piece ke saath pair karo. P par unke fields ke wire ke along components barabar aur opposite senses mein hain, jo cancel ho jaate hain, aur sirf woh component bachta hai jo wire se seedha door point karta hai. Isliye $𝐄$ radial hai: $λ > 0$ ho to bahar ki taraf, $λ < 0$ ho to andar ki taraf.

DerivationLambi charged wire ka field

Is symmetry ke hisaab se Gaussian surface chuno: radius $r$ aur length $l$ ka cylinder jiske axis ke along wire hai (Fig. 1.22).

Fig. 1.22Symmetry se field radial hai, isliye sirf curved surface se flux guzarta hai.NCERT Fig. 1.26 · p. 33

Plus signs se bhari ek lambi seedhi wire figure ke aar-paar jaati hai. Radius r aur length l ka ek dashed Gaussian cylinder uske ek hisse ko gherta hai. Field arrows curved surface se seedhe bahar, wire ke perpendicular, point karte hain, jabki flat end faces ko koi arrow cross nahi karta.

Field radial hai, isliye yeh flat end faces ke parallel chalta hai aur unhe koi flux nahi deta. Curved surface par $𝐄$ outward normal ke along hai aur har jagah uska size same hai (har point distance $r$ par hai). Isliye flux $E$ guna curved area hai:

ϕ = E \times 2 π r l

Cylinder wire ki length $l$ gherta hai, jis par charge $λ l$ hai. Gauss's law deta hai

E \times 2 π r l = \frac{λ l}{ε_{0}} \Rightarrow E = \frac{λ}{2 π ε_{0} r}

Vector form mein, jahan $\hat{𝐧}$ P se guzarte aur wire ke perpendicular plane mein wire se radially door point karne wala unit vector hai,

Lambi charged wire

𝐄 = \frac{λ}{2 π ε_{0} r} \hat{𝐧}

(1.32)

Line charge ka field $1 / r$ ki tarah girta hai: distance double karo to field aadha, jo point charge ke $1 / r^{2}$ se dheema girna hai. Do savdhaaniyan. $𝐄 = E \hat{𝐧}$ mein scalar $E$ , $λ$ ka sign rakhta hai: negatively charged wire ke liye $E < 0$ aur field andar ki taraf, $\hat{𝐧}$ ke against, point karta hai, jabki uska size $∣ 𝐄 ∣ = ∣ λ ∣ / 2 π ε_{0} r$ hai. Aur argument ke liye infinite wire chahiye thi, kyunki tabhi symmetry wire ke along field component ko rokti hai. Length $L$ ki asli wire ke liye Eq. (1.32) uske beech ke paas $L$ se kaafi chhoti distances par achha kaam karta hai, aur ends ke paas fail ho jaata hai.

1.14.2 Field due to a uniformly charged infinite plane sheet

Ab uniform surface charge density $σ$ wali ek infinite flat sheet lo, aur $x$ -axis use normal ho. Sheet ko uske apne plane mein khiskane, ya $x$ -axis ke around ghumaane se kuch nahi badalta. Isliye $𝐄$ , $y$ ya $z$ par depend nahi kar sakta, aur sheet ke along uska koi component nahi ho sakta. $σ > 0$ ke liye yeh dono taraf sheet se seedha door point karta hai.

DerivationInfinite charged sheet ka field

Ek Gaussian 'pillbox' use karo: ek chhota cylinder jo sheet ko right angle par cheerta hai, aur jiske dono taraf area $A$ ke flat end faces hain (Fig. 1.23).

Fig. 1.23Flux sirf dono end faces se nikalta hai.NCERT Fig. 1.27 · p. 34

Plus signs se dhaki ek patli vertical sheet edge-on dikhti hai. Ek dashed cylinder use right angle par cheerta hai, dono taraf area A ke flat end faces hain, jin par face 1 aur face 2 likha hai. Field arrows dono taraf sheet se door point karte hain, aur dono end faces se seedhe bahar nikalte hain.

Field curved side ke parallel chalta hai, isliye us side se koi flux nahi. Face 1 par outward normal $- x$ ke along point karta hai aur $𝐄$ bhi; face 2 par dono $+ x$ ke along point karte hain. Har end face $+ E A$ deta hai, isliye total flux $2 E A$ hai. Ghera hua charge cylinder ke andar sheet ka patch hai, $σ A$ :

2 E A = \frac{σ A}{ε_{0}} \Rightarrow E = \frac{σ}{2 ε_{0}}

Vector form mein, jahan $\hat{𝐧}$ har side par sheet se door point karne wala unit normal hai,

Infinite charged sheet

𝐄 = \frac{σ}{2 ε_{0}} \hat{𝐧}

(1.33)

Hairani ki baat hai ki $E$ sheet se distance par depend nahi karta. Iski wajah samajhne ka ek tarika: sheet ko do guna door se dekho. Uska har patch ab chaar guna kamzor hai, lekin tumhari aankh se fixed solid angle ka cone sheet ka chaar guna hissa gherta hai, isliye kuch nahi badalta. Isliye infinite sheet ka field har side par uniform hai: $σ > 0$ ho to sheet se door, $σ < 0$ ho to uski taraf. Sheet ko cross karte waqt $𝐄$ ka $x$ -component $- σ / 2 ε_{0}$ se $+ σ / 2 ε_{0}$ par jump karta hai. Surface charge field ko discontinuous bana deta hai; volume charge (jaise Example Example 1.12 mein) nahi. Asli, finite sheet aisa behaviour sirf apne size ke comparison mein chhoti distances par, edges se door, dikhati hai.

1.14.3 Field due to a uniformly charged thin spherical shell

Aakhir mein radius $R$ ka ek patla spherical shell lo jis par charge $q$ uniformly phaila hai, isliye $σ = q / 4 π R^{2}$ . Arrangement apne centre O ke about har direction se same dikhta hai, isliye kisi bhi point P par field radial hona chahiye aur uska size sirf distance $r = O P$ par depend kar sakta hai. Natural Gaussian surface O par centred radius $r$ ka sphere hai (Fig. 1.24).

DerivationCharged spherical shell ka field

Shell ke bahar ( $r > R$ ). P se guzarte Gaussian sphere par field ka size har jagah same $E$ hai aur woh outward normal ke along point karta hai, isliye flux $E \times 4 π r^{2}$ hai. Sphere poora charge gherta hai, $q = σ \times 4 π R^{2}$ :

E \times 4 π r^{2} = \frac{q}{ε_{0}} \Rightarrow E = \frac{q}{4 π ε_{0} r^{2}} = \frac{σ R^{2}}{ε_{0} r^{2}}

Charged shell ke bahar

𝐄 = \frac{q}{4 π ε_{0} r^{2}} \hat{𝐫} (r \geq R)

(1.34)

Shell ke andar ( $r < R$ ). Gaussian sphere se flux phir $E \times 4 π r^{2}$ hai, lekin ab sphere koi charge nahi gherta:

Charged shell ke andar

E \times 4 π r^{2} = 0 \Rightarrow E = 0 (r < R)

(1.35)

1/2Gaussian sphere poora $q$ gherta hai: field O par rakhe point charge jaisa hai.

Fig. 1.24Bahar se shell apne centre par rakhe point charge jaisa behave karta hai; andar field zero ho jaata hai.NCERT Fig. 1.28 · p. 35

Radius R ke patle spherical shell par chaaron taraf barabar plus signs hain. Pehle step mein ek bada dashed Gaussian sphere bahar ke point P se guzarta hai; field arrows radially bahar point karte hain aur label E equals q over four pi epsilon zero r squared batata hai. Doosre step mein ek chhota dashed Gaussian sphere shell ke andar, andar ke point P se guzarta hai; yeh koi charge nahi gherta, bahar ke arrows halke pad jaate hain, aur label par E equals zero likha hai.

To bahar se uniformly charged shell bilkul apne centre par rakhe point charge jaisa dikhta hai; Newton ne spherical planet ki gravity ke baare mein yahi prove kiya tha. Andar field har point par zero hai, sirf centre par nahi. Kyunki yeh conclusion inverse-square law par tika hai, charged khokhle conductor ke andar field dhoondhne wale experiments is baat ke sabse sensitive tests mein se hain ki Coulomb's law mein exponent bilkul $2$ hai.

Example 1.12NCERT Example 1.12 · p. 36

Charged ball ke andar aur bahar

Insulating material ki ek ball, radius $R = 5.0 c m$ , par charge $Q = + 2.0 μ C$ uske volume mein uniformly phaila hai. Uske centre se distance $r$ par field ka expression nikaalo, andar aur bahar dono, aur use $r = 2.0 c m$ , $5.0 c m$ aur $10 c m$ par evaluate karo. ( $k = 9 \times 10^{9} N m^{2} C^{- 2}$ )

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

Quick check

4 mein se 1

Ek lambi seedhi wire par

λ = 2.0 μ C / m

hai. Usse

30 c m

par field N/C mein kitna hai? (jaise 4.5e4)

N/C

Answer dekho

Answer 120000 N/C

E = λ / (2 π ε_{0} r) = 2 k λ / r = 2 (9 \times 10^{9}) (2.0 \times 10^{- 6}) / 0.30 = 1.2 \times 10^{5} N / C

Table 1.2Is chapter mein mili quantitiesNCERT p. 40

Quantity	Symbol	SI unit	Dimensions
Electric field	$𝐄$	$N / C$ (ya $V / m$ )	$[M L T^{- 3} A^{- 1}]$
Electric flux	$ϕ$	$N m^{2} / C$ (ya $V m$ )	$[M L^{3} T^{- 3} A^{- 1}]$
Dipole moment	$𝐩$	$C m$	$[L T A]$
Linear charge density	$λ$	$C / m$	$[L^{- 1} T A]$
Surface charge density	$σ$	$C / m^{2}$	$[L^{- 2} T A]$
Volume charge density	$ρ$	$C / m^{3}$	$[L^{- 3} T A]$
Permittivity of free space	$ε_{0}$	$C^{2} N^{- 1} m^{- 2}$	$[M^{- 1} L^{- 3} T^{4} A^{2}]$
Vector area element	$Δ 𝐒$	$m^{2}$	$[L^{2}]$

Flash cards · 12

Formula flashcards

1 / 12

Recap

Summary aur key points

Charge do kinds mein aata hai, positive aur negative: like charges repel karte hain, unlike charges attract karte hain. Rub karna electrons transfer karke bodies ko charge karta hai; convention se silk se rub kiya glass positive hai (Section 1.2).
Conductors charge ko aazaadi se move karne dete hain; insulators use ek jagah pakde rakhte hain. Earthing conductor se charge nikaal deti hai (Section 1.3).
Charge additive hai (sign wala scalar), har isolated system mein conserved hai aur quantised hai: $q = n e$ , jahan $e = 1.602 \times 10^{- 19} C$ (Section 1.4).
Coulomb's law: $𝐅_{21} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{21}^{2}} {\hat{𝐫}}_{21}$ , jahan $k = 1 / 4 π ε_{0} \approx 9 \times 10^{9} N m^{2} C^{- 2}$ . Dono charges par forces barabar aur opposite hain (Section 1.5).
Electron aur proton ke beech electrical attraction gravitational attraction se lagbhag $10^{39}$ guna zyada hai (Example 1.3).
Superposition: har pair force doosre charges se affect nahi hota, aur total force pair forces ka vector sum hai (Section 1.6).
Electric field per unit positive test charge force hai, $𝐄 = 𝐅 / q_{0}$ . Point charge ka field radial hai aur $1 / r^{2}$ ki tarah girta hai; kai charges ke fields vectors ki tarah judte hain (Section 1.7).
Field lines positive se negative charges tak jaati hain, kabhi cross nahi karti, electrostatics mein kabhi closed loops nahi banati, aur jahan field strong ho wahan ghani hoti hain (Section 1.8).
Chhote patch se flux $Δ ϕ = 𝐄 \cdot Δ 𝐒$ hai; closed surfaces ke liye normal bahar ki taraf point karta hai (Section 1.9).
Dipole ( $𝐩 = 2 q a$ , $- q$ se $+ q$ ki taraf) ka far field $1 / r^{3}$ ki tarah girta hai: axis par $2 𝐩 / (4 π ε_{0} r^{3})$ aur equatorial plane par $- 𝐩 / (4 π ε_{0} r^{3})$ (Section 1.10).
Uniform field mein dipole par koi net force nahi lagta lekin torque $𝝉 = 𝐩 \times 𝐄$ lagta hai; non-uniform field mein net force bhi lagta hai, isi wajah se charged objects neutral objects ko attract karte hain (Section 1.11).
Continuous charge ko densities $λ$ , $σ$ aur $ρ$ se describe karte hain; Coulomb's law aur superposition ab bhi field dete hain (Section 1.12).
Gauss's law: kisi bhi closed surface se bahar net flux $q_{enc} / ε_{0}$ hai (Section 1.13).
Symmetry ke saath Gauss's law lambi wire ke liye $E = λ / 2 π ε_{0} r$ , infinite sheet ke liye $E = σ / 2 ε_{0}$ , aur charged shell ke bahar $q / 4 π ε_{0} r^{2}$ lekin andar zero deta hai (Section 1.14).

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 1.1	Practice Q1
NCERT Exercises 1.2	Practice Q2
NCERT Exercises 1.3	Practice Q3
NCERT Exercises 1.4	Practice Q4
NCERT Exercises 1.5	Practice Q5
NCERT Exercises 1.6	Practice Q6
NCERT Exercises 1.7	Practice Q7
NCERT Exercises 1.8	Practice Q8
NCERT Exercises 1.9	Practice Q9
NCERT Exercises 1.10	Practice Q10
NCERT Exercises 1.11	Practice Q11
NCERT Exercises 1.12	Practice Q12
NCERT Exercises 1.13	Practice Q13
NCERT Exercises 1.14	Practice Q14
NCERT Exercises 1.15	Practice Q15
NCERT Exercises 1.16	Practice Q16
NCERT Exercises 1.17	Practice Q17
NCERT Exercises 1.18	Practice Q18
NCERT Exercises 1.19	Practice Q19
NCERT Exercises 1.20	Practice Q20
NCERT Exercises 1.21	Practice Q21
NCERT Exercises 1.22	Practice Q22
NCERT Exercises 1.23	Practice Q23

Quick check: MCQs

1.2 Electric Charge4 questions
1.3 Conductors and Insulators3 questions
1.4.3 Quantisation of charge4 questions
1.5 Coulomb's Law4 questions
1.6 Forces between Multiple Charges3 questions
1.7.2 Physical significance of electric field4 questions
1.8 Electric Field Lines3 questions
1.9 Electric Flux3 questions
1.10.2 Physical significance of dipoles4 questions
1.11 Dipole in a Uniform External Field3 questions
1.12 Continuous Charge Distribution3 questions
1.13 Gauss's Law3 questions
1.14.3 Field due to a uniformly charged thin spherical shell4 questions

Glossary · 15

Key terms (English + Hindi)

charge densityआवेश घनत्वaavesh ghanatva: Per unit length ( $λ$ ), area ( $σ$ ) ya volume ( $ρ$ ) charge.Chapter mein pehli baar yahan aaya →
conductorचालकchaalak: Aisa material jismein charge aazaadi se move karta hai, jaise metal.Chapter mein pehli baar yahan aaya →
conservation of chargeआवेश संरक्षणaavesh sanrakshan: Isolated system ka total charge kabhi nahi badalta.Chapter mein pehli baar yahan aaya →
Coulomb's lawकूलॉम का नियमkooloam ka niyam: Do point charges ke beech force $q_{1} q_{2}$ ke aur $1 / r^{2}$ ke proportional hota hai, aur unhe jodne wali line ke along lagta hai.Chapter mein pehli baar yahan aaya →

dipole momentद्विध्रुव आघूर्णdvidhruv aaghoorn: $𝐩 = 2 q a$ , $- q$ se $+ q$ ki taraf; SI unit C m.Chapter mein pehli baar yahan aaya →

electric chargeविद्युत आवेशvidyut aavesh: Matter ki ek property jiski wajah se woh electric forces lagata aur mehsoos karta hai; yeh positive ya negative hota hai, aur like charges repel karte hain.Chapter mein pehli baar yahan aaya →
electric dipoleविद्युत द्विध्रुवvidyut dvidhruv: Do barabar aur opposite charges $\pm q$ jo chhoti distance $2 a$ par rakhe hain.Chapter mein pehli baar yahan aaya →
electric fieldविद्युत क्षेत्रvidyut kshetra: Kisi point par per unit positive test charge force, $𝐄 = 𝐅 / q_{0}$ , N/C mein.Chapter mein pehli baar yahan aaya →
electric field lineविद्युत क्षेत्र रेखाvidyut kshetra rekha: Aisa curve jiska tangent field direction deta hai; ghani lines ka matlab strong field.Chapter mein pehli baar yahan aaya →
electric fluxविद्युत फ्लक्सvidyut flux: Surface par $\sum 𝐄 \cdot Δ 𝐒$ ; use cross karne wali field lines ka measure.Chapter mein pehli baar yahan aaya →

Gaussian surfaceगाउसीय पृष्ठgaussiya prishth: Gauss's law lagaane ke liye use hone wali imaginary closed surface.Chapter mein pehli baar yahan aaya →

insulatorविद्युतरोधीvidyutrodhi: Aisa material jismein charges aazaadi se move nahi kar sakte, jaise glass ya plastic.Chapter mein pehli baar yahan aaya →

polar moleculeध्रुवीय अणुdhruveey anu: Permanent dipole moment wala molecule, jaise paani.Chapter mein pehli baar yahan aaya →

quantisation of chargeआवेश का क्वांटमीकरणaavesh ka kvaantameekaran: Har free charge $e = 1.6 \times 10^{- 19} C$ ka whole-number multiple hota hai: $q = n e$ .Chapter mein pehli baar yahan aaya →

superposition principleअध्यारोपण सिद्धांतadhyaropan siddhaant: Kai charges se total force (ya field) alag-alag contributions ka vector sum hota hai.Chapter mein pehli baar yahan aaya →

Electric Charges and Fields: Class 12 Physics Chapter 1 Hinglish mein

Is chapter mein aap kya seekhenge

1.1 Introduction

1.2 Electric Charge

1.3 Conductors and Insulators

1.4 Basic Properties of Electric Charge

1.4.1 Additivity of charges

1.4.2 Charge conserved hota hai

1.4.3 Quantisation of charge

1.5 Coulomb's Law

1.6 Forces between Multiple Charges

1.7 Electric Field

1.7.1 Electric field due to a system of charges

1.7.2 Physical significance of electric field

1.8 Electric Field Lines

1.9 Electric Flux

1.10 Electric Dipole

1.10.1 The field of an electric dipole

1.10.2 Physical significance of dipoles

1.11 Dipole in a Uniform External Field

1.12 Continuous Charge Distribution

1.13 Gauss's Law

1.14 Applications of Gauss's Law

1.14.1 Field due to an infinitely long straight uniformly charged wire

1.14.2 Field due to a uniformly charged infinite plane sheet

1.14.3 Field due to a uniformly charged thin spherical shell

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