CLASS 12 PHYSICS · CHAPTER 1 · READ
Electric Charges and Fieldsवैद्युत आवेश तथा क्षेत्रElectric Charges and Fields
Charge and its properties → Coulomb’s law → many charges → electric field → field lines → flux → dipole → dipole in a field → continuous charges → Gauss’s law and its uses. NCERT order, simple words, with live simulations.आवेश और उसके गुण → कूलॉम नियम → अनेक आवेश → विद्युत क्षेत्र → क्षेत्र रेखाएँ → फ्लक्स → द्विध्रुव → क्षेत्र में द्विध्रुव → संतत आवेश → गाउस नियम और उसके उपयोग। NCERT के क्रम में, आसान भाषा, लाइव सिमुलेशन के साथ।Charge aur uske properties → Coulomb’s law → kai charges → electric field → field lines → flux → dipole → field mein dipole → continuous charges → Gauss’s law aur uske uses. NCERT order, simple words, live simulations ke saath.
Board Notes: Pass to Topperबोर्ड नोट्स: पास से टॉपर तकBoard Notes: Pass se Topper tak🔥 Top topic came in 7 of 7 years · see what to learn first🔥 सबसे ऊपर वाला टॉपिक 7 में से 7 वर्ष आया · देखें पहले क्या पढ़ें🔥 Top topic 7 mein se 7 saal aaya · dekhiye pehle kya padhna haiDidn’t understand a paragraph? Tap it (or its Explain button) → Hindi / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या समझें बटन पर) टैप करें → अंग्रेज़ी / हिंग्लिश / सरल अंग्रेज़ी।Koi paragraph samajh nahi aaya? Us par (ya Samjho button par) tap kijiye → Hindi / English / Simple English.
Try it yourself — live simulationsख़ुद करके देखें — लाइव सिमुलेशनKhud karke dekhiye — live simulations
Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams
▶ Animate: dipole turning in a uniform field▶ एनिमेट: एकसमान क्षेत्र में घूमता द्विध्रुव▶ Animate: uniform field mein ghoomta dipoleAnimated · 3D
▶ Animate: counting flux through a closed surface▶ एनिमेट: बंद पृष्ठ से फ्लक्स की गिनती▶ Animate: closed surface se flux ki gintiAnimatedIn this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
- 1.1IntroductionभूमिकाIntroduction
- 1.2Electric Chargeवैद्युत आवेशElectric Charge
- 1.3Conductors and Insulatorsचालक तथा विद्युतरोधीConductors aur Insulators
- 1.4Basic Properties of Electric Chargeवैद्युत आवेश के मूल गुणElectric Charge ke Basic Properties
- 1.4.1Additivity of chargesआवेशों की योज्यताCharges ki Additivity
- 1.4.2Charge is conservedआवेश संरक्षित हैCharge conserved hai
- 1.4.3Quantisation of chargeआवेश का क्वांटमीकरणCharge ka Quantisation
- 1.5Coulomb’s Lawकूलॉम नियमCoulomb’s Law
- 1.6Forces between Multiple Chargesबहुल आवेशों के बीच बलKai Charges ke beech Force
- 1.7Electric Fieldविद्युत क्षेत्रElectric Field
- 1.7.1Electric field due to a system of chargesआवेशों के निकाय के कारण विद्युत क्षेत्रCharges ke system ka Electric Field
- 1.7.2Physical significance of electric fieldविद्युत क्षेत्र का भौतिक महत्वElectric Field ka Physical Meaning
- 1.8Electric Field Linesविद्युत क्षेत्र रेखाएँElectric Field Lines
- 1.9Electric Fluxविद्युत फ्लक्सElectric Flux
- 1.10Electric Dipoleविद्युत द्विध्रुवElectric Dipole
- 1.10.1The field of an electric dipoleविद्युत द्विध्रुव का क्षेत्रElectric Dipole ka Field
- 1.10.2Physical significance of dipolesद्विध्रुवों का भौतिक महत्वDipoles ka Physical Meaning
- 1.11Dipole in a Uniform External Fieldएकसमान बाह्य क्षेत्र में द्विध्रुवUniform External Field mein Dipole
- 1.12Continuous Charge Distributionसंतत आवेश वितरणContinuous Charge Distribution
- 1.13Gauss’s Lawगाउस नियमGauss’s Law
- 1.14Applications of Gauss’s Lawगाउस नियम के अनुप्रयोगGauss’s Law ke Applications
- 1.14.1Field due to an infinitely long straight uniformly charged wireअनंत लंबे सीधे एकसमान आवेशित तार के कारण क्षेत्रInfinitely long straight charged wire ka Field
- 1.14.2Field due to a uniformly charged infinite plane sheetएकसमान आवेशित अनंत समतल चादर के कारण क्षेत्रInfinite charged plane sheet ka Field
- 1.14.3Field due to a uniformly charged thin spherical shellएकसमान आवेशित पतले गोलीय कोश के कारण क्षेत्रThin charged spherical shell ka Field
1.1IntroductionभूमिकाIntroduction
Have you ever felt a tiny spark or crackle while taking off a sweater in winter, or a small shock when touching a car door or a metal handle after walking on a carpet? Lightning in the sky is the same thing on a giant scale. All of these happen because of electric charges that collect on our body and on objects. The physics of charges at rest is called electrostatics — the study of the forces, fields and potentials that come from static charges.क्या आपने कभी सर्दियों में स्वेटर उतारते समय चटकने की आवाज़ या छोटी-सी चिंगारी देखी है, या कालीन पर चलने के बाद कार के दरवाज़े या धातु के हैंडल को छूने पर हल्का झटका महसूस किया है? आकाश में बिजली कड़कना भी यही है, बस बहुत बड़े पैमाने पर। यह सब हमारे शरीर और वस्तुओं पर जमा होने वाले विद्युत आवेशों के कारण होता है। स्थिर (विरामावस्था में) आवेशों की भौतिकी को स्थिरवैद्युतिकी कहते हैं — स्थिर आवेशों से उत्पन्न बलों, क्षेत्रों और विभवों का अध्ययन।Kya aapne kabhi sardiyon mein sweater utaarte waqt chatakne ki awaaz ya chhoti si spark dekhi hai, ya carpet par chalne ke baad car ke darwaaze ya metal handle ko chhoone par halka jhatka mehsoos kiya hai? Aasmaan mein bijli kadakna bhi yahi hai, bas bahut bade scale par. Ye sab hamare body aur cheezon par jama hone wale electric charges ki wajah se hota hai. Rest wale (static) charges ki physics ko electrostatics kehte hain — static charges se bane forces, fields aur potentials ki padhai.Sparks from a sweater, a shock from a car door and lightning all come from electric charge. Electrostatics studies charges that are not moving.
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1.2Electric Chargeवैद्युत आवेशElectric Charge
Around 600 BC, the Greek thinker Thales of Miletus noticed that amber rubbed with wool or silk attracts light things like bits of paper. The word electricity comes from the Greek word elektron, meaning amber. Many pairs of materials behave like this when rubbed together.लगभग 600 ईसा पूर्व यूनान के थेल्स (मिलेटस) ने देखा कि ऊन या रेशम से रगड़ा गया अंबर (तृणमणि) काग़ज़ के टुकड़ों जैसी हल्की चीज़ों को आकर्षित करता है। Electricity शब्द यूनानी शब्द इलेक्ट्रॉन से बना है, जिसका अर्थ है अंबर। बहुत से पदार्थों के जोड़े रगड़ने पर ऐसा व्यवहार करते हैं।Lagbhag 600 BC mein Greek thinker Thales of Miletus ne dekha ki wool ya silk se ragda hua amber kaagaz ke tukdon jaisi halki cheezon ko kheenchta hai. Electricity shabd Greek word elektron se aaya hai, jiska matlab hai amber. Bahut se materials ke jode ragadne par aisa karte hain.About 600 BC, a Greek thinker named Thales of Miletus saw something strange. When amber (a hard, yellow tree resin) was rubbed with wool or silk, it pulled light things like small bits of paper towards it. The word electricity comes from the Greek word elektron, which means amber. Many other pairs of materials also do this when we rub them together.
Simple experiments show that there are only two kinds of charge. Rub two glass rods with silk and hang them near each other — they repel. Rub a plastic rod with fur and bring it near the glass rod — they attract. So: like charges repel and unlike charges attract (see the diagram). By convention (named by Benjamin Franklin), the charge on a glass rod rubbed with silk is called positive, and the charge on a plastic rod rubbed with fur is negative. A body with no net charge is electrically neutral.सरल प्रयोग दिखाते हैं कि आवेश केवल दो प्रकार के होते हैं। रेशम से रगड़ी दो काँच की छड़ों को पास-पास लटकाएँ — वे प्रतिकर्षित करती हैं। फर से रगड़ी प्लास्टिक की छड़ को काँच की छड़ के पास लाएँ — वे आकर्षित करती हैं। अतः सजातीय आवेश प्रतिकर्षित और विजातीय आवेश आकर्षित करते हैं (चित्र देखें)। परिपाटी के अनुसार (नाम बेंजामिन फ्रैंकलिन ने दिए) रेशम से रगड़ी काँच की छड़ का आवेश धनात्मक और फर से रगड़ी प्लास्टिक की छड़ का आवेश ऋणात्मक कहलाता है। जिस वस्तु पर कोई नेट आवेश नहीं है, वह वैद्युत उदासीन है।Simple experiments dikhaate hain ki charge sirf do type ke hote hain. Silk se ragdi do glass rods ko paas-paas latkao — wo ek-doosre ko repel karti hain. Fur se ragdi plastic rod ko glass rod ke paas laao — wo attract karti hain. Yaani like charges repel, unlike charges attract (diagram dekhiye). Convention ke hisaab se (naam Benjamin Franklin ne diye) silk se ragdi glass rod ka charge positive aur fur se ragdi plastic rod ka charge negative kehlaata hai. Jis cheez par koi net charge nahi, wo electrically neutral hai.Only two kinds of charge exist: positive and negative. Same kinds push apart; opposite kinds pull together.
What really happens on rubbing? Electrons move from one body to the other. The body that loses electrons becomes positive; the body that gains them becomes negative. No charge is created — it is only transferred. For example, when a glass rod is rubbed with silk, some electrons go from the glass to the silk.रगड़ने पर असल में क्या होता है? इलेक्ट्रॉन एक वस्तु से दूसरी वस्तु पर चले जाते हैं। जो वस्तु इलेक्ट्रॉन खोती है वह धनावेशित हो जाती है; जो पाती है वह ऋणावेशित। कोई आवेश बनता नहीं — केवल स्थानांतरित होता है। जैसे, रेशम से काँच की छड़ रगड़ने पर कुछ इलेक्ट्रॉन काँच से रेशम पर चले जाते हैं।Ragadne par asal mein kya hota hai? Electrons ek body se doosri body par chale jaate hain. Jo body electrons khoti hai wo positive ho jaati hai; jo paati hai wo negative. Koi charge banta nahi — sirf transfer hota hai. Jaise silk se glass rod ragadne par kuch electrons glass se silk par chale jaate hain.What really happens when we rub two things? Electrons move from one body to the other. The body that loses electrons becomes positive. The body that gets extra electrons becomes negative. No new charge is made — charge only moves (is transferred) from one body to another. For example, when we rub a glass rod with silk, some electrons go from the glass to the silk.
A gold-leaf electroscope detects charge. Touch its metal cap with a charged rod: charge flows down the metal rod to the two thin gold leaves. Both leaves get the same kind of charge, so they repel and spread apart. The wider they open, the more charge there is.स्वर्णपत्र विद्युतदर्शी आवेश का पता लगाता है। इसकी धातु की टोपी को आवेशित छड़ से छुएँ: आवेश धातु की छड़ से होते हुए दो पतली स्वर्ण पत्तियों तक पहुँचता है। दोनों पत्तियों पर एक ही प्रकार का आवेश आता है, इसलिए वे प्रतिकर्षित होकर फैल जाती हैं। जितना अधिक फैलाव, उतना अधिक आवेश।Gold-leaf electroscope charge ka pata lagata hai. Iske metal cap ko charged rod se chhuo: charge metal rod se hote hue do patli gold leaves tak pahunchta hai. Dono leaves par same type ka charge aata hai, isliye wo repel karke phail jaati hain. Jitna zyada phailaav, utna zyada charge.A gold-leaf electroscope is a simple instrument that tells us if a body has charge. Touch its metal cap (the round top) with a charged rod. The charge flows down the metal rod into two thin gold leaves. Both leaves now get the same kind of charge, so they push each other away and spread apart. The wider the leaves open, the more charge there is.
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1.3Conductors and Insulatorsचालक तथा विद्युतरोधीConductors aur Insulators
Conductors allow electricity (charge) to pass through them easily, because they have electrons that are comparatively free to move. Metals, the human and animal body, and the earth are conductors. Insulators offer high resistance: most non-metals like glass, porcelain, plastic, nylon and wood. (A third group, semiconductors, lies in between — Chapter 14.)चालक विद्युत (आवेश) को आसानी से अपने अंदर से जाने देते हैं, क्योंकि उनमें अपेक्षाकृत स्वतंत्र रूप से गति कर सकने वाले इलेक्ट्रॉन होते हैं। धातुएँ, मनुष्य व जंतुओं का शरीर और पृथ्वी चालक हैं। विद्युतरोधी (कुचालक) अधिक प्रतिरोध देते हैं: अधिकांश अधातुएँ जैसे काँच, पोर्सिलेन, प्लास्टिक, नायलॉन और लकड़ी। (तीसरा समूह अर्धचालक बीच में आता है — अध्याय 14।)Conductors electricity (charge) ko aasani se apne andar se jaane dete hain, kyunki unmein comparatively free electrons hote hain jo move kar sakte hain. Metals, insaan aur jaanwaron ka body, aur earth conductors hain. Insulators zyada resistance dete hain: zyadatar non-metals jaise glass, porcelain, plastic, nylon aur wood. (Teesra group semiconductors beech mein aata hai — Chapter 14.)Conductors let electricity (charge) pass through them easily. This is because they have electrons that are quite free to move. Metals, the human and animal body, and the earth are conductors. Insulators do not let charge pass easily (they have high resistance). Most non-metals are insulators, like glass, porcelain, plastic, nylon and wood. A third group, semiconductors, comes in between these two (you will study them in Chapter 14).
Charge given to a conductor spreads over its whole surface at once; charge put on an insulator stays where it was put. That is why a plastic comb gets charged when you comb dry hair, but a metal spoon held in your hand does not: the charge on the metal leaks through your body to the ground. Hold the metal by a plastic or wooden handle and rub it without touching the metal — now it does show charging.चालक को दिया गया आवेश तुरंत उसके पूरे पृष्ठ पर फैल जाता है; विद्युतरोधी पर रखा आवेश वहीं रहता है जहाँ रखा गया। इसीलिए सूखे बालों में कंघी करने पर प्लास्टिक की कंघी आवेशित हो जाती है, पर हाथ में पकड़ा धातु का चम्मच नहीं: धातु का आवेश हमारे शरीर से होकर ज़मीन में चला जाता है। धातु को प्लास्टिक या लकड़ी के हत्थे से पकड़कर (धातु को छुए बिना) रगड़ें — तब वह आवेशित होती दिखती है।Conductor ko diya gaya charge turant uski poori surface par phail jaata hai; insulator par rakha charge wahin rehta hai jahan rakha gaya. Isiliye sookhe baalon mein kanghi karne par plastic comb charge ho jaati hai, par haath mein pakda metal spoon nahi: metal ka charge hamare body se hokar zameen mein chala jaata hai. Metal ko plastic ya wooden handle se pakadkar (metal ko chhue bina) ragdo — tab wo charged hota dikhta hai.If you put charge on a conductor, it spreads over the whole surface at once. If you put charge on an insulator, it stays at the same spot. This is why a plastic comb gets charged when you comb dry hair. But a metal spoon held in your hand does not get charged. The charge on the metal escapes through your body into the ground. Now hold the metal with a plastic or wooden handle and rub it without touching the metal. This time, it does get charged.
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1.4Basic Properties of Electric Chargeवैद्युत आवेश के मूल गुणElectric Charge ke Basic Properties
When the sizes of charged bodies are very small compared with the distances between them, we treat them as point charges — all the charge is taken to be at one point.जब आवेशित वस्तुओं का आकार उनके बीच की दूरी की तुलना में बहुत छोटा हो, तो उन्हें बिंदु आवेश मानते हैं — पूरा आवेश एक बिंदु पर माना जाता है।Jab charged objects ka size unke beech ki doori ke comparison mein bahut chhota ho, to unhe point charges maante hain — poora charge ek point par maana jaata hai.Sometimes the charged bodies are very small compared with the distance between them. Then we treat each one as a point charge. This means we imagine that all its charge sits at one single point.
1.4.1Additivity of chargesआवेशों की योज्यताCharges ki Additivity
Charges add up like ordinary numbers (scalars), taking their signs into account. The total charge of a system of charges q₁, q₂, …, qₙ is q₁ + q₂ + … + qₙ. Example: +1, +2, −3, +4 and −5 units add to −1 unit. Like mass, charge has no direction; unlike mass, charge can be positive or negative.आवेश साधारण संख्याओं (अदिशों) की तरह जुड़ते हैं, उनके चिह्नों के साथ। आवेशों q₁, q₂, …, qₙ के निकाय का कुल आवेश q₁ + q₂ + … + qₙ है। उदाहरण: +1, +2, −3, +4 और −5 इकाई का योग −1 इकाई। द्रव्यमान की तरह आवेश की कोई दिशा नहीं होती; पर द्रव्यमान के विपरीत आवेश धनात्मक या ऋणात्मक हो सकता है।Charges normal numbers (scalars) ki tarah judte hain, apne signs ke saath. Charges q₁, q₂, …, qₙ ke system ka total charge q₁ + q₂ + … + qₙ hai. Example: +1, +2, −3, +4 aur −5 unit ka sum −1 unit. Mass ki tarah charge ki koi direction nahi hoti; par mass se alag, charge positive ya negative ho sakta hai.Charges add up like ordinary numbers (scalars). We only have to take care of their plus or minus signs. If a system has charges q₁, q₂, …, qₙ, its total charge is q₁ + q₂ + … + qₙ. Example: charges of +1, +2, −3, +4 and −5 units add up to −1 unit. Like mass, charge has no direction. But unlike mass, charge can be positive or negative.
1.4.2Charge is conservedआवेश संरक्षित हैCharge conserved hai
The total charge of an isolated system never changes. Charges can move from one body to another, but net charge is neither created nor destroyed. Even when nature creates charged particles, it makes them in balanced amounts: a neutron turns into a proton and an electron, with total charge zero before and after.किसी विलगित निकाय का कुल आवेश कभी नहीं बदलता। आवेश एक वस्तु से दूसरी पर जा सकता है, पर नेट आवेश न बनता है न नष्ट होता है। जब प्रकृति आवेशित कण बनाती भी है, तो संतुलित मात्रा में: एक न्यूट्रॉन, प्रोटॉन और इलेक्ट्रॉन में बदलता है, कुल आवेश पहले और बाद में शून्य।Kisi isolated system ka total charge kabhi nahi badalta. Charge ek body se doosri par ja sakta hai, par net charge na banta hai na khatam hota hai. Jab nature charged particles banati bhi hai, to balanced amount mein: ek neutron, proton aur electron mein badalta hai, total charge pehle aur baad mein zero.In an isolated system (a group of bodies that does not share charge with anything outside), the total charge never changes. Charges can move from one body to another. But the net (total) charge is never made and never destroyed. Even when nature makes new charged particles, it makes them in balanced amounts. For example, a neutron can turn into a proton and an electron. The total charge is zero before and zero after.
1.4.3Quantisation of chargeआवेश का क्वांटमीकरणCharge ka Quantisation
Every free charge is a whole-number multiple of a basic unit e: q = ne, where n = 0, ±1, ±2, ±3, … The electron carries −e and the proton +e. This is called quantisation of charge. Faraday’s laws of electrolysis first hinted at it; Millikan demonstrated it experimentally in 1912. A body with n₂ protons and n₁ electrons has charge (n₂ − n₁)e.हर मुक्त आवेश एक मूल इकाई e का पूर्णांक गुणज होता है: q = ne, जहाँ n = 0, ±1, ±2, ±3, … इलेक्ट्रॉन पर −e और प्रोटॉन पर +e आवेश है। इसे आवेश का क्वांटमीकरण कहते हैं। फैराडे के विद्युत अपघटन के नियमों ने पहला संकेत दिया; मिलिकन ने 1912 में प्रयोग से इसे सिद्ध किया। n₂ प्रोटॉन और n₁ इलेक्ट्रॉन वाली वस्तु का आवेश (n₂ − n₁)e है।Har free charge ek basic unit e ka whole-number multiple hota hai: q = ne, jahan n = 0, ±1, ±2, ±3, … Electron par −e aur proton par +e. Ise quantisation of charge kehte hain. Faraday ke electrolysis laws ne pehla hint diya; Millikan ne 1912 mein experiment se prove kiya. n₂ protons aur n₁ electrons wali body ka charge (n₂ − n₁)e hai.Charge comes in packets of e. You can have 1e, 2e, 3e… but never 1.5e.
One coulomb is the charge that flows through a wire in 1 s when the current is 1 A. Because e is so tiny, an everyday charge of 1 μC already contains about 10¹³ electrons’ worth of charge. At that big (macroscopic) scale the steps of e are too small to notice and charge looks continuous — just as a dotted line seen from far looks continuous. At the microscopic scale (a few tens or hundreds of e), quantisation cannot be ignored.एक कूलॉम वह आवेश है जो 1 A धारा होने पर 1 s में तार से बहता है। e इतना छोटा है कि रोज़मर्रा के 1 μC आवेश में ही लगभग 10¹³ इलेक्ट्रॉनों जितना आवेश होता है। इस बड़े (स्थूल) पैमाने पर e के कदम दिखते नहीं और आवेश संतत लगता है — जैसे दूर से देखी बिंदुदार रेखा सतत दिखती है। सूक्ष्म पैमाने पर (कुछ दसियों या सैकड़ों e) क्वांटमीकरण की अनदेखी नहीं की जा सकती।Ek coulomb wo charge hai jo 1 A current par 1 s mein wire se behta hai. e itna chhota hai ki roz ke 1 μC charge mein hi lagbhag 10¹³ electrons jitna charge hota hai. Is bade (macroscopic) scale par e ke steps dikhte nahi aur charge continuous lagta hai — jaise door se dekhi dotted line continuous dikhti hai. Microscopic scale par (kuch das ya sau e) quantisation ignore nahi kar sakte.One coulomb is the charge that flows through a wire in 1 s when the current is 1 A. The charge e of one electron is very, very small. So even a small everyday charge of 1 μC already has about 10¹³ electrons’ worth of charge. At this big (macroscopic) level, the tiny steps of e are too small to notice. So charge looks continuous (smooth, with no steps) — just like a dotted line looks like a full line from far away. But at the tiny (microscopic) level, with only a few tens or hundreds of e, we cannot ignore quantisation (the fact that charge comes only in steps of e).
If 10⁹ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?यदि किसी वस्तु से दूसरी वस्तु पर हर सेकंड 10⁹ इलेक्ट्रॉन जाते हैं, तो दूसरी वस्तु पर कुल 1 C आवेश आने में कितना समय लगेगा?Agar ek body se doosri body par har second 10⁹ electrons jaate hain, to doosri body par total 1 C charge aane mein kitna time lagega?Suppose 10⁹ electrons leave a body and go to another body every second. How long will it take for the other body to get a total charge of 1 C?
Show solutionहल देखेंSolution dekhiye
Charge per second = 10⁹ × 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁰ C/s. Time = 1 C ÷ (1.6 × 10⁻¹⁰ C/s) = 6.25 × 10⁹ s ≈ 198 years (about 200 years). So 1 coulomb is a very large unit for everyday purposes.प्रति सेकंड आवेश = 10⁹ × 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁰ C/s। समय = 1 C ÷ (1.6 × 10⁻¹⁰ C/s) = 6.25 × 10⁹ s ≈ 198 वर्ष (लगभग 200 वर्ष)। अतः 1 कूलॉम रोज़मर्रा के लिए बहुत बड़ा मात्रक है।Charge per second = 10⁹ × 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁰ C/s. Time = 1 C ÷ (1.6 × 10⁻¹⁰ C/s) = 6.25 × 10⁹ s ≈ 198 saal (lagbhag 200 saal). Yaani 1 coulomb roz ke kaam ke liye bahut badi unit hai.First find the charge that moves each second: 10⁹ × 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁰ C/s. Then divide the total charge by this rate. Time = 1 C ÷ (1.6 × 10⁻¹⁰ C/s) = 6.25 × 10⁹ s. This is about 198 years (nearly 200 years). So 1 coulomb is a very big unit for everyday use.
How much positive and negative charge is there in a cup of water (take 250 g)?एक कप पानी (250 g मानें) में कितना धनात्मक और कितना ऋणात्मक आवेश है?Ek cup paani (250 g maano) mein kitna positive aur kitna negative charge hai?A cup of water has 250 g of water. How much positive charge and how much negative charge is there in it?
Show solutionहल देखेंSolution dekhiye
Moles of water = 250/18. Molecules = (250/18) × 6.02 × 10²³. Each H₂O molecule has 10 electrons and 10 protons. Charge of each kind = (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10⁻¹⁹ C ≈ 1.34 × 10⁷ C. The positive and negative totals are equal, so the water is neutral.पानी के मोल = 250/18। अणु = (250/18) × 6.02 × 10²³। हर H₂O अणु में 10 इलेक्ट्रॉन और 10 प्रोटॉन। हर प्रकार का आवेश = (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10⁻¹⁹ C ≈ 1.34 × 10⁷ C। धन और ऋण आवेश बराबर हैं, इसलिए पानी उदासीन है।Paani ke moles = 250/18. Molecules = (250/18) × 6.02 × 10²³. Har H₂O molecule mein 10 electrons aur 10 protons. Har type ka charge = (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10⁻¹⁹ C ≈ 1.34 × 10⁷ C. Positive aur negative barabar hain, isliye paani neutral hai.Step 1: Number of moles of water = 250/18. Step 2: Number of molecules = (250/18) × 6.02 × 10²³. Step 3: Each H₂O molecule has 10 electrons and 10 protons. Step 4: So the charge of each kind = (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10⁻¹⁹ C ≈ 1.34 × 10⁷ C. The total positive charge and the total negative charge are equal. That is why the water is neutral (has no net charge).
Didn’t get this part?समझ नहीं आया?Samajh nahi aaya?
1.5Coulomb’s Lawकूलॉम नियमCoulomb’s Law
Coulomb’s law tells us exactly how strong the force between two point charges is. Coulomb found that the force between two point charges q₁ and q₂ separated by a distance r (in vacuum) is directly proportional to the product of the charges, inversely proportional to the square of the distance, and acts along the line joining them.कूलॉम नियम बताता है कि दो बिंदु आवेशों के बीच बल ठीक कितना होता है। कूलॉम ने पाया कि r दूरी पर रखे दो बिंदु आवेशों q₁ और q₂ के बीच (निर्वात में) बल आवेशों के गुणनफल के अनुक्रमानुपाती, दूरी के वर्ग के व्युत्क्रमानुपाती होता है, और उन्हें मिलाने वाली रेखा के अनुदिश लगता है।Coulomb’s law batata hai ki do point charges ke beech force exactly kitna hota hai. Coulomb ne paaya ki r doori par rakhe do point charges q₁ aur q₂ ke beech (vacuum mein) force charges ke product ke directly proportional, doori ke square ke inversely proportional hota hai, aur unhe jodne wali line ke along lagta hai.Bigger charges → bigger force. Double the distance → force becomes one-fourth.
Coulomb measured this force with a torsion balance. If q₁ = q₂ = 1 C and r = 1 m, then F = 9 × 10⁹ N — an enormous force. This shows again that 1 C is a huge charge; in electrostatics we usually meet μC or nC.कूलॉम ने यह बल ऐंठन तुला (मरोड़ तुला) से मापा। यदि q₁ = q₂ = 1 C और r = 1 m, तो F = 9 × 10⁹ N — बहुत बड़ा बल। इससे फिर दिखता है कि 1 C बहुत बड़ा आवेश है; स्थिरवैद्युतिकी में प्रायः μC या nC मिलते हैं।Coulomb ne ye force torsion balance se naapa. Agar q₁ = q₂ = 1 C aur r = 1 m, to F = 9 × 10⁹ N — bahut bada force. Isse phir dikhta hai ki 1 C bahut bada charge hai; electrostatics mein aksar μC ya nC milte hain.Coulomb measured this force using a torsion balance (a device that measures a small force from how much a thin wire twists). Take q₁ = q₂ = 1 C and r = 1 m. Then the force F = 9 × 10⁹ N. This is a huge force. It shows again that 1 C is a very large charge. In electrostatics, we usually deal with much smaller charges like μC (microcoulomb) or nC (nanocoulomb).
Vector form. Let q₁ and q₂ be at position vectors r₁ and r₂. Write r₂₁ = r₂ − r₁ (from q₁ to q₂) see r₂₁ and r̂₂₁ for its unit vector. Then the force on q₂ due to q₁ is F₂₁ = k (q₁q₂ / r₂₁²) r̂₂₁. This one formula works for every sign: if q₁q₂ > 0 the force on q₂ points away from q₁ (repulsion); if q₁q₂ < 0 it points towards q₁ (attraction). Also F₁₂ = −F₂₁, which agrees with Newton’s third law (see the forces).सदिश रूप। मान लें q₁ और q₂ स्थिति सदिश r₁ और r₂ पर हैं। r₂₁ = r₂ − r₁ (q₁ से q₂ की ओर) r₂₁ देखें और इसका एकांक सदिश r̂₂₁ लें। तब q₁ के कारण q₂ पर बल F₂₁ = k (q₁q₂ / r₂₁²) r̂₂₁। यह एक सूत्र हर चिह्न के लिए सही है: यदि q₁q₂ > 0 तो q₂ पर बल q₁ से दूर (प्रतिकर्षण); यदि q₁q₂ < 0 तो q₁ की ओर (आकर्षण)। साथ ही F₁₂ = −F₂₁, जो न्यूटन के तीसरे नियम से मेल खाता है (बल देखें)।Vector form. Maano q₁ aur q₂ position vectors r₁ aur r₂ par hain. r₂₁ = r₂ − r₁ (q₁ se q₂ ki taraf) r₂₁ dekhiye aur iska unit vector r̂₂₁ lo. Tab q₁ ki wajah se q₂ par force F₂₁ = k (q₁q₂ / r₂₁²) r̂₂₁. Ye ek formula har sign ke liye sahi hai: agar q₁q₂ > 0 to q₂ par force q₁ se door (repulsion); agar q₁q₂ < 0 to q₁ ki taraf (attraction). Saath hi F₁₂ = −F₂₁, jo Newton ke third law se match karta hai (forces dekhiye).Vector form (the form that also gives direction). Let q₁ and q₂ be at position vectors r₁ and r₂. Write r₂₁ = r₂ − r₁. This vector points from q₁ to q₂. Its unit vector (a vector of length one that shows only direction) is r̂₂₁. Then the force on q₂ due to q₁ is F₂₁ = k (q₁q₂ / r₂₁²) r̂₂₁. This one formula works for all signs of charge. If q₁q₂ > 0, the force on q₂ points away from q₁ (they push apart — repulsion). If q₁q₂ < 0, the force points towards q₁ (they pull together — attraction). Also, F₁₂ = −F₂₁: the two forces are equal and opposite. This matches Newton’s third law. 👉 see r₂₁ 👉 (see the forces)
Change the two charges and the distance. Watch the force arrows: same signs push apart, opposite signs pull together — and halving r makes F four times bigger.दोनों आवेश और दूरी बदलें। बल के तीर देखें: समान चिह्न दूर धकेलते हैं, विपरीत चिह्न खींचते हैं — और r आधा करने पर F चार गुना हो जाता है।Dono charges aur doori badliye. Force ke arrows dekhiye: same sign door dhakelte hain, opposite sign kheenchte hain — aur r aadha karne par F chaar guna ho jaata hai.
Electric vs gravitational force. For an electron and a proton, the ratio of the Coulomb force to the gravitational force is about 2.4 × 10³⁹; for two protons it is about 1.3 × 10³⁶. Both follow an inverse-square law, but electric forces are enormously stronger. That is why the electron in an atom feels a huge acceleration (about 10²² m s⁻² at 1 Å from a proton) and gravity is negligible in atomic physics.विद्युत बनाम गुरुत्व बल। इलेक्ट्रॉन और प्रोटॉन के लिए कूलॉम बल और गुरुत्व बल का अनुपात लगभग 2.4 × 10³⁹ है; दो प्रोटॉनों के लिए लगभग 1.3 × 10³⁶। दोनों व्युत्क्रम-वर्ग नियम मानते हैं, पर विद्युत बल बहुत अधिक प्रबल है। इसीलिए परमाणु में इलेक्ट्रॉन का त्वरण बहुत बड़ा होता है (प्रोटॉन से 1 Å पर लगभग 10²² m s⁻²) और परमाणु भौतिकी में गुरुत्व नगण्य है।Electric vs gravitational force. Electron aur proton ke liye Coulomb force aur gravity force ka ratio lagbhag 2.4 × 10³⁹ hai; do protons ke liye lagbhag 1.3 × 10³⁶. Dono inverse-square law follow karte hain, par electric force bahut zyada strong hai. Isiliye atom mein electron ka acceleration bahut bada hota hai (proton se 1 Å par lagbhag 10²² m s⁻²) aur atomic physics mein gravity negligible hai.Electric force vs gravitational force. For an electron and a proton, the electric (Coulomb) force is about 2.4 × 10³⁹ times the gravitational force. For two protons, it is about 1.3 × 10³⁶ times. Both forces follow an inverse-square law (force falls as 1/r²). But the electric force is far, far stronger. So an electron in an atom gets a huge acceleration (about 10²² m s⁻² at 1 Å from a proton). In atomic physics, gravity is so small that we can ignore it.
Where Coulomb’s law applies: point charges at rest (electrostatics). In a material medium the force becomes smaller by a factor K (the dielectric constant): F = q₁q₂/(4πε₀K r²). You will meet K properly in Chapter 2.कूलॉम नियम कहाँ लागू: विरामावस्था वाले बिंदु आवेश (स्थिरवैद्युतिकी)। किसी माध्यम में बल K (परावैद्युतांक) गुना कम हो जाता है: F = q₁q₂/(4πε₀K r²)। K को आप अध्याय 2 में विस्तार से पढ़ेंगे।Coulomb’s law kahan lagta hai: rest wale point charges (electrostatics). Kisi medium mein force K (dielectric constant) guna kam ho jaata hai: F = q₁q₂/(4πε₀K r²). K ko aap Chapter 2 mein detail mein padhenge.Where Coulomb’s law works: for point charges that are at rest (this is called electrostatics). If the charges are inside a material (a medium), the force becomes smaller by a factor K. K is called the dielectric constant. Then F = q₁q₂/(4πε₀K r²). You will study K properly in Chapter 2.
A charged metal sphere A is hung from a nylon thread. An identical charged sphere B (on an insulating handle) is brought so that their centres are 10 cm apart, and the repulsion F is noted. A and B are then touched by identical uncharged spheres C and D, which are removed. B is now brought to 5.0 cm from A. What is the new force?नायलॉन के धागे से लटके आवेशित धातु के गोले A के पास, समान आवेशित गोला B (विद्युतरोधी हत्थे पर) इतना लाया जाता है कि केंद्रों के बीच 10 cm दूरी हो; प्रतिकर्षण बल F नोट किया जाता है। फिर A और B को समान अनावेशित गोलों C और D से छुआकर C, D हटा लिए जाते हैं। अब B को A से 5.0 cm पर लाया जाता है। नया बल कितना होगा?Nylon thread se latke charged metal sphere A ke paas, waisa hi charged sphere B (insulating handle par) itna laaya jaata hai ki centres ke beech 10 cm doori ho; repulsion F note hota hai. Phir A aur B ko same uncharged spheres C aur D se chhua kar C, D hata liye jaate hain. Ab B ko A se 5.0 cm par laate hain. Naya force kitna hoga?A charged metal sphere A hangs from a nylon thread. Another charged sphere B, of the same size, is held on an insulating handle. B is brought near A so that their centres are 10 cm apart, and the push (repulsion) F between them is noted. Then A and B are each touched with an identical uncharged sphere, C and D, which are then taken away. Now B is brought to 5.0 cm from A. What is the new force?
Show solutionहल देखेंSolution dekhiye
Touching an identical uncharged sphere shares the charge equally, so A now has q/2 and B has q′/2. New force F′ = k (q/2)(q′/2)/(r/2)² = k qq′/r² = F. The force stays the same: the charges became one-half each (product ÷ 4) but the distance halved (r² ÷ 4).समान अनावेशित गोले को छूने पर आवेश बराबर बँट जाता है, इसलिए अब A पर q/2 और B पर q′/2। नया बल F′ = k (q/2)(q′/2)/(r/2)² = k qq′/r² = F। बल वही रहता है: आवेश आधे हुए (गुणनफल ÷ 4) पर दूरी भी आधी हुई (r² ÷ 4)।Same uncharged sphere ko chhoone par charge barabar bant jaata hai, isliye ab A par q/2 aur B par q′/2. Naya force F′ = k (q/2)(q′/2)/(r/2)² = k qq′/r² = F. Force wahi rehta hai: charges aadhe hue (product ÷ 4) par doori bhi aadhi hui (r² ÷ 4).When a sphere touches an identical uncharged sphere, the charge is shared equally. So A now has q/2 and B has q′/2. The new force F′ = k (q/2)(q′/2)/(r/2)² = k qq′/r² = F. So the force does not change. Why? Each charge became half, so their product became 4 times smaller. But the distance also became half, so r² became 4 times smaller. The two effects cancel.
Two point charges +2 μC and −3 μC are 30 cm apart in vacuum. Find the size and nature of the force between them.निर्वात में +2 μC और −3 μC के दो बिंदु आवेश 30 cm दूर हैं। उनके बीच बल का परिमाण और प्रकृति ज्ञात करें।Vacuum mein +2 μC aur −3 μC ke do point charges 30 cm door hain. Unke beech force ka magnitude aur nature bataiye.Two point charges, +2 μC and −3 μC, are 30 cm apart in vacuum (empty space). Find how big the force between them is, and whether it pulls or pushes.
Show solutionहल देखेंSolution dekhiye
F = k|q₁q₂|/r² = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.30)² = (9 × 10⁹ × 6 × 10⁻¹²)/0.09 = 0.6 N. The charges have opposite signs, so the force is attractive, along the line joining them.F = k|q₁q₂|/r² = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.30)² = (9 × 10⁹ × 6 × 10⁻¹²)/0.09 = 0.6 N। आवेशों के चिह्न विपरीत हैं, इसलिए बल आकर्षी है, उन्हें मिलाने वाली रेखा के अनुदिश।F = k|q₁q₂|/r² = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.30)² = (9 × 10⁹ × 6 × 10⁻¹²)/0.09 = 0.6 N. Charges ke signs opposite hain, isliye force attractive hai, unhe jodne wali line ke along.Use F = k|q₁q₂|/r². Put in the values: F = (9 × 10⁹)(2 × 10⁻⁶)(3 × 10⁻⁶)/(0.30)² = (9 × 10⁹ × 6 × 10⁻¹²)/0.09 = 0.6 N. One charge is positive and the other is negative (opposite signs). So the force is attractive (they pull each other). It acts along the line that joins the two charges.
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1.6Forces between Multiple Chargesबहुल आवेशों के बीच बलKai Charges ke beech Force
What if there are many charges? Experiments show the superposition principle: the force between any two charges is not affected by the other charges. So the total force on a charge is the vector sum of the forces due to each of the other charges, each found from Coulomb’s law as if the others were absent.यदि कई आवेश हों तो? प्रयोग अध्यारोपण सिद्धांत दिखाते हैं: किन्हीं दो आवेशों के बीच बल अन्य आवेशों से प्रभावित नहीं होता। अतः किसी आवेश पर कुल बल, बाकी हर आवेश के कारण लगे बलों का सदिश योग है; हर बल कूलॉम नियम से ऐसे निकाला जाता है मानो बाकी आवेश हों ही नहीं।Agar kai charges hon to? Experiments superposition principle dikhaate hain: kinhi do charges ke beech force baaki charges se affect nahi hota. Isliye kisi charge par total force, baaki har charge ki wajah se lage forces ka vector sum hai; har force Coulomb’s law se aise nikaalte hain jaise baaki charges hon hi nahi.Find each pair force separately, then add them as vectors (arrows).
Three equal charges q are at the corners of an equilateral triangle of side l. What is the force on a charge Q (same sign as q) placed at the centroid?भुजा l वाले समबाहु त्रिभुज के कोनों पर तीन बराबर आवेश q हैं। केन्द्रक पर रखे आवेश Q (q के समान चिह्न) पर बल कितना है?Side l wale equilateral triangle ke corners par teen barabar charges q hain. Centroid par rakhe charge Q (q jaisa sign) par force kitna hai?Three equal charges q sit at the three corners of an equilateral triangle (all sides equal) with side l. A charge Q, with the same sign as q, is placed at the centroid (the centre point of the triangle). What is the force on Q?
Show solutionहल देखेंSolution dekhiye
Each corner is at distance l/√3 from the centroid, so each force has the same size 3Qq/(4πε₀l²), and the three forces point along the three lines away from the corners, 120° apart. Three equal vectors at 120° add to zero. Symmetry also says so: rotate the triangle by 120° and nothing changes, so the net force cannot point in any direction.हर कोना केन्द्रक से l/√3 दूरी पर है, इसलिए हर बल का परिमाण 3Qq/(4πε₀l²) है और तीनों बल 120° के अंतर पर हैं। 120° पर तीन बराबर सदिशों का योग शून्य होता है। सममिति भी यही कहती है: त्रिभुज को 120° घुमाने पर कुछ नहीं बदलता, इसलिए नेट बल किसी दिशा में नहीं हो सकता।Har corner centroid se l/√3 door hai, isliye har force ka magnitude 3Qq/(4πε₀l²) hai aur teeno forces 120° ke gap par hain. 120° par teen barabar vectors ka sum zero hota hai. Symmetry bhi yahi kehti hai: triangle ko 120° ghumaane par kuch nahi badalta, isliye net force kisi direction mein nahi ho sakta.Each corner is at the same distance, l/√3, from the centroid. So each corner pushes Q with a force of the same size, 3Qq/(4πε₀l²). The three forces point away from the three corners, and they are 120° apart. Three equal vectors at 120° to each other add up to zero. Symmetry tells us the same thing. If we turn the triangle by 120°, it looks exactly the same. So the net force cannot point in any one direction — it must be zero.
Charges q, q and −q are at the corners A, B, C of an equilateral triangle of side l. Find the force on each charge.भुजा l के समबाहु त्रिभुज के कोनों A, B, C पर आवेश q, q और −q हैं। हर आवेश पर बल ज्ञात करें।Side l ke equilateral triangle ke corners A, B, C par charges q, q aur −q hain. Har charge par force nikaalo.Charges q, q and −q are placed at the corners A, B and C of an equilateral triangle with side l. Find the force on each charge.
Show solutionहल देखेंSolution dekhiye
Each pair force has size F = q²/(4πε₀l²). On q at A: repulsion from B (along BA) and attraction to C (along AC), two forces of size F at 120° → resultant F, parallel to BC. On q at B: similarly F, parallel to AC. On −q at C: two attractions of size F at 60° → resultant √3 F, along the bisector of angle C. Check: F₁ + F₂ + F₃ = 0, as Newton’s third law demands.हर युग्म बल का परिमाण F = q²/(4πε₀l²)। A पर q: B से प्रतिकर्षण (BA के अनुदिश) और C की ओर आकर्षण (AC के अनुदिश), 120° पर दो बल F → परिणामी F, BC के समांतर। B पर q: इसी प्रकार F, AC के समांतर। C पर −q: 60° पर दो आकर्षण F → परिणामी √3 F, कोण C के समद्विभाजक के अनुदिश। जाँच: F₁ + F₂ + F₃ = 0, जैसा न्यूटन का तीसरा नियम कहता है।Har pair force ka magnitude F = q²/(4πε₀l²). A par q: B se repulsion (BA ke along) aur C ki taraf attraction (AC ke along), 120° par do forces F → resultant F, BC ke parallel. B par q: isi tarah F, AC ke parallel. C par −q: 60° par do attractions F → resultant √3 F, angle C ke bisector ke along. Check: F₁ + F₂ + F₃ = 0, jaisa Newton ka third law kehta hai.Every pair of charges pulls or pushes with a force of size F = q²/(4πε₀l²). On q at A: B pushes it (along BA) and C pulls it (along AC). These are two forces of size F at 120° to each other, so the result is F, parallel to side BC. On q at B: in the same way, the result is F, parallel to AC. On −q at C: both A and B pull it, two forces of size F at 60° to each other. So the result is √3 F, along the line that cuts angle C into two equal halves. Check: F₁ + F₂ + F₃ = 0. The three forces add to zero, as Newton’s third law says they must.
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1.7Electric Fieldविद्युत क्षेत्रElectric Field
A charge Q changes the space around it: any other charge placed nearby feels a force. We call this effect the electric field of Q. The field at a point is the force per unit positive test charge placed there. Q is the source charge; the small charge q we use to measure the field is the test charge (kept tiny so it does not disturb Q).आवेश Q अपने आसपास के स्थान को बदल देता है: पास रखा कोई भी आवेश बल अनुभव करता है। इस प्रभाव को Q का विद्युत क्षेत्र कहते हैं। किसी बिंदु पर क्षेत्र, वहाँ रखे एकांक धनात्मक परीक्षण आवेश पर बल है। Q स्रोत आवेश है; क्षेत्र मापने के लिए लिया गया छोटा आवेश q परीक्षण आवेश है (इतना छोटा कि Q को विचलित न करे)।Charge Q apne aas-paas ki space ko badal deta hai: paas rakha koi bhi charge force feel karta hai. Is effect ko Q ka electric field kehte hain. Kisi point par field, wahan rakhe unit positive test charge par force hai. Q source charge hai; field naapne ke liye liya gaya chhota charge q test charge hai (itna chhota ki Q ko disturb na kare).Electric field = the push or pull that 1 coulomb of positive charge would feel at that point.
Notice: E depends only on the source charge Q and the distance r, not on the test charge. The force on any charge q placed in a field is F = qE: along E for a positive charge, opposite to E for a negative charge. Points at the same distance from a point charge have fields of the same size — the field has spherical symmetry (see how it weakens with distance).ध्यान दें: E केवल स्रोत आवेश Q और दूरी r पर निर्भर है, परीक्षण आवेश पर नहीं। क्षेत्र में रखे किसी आवेश q पर बल F = qE है: धनावेश के लिए E की दिशा में, ऋणावेश के लिए E के विपरीत। बिंदु आवेश से समान दूरी वाले बिंदुओं पर क्षेत्र का परिमाण समान है — क्षेत्र गोलीय सममित है (दूरी के साथ घटना देखें)।Dhyaan do: E sirf source charge Q aur doori r par depend karta hai, test charge par nahi. Field mein rakhe kisi charge q par force F = qE hai: positive charge ke liye E ki direction mein, negative ke liye E ke opposite. Point charge se same doori wale points par field same hota hai — field spherically symmetric hai (doori ke saath ghatna dekhiye).Notice: E depends only on the source charge Q and the distance r. It does not depend on the test charge. Any charge q placed in a field feels a force F = qE. For a positive charge, the force is along E. For a negative charge, it is opposite to E. All points at the same distance from a point charge have fields of the same size. So the field is the same in every direction around the charge — it has spherical symmetry. 👉 (see how it weakens with distance)
1.7.1Electric field due to a system of chargesआवेशों के निकाय के कारण विद्युत क्षेत्रCharges ke system ka Electric Field
Fields also obey superposition: the field at a point due to many charges is the vector sum of the fields due to each charge: E = E₁ + E₂ + … + Eₙ, where each Eᵢ = k qᵢ/rᵢ² along r̂ᵢ (pointing from qᵢ to the point).क्षेत्र भी अध्यारोपण मानते हैं: कई आवेशों के कारण किसी बिंदु पर क्षेत्र, हर आवेश के क्षेत्रों का सदिश योग है: E = E₁ + E₂ + … + Eₙ, जहाँ हर Eᵢ = k qᵢ/rᵢ², r̂ᵢ के अनुदिश (qᵢ से उस बिंदु की ओर)।Fields bhi superposition follow karte hain: kai charges ki wajah se kisi point par field, har charge ke fields ka vector sum hai: E = E₁ + E₂ + … + Eₙ, jahan har Eᵢ = k qᵢ/rᵢ², r̂ᵢ ke along (qᵢ se us point ki taraf).Fields also follow the rule of superposition (adding up). When there are many charges, the field at a point is the vector sum of the fields from each charge. That is, E = E₁ + E₂ + … + Eₙ. Each Eᵢ = k qᵢ/rᵢ² along r̂ᵢ, which points from the charge qᵢ to the point.
1.7.2Physical significance of electric fieldविद्युत क्षेत्र का भौतिक महत्वElectric Field ka Physical Meaning
Why bother with a field instead of just force? (1) The field tells us the force on any charge we may place there, without repeating Coulomb’s law. (2) More deeply, when charges move or accelerate, their effects do not reach other charges instantly; they travel at the speed of light. The field carries this effect through space and has a real existence of its own — this idea becomes essential for electromagnetic waves (Chapter 8).केवल बल के बजाय क्षेत्र क्यों? (1) क्षेत्र बताता है कि वहाँ रखे किसी भी आवेश पर कितना बल होगा, बार-बार कूलॉम नियम लगाए बिना। (2) गहराई से, जब आवेश गति या त्वरण करते हैं, तो उनका प्रभाव दूसरे आवेशों तक तुरंत नहीं पहुँचता; प्रकाश की चाल से जाता है। यह प्रभाव क्षेत्र ही स्थान में ले जाता है, और क्षेत्र का अपना वास्तविक अस्तित्व है — यह विचार वैद्युतचुंबकीय तरंगों (अध्याय 8) के लिए ज़रूरी है।Sirf force ke bajaay field kyun? (1) Field batata hai ki wahan rakhe kisi bhi charge par kitna force hoga, baar-baar Coulomb’s law lagaaye bina. (2) Gehraai se, jab charges move ya accelerate karte hain, unka effect doosre charges tak turant nahi pahunchta; light ki speed se jaata hai. Ye effect field hi space mein le jaata hai, aur field ka apna real existence hai — ye idea EM waves (Chapter 8) ke liye zaroori hai.Why use a field at all, and not just the force? (1) If we know the field at a point, we know the force on any charge we put there. We do not need to use Coulomb’s law again and again. (2) There is a deeper reason. When charges move or speed up, their effect does not reach other charges at once. It travels at the speed of light. The field carries this effect through space, so the field is something real on its own. This idea is very important for electromagnetic waves (Chapter 8).
An electron falls through 1.5 cm in a uniform field of 2.0 × 10⁴ N C⁻¹ (field upward). Then the field is reversed and a proton falls through the same distance. Find the time of fall in each case (ignore gravity).एक इलेक्ट्रॉन 2.0 × 10⁴ N C⁻¹ के एकसमान क्षेत्र (ऊपर की ओर) में 1.5 cm गिरता है। फिर क्षेत्र उलटकर एक प्रोटॉन उतनी ही दूरी गिरता है। दोनों में गिरने का समय ज्ञात करें (गुरुत्व की उपेक्षा करें)।Ek electron 2.0 × 10⁴ N C⁻¹ ke uniform field (upar ki taraf) mein 1.5 cm girta hai. Phir field ulta karke ek proton utni hi doori girta hai. Dono mein girne ka time nikaalo (gravity ignore kijiye).An electron falls through a distance of 1.5 cm in a uniform (same everywhere) electric field of 2.0 × 10⁴ N C⁻¹. The field points upward. Then the field is turned the other way, and a proton falls through the same distance. Find the time of fall in each case. Ignore gravity.
Show solutionहल देखेंSolution dekhiye
Acceleration a = eE/m and, from rest, h = ½at², so t = √(2hm/(eE)). Electron (m = 9.11 × 10⁻³¹ kg): t ≈ 2.9 × 10⁻⁹ s. Proton (m = 1.67 × 10⁻²⁷ kg): t ≈ 1.3 × 10⁻⁷ s. The heavier proton takes longer — unlike free fall under gravity, where time does not depend on mass. Ignoring gravity is fine: the proton’s acceleration eE/m ≈ 1.9 × 10¹² m s⁻², far bigger than g.त्वरण a = eE/m और विराम से h = ½at², अतः t = √(2hm/(eE))। इलेक्ट्रॉन (m = 9.11 × 10⁻³¹ kg): t ≈ 2.9 × 10⁻⁹ s। प्रोटॉन (m = 1.67 × 10⁻²⁷ kg): t ≈ 1.3 × 10⁻⁷ s। भारी प्रोटॉन अधिक समय लेता है — गुरुत्व में मुक्त पतन से अलग, जहाँ समय द्रव्यमान पर निर्भर नहीं। गुरुत्व की उपेक्षा ठीक है: प्रोटॉन का त्वरण eE/m ≈ 1.9 × 10¹² m s⁻², g से बहुत अधिक।Acceleration a = eE/m aur rest se h = ½at², isliye t = √(2hm/(eE)). Electron (m = 9.11 × 10⁻³¹ kg): t ≈ 2.9 × 10⁻⁹ s. Proton (m = 1.67 × 10⁻²⁷ kg): t ≈ 1.3 × 10⁻⁷ s. Bhaari proton zyada time leta hai — gravity ke free fall se alag, jahan time mass par depend nahi karta. Gravity ignore karna theek hai: proton ka acceleration eE/m ≈ 1.9 × 10¹² m s⁻², g se bahut zyada.The acceleration is a = eE/m. The particle starts from rest, so h = ½at². This gives t = √(2hm/(eE)). For the electron (m = 9.11 × 10⁻³¹ kg): t ≈ 2.9 × 10⁻⁹ s. For the proton (m = 1.67 × 10⁻²⁷ kg): t ≈ 1.3 × 10⁻⁷ s. The proton is heavier, so it takes longer to fall. This is different from free fall under gravity, where the time does not depend on mass. Ignoring gravity is fine here: the proton’s acceleration eE/m ≈ 1.9 × 10¹² m s⁻², which is far bigger than g.
Charges +10⁻⁸ C and −10⁻⁸ C are 0.1 m apart. Find the field at the midpoint A between them.+10⁻⁸ C और −10⁻⁸ C के आवेश 0.1 m दूर हैं। उनके बीच मध्य-बिंदु A पर क्षेत्र ज्ञात करें।+10⁻⁸ C aur −10⁻⁸ C ke charges 0.1 m door hain. Unke beech midpoint A par field nikaalo.Two charges, +10⁻⁸ C and −10⁻⁸ C, are 0.1 m apart. Find the electric field at the point A exactly halfway between them.
Show solutionहल देखेंSolution dekhiye
Each charge is 0.05 m from A. Field due to each: E = (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹. The field of the + charge points away from it (towards −), and the field of the − charge points towards it (also towards −). Same direction, so they add: E_A = 7.2 × 10⁴ N C⁻¹, towards the negative charge.हर आवेश A से 0.05 m पर है। हर एक के कारण क्षेत्र: E = (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹। + आवेश का क्षेत्र उससे दूर (− की ओर) और − आवेश का क्षेत्र उसकी ओर (यानी फिर − की ओर) है। दिशा एक, इसलिए जुड़ते हैं: E_A = 7.2 × 10⁴ N C⁻¹, ऋण आवेश की ओर।Har charge A se 0.05 m par hai. Har ek ka field: E = (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹. + charge ka field usse door (− ki taraf) aur − charge ka field uski taraf (yaani phir − ki taraf). Direction same, isliye jud jaate hain: E_A = 7.2 × 10⁴ N C⁻¹, negative charge ki taraf.Each charge is 0.05 m from A. The field due to each one is E = (9 × 10⁹)(10⁻⁸)/(0.05)² = 3.6 × 10⁴ N C⁻¹. The field of the + charge points away from it, that is, towards the − charge. The field of the − charge points towards itself, which is also towards the − charge. Both fields point the same way, so we simply add them: E_A = 7.2 × 10⁴ N C⁻¹, towards the negative charge.
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1.8Electric Field Linesविद्युत क्षेत्र रेखाएँElectric Field Lines
An electric field line is a curve drawn so that the tangent at every point gives the direction of the field there. Field lines give a picture of the whole field at a glance. Where lines are crowded, the field is strong; where they are spread out, it is weak. In a uniform field the lines are parallel and equally spaced. Field lines are really in 3D; the pictures show a slice.विद्युत क्षेत्र रेखा ऐसा वक्र है जिसके हर बिंदु पर स्पर्शरेखा उस बिंदु पर क्षेत्र की दिशा बताती है। क्षेत्र रेखाएँ एक नज़र में पूरे क्षेत्र का चित्र देती हैं। जहाँ रेखाएँ घनी हैं, क्षेत्र प्रबल है; जहाँ दूर-दूर हैं, दुर्बल। एकसमान क्षेत्र में रेखाएँ समांतर और बराबर दूरी पर होती हैं। क्षेत्र रेखाएँ वास्तव में 3D में होती हैं; चित्र एक काट दिखाते हैं।Electric field line aisa curve hai jiski har point par tangent us point par field ki direction batati hai. Field lines ek nazar mein poore field ki picture deti hain. Jahan lines ghani hain, field strong hai; jahan door-door hain, weak. Uniform field mein lines parallel aur barabar doori par hoti hain. Field lines asal mein 3D mein hoti hain; pictures ek slice dikhati hain.Field lines are arrows that show which way a + charge would be pushed. Close lines = strong field.
Why is the number of lines a good measure of strength? For a point charge, the same number of lines that cross a small area at distance r spread over four times the area at 2r (area grows as r²). So the line density falls as 1/r² — exactly like E. That is why field-line pictures work.रेखाओं की संख्या प्रबलता का अच्छा माप क्यों है? बिंदु आवेश के लिए, दूरी r पर छोटे क्षेत्रफल से जितनी रेखाएँ गुज़रती हैं, वही 2r पर चार गुना क्षेत्रफल पर फैल जाती हैं (क्षेत्रफल r² के साथ बढ़ता है)। अतः रेखाओं का घनत्व 1/r² से घटता है — ठीक E की तरह। इसीलिए क्षेत्र रेखा चित्र काम करते हैं।Lines ki ginti strength ka achha measure kyun hai? Point charge ke liye, doori r par chhote area se jitni lines guzarti hain, wahi 2r par chaar guna area par phail jaati hain (area r² ke saath badhta hai). Isliye lines ki density 1/r² se ghatati hai — bilkul E ki tarah. Isiliye field-line pictures kaam karti hain.Why does the number of field lines show how strong the field is? Think of a point charge. The lines that pass through a small area at distance r spread out over four times that area at distance 2r (because area grows as r²). So the number of lines per area falls as 1/r². This is exactly how E falls. That is why pictures of field lines work.
- Field lines start on positive charges and end on negative charges (or start/end at infinity for a single charge).क्षेत्र रेखाएँ धनावेश से शुरू होकर ऋणावेश पर समाप्त होती हैं (अकेले आवेश के लिए अनंत से/तक)।Field lines positive charge se shuru hokar negative charge par khatam hoti hain (akele charge ke liye infinity se/tak).Field lines begin on positive charges and finish on negative charges. For a single charge, they may begin or finish at infinity (very far away).
- In a charge-free region, field lines are continuous curves without any breaks.आवेश-रहित क्षेत्र में क्षेत्र रेखाएँ बिना टूटे सतत वक्र होती हैं।Charge-free region mein field lines bina toote continuous curves hoti hain.In a region with no charge, field lines are smooth, unbroken curves (they have no gaps or breaks).
- Two field lines can never cross — at a crossing point the field would have two directions, which is impossible.दो क्षेत्र रेखाएँ कभी एक-दूसरे को नहीं काटतीं — कटान बिंदु पर क्षेत्र की दो दिशाएँ होतीं, जो असंभव है।Do field lines kabhi ek-doosre ko cross nahi karti — crossing point par field ki do directions hoti, jo impossible hai.Two field lines can never cross each other. If they did, the field at the crossing point would point in two directions at once, which is impossible.
- Electrostatic field lines do not form closed loops (this follows because the electrostatic field is conservative — Chapter 2).स्थिरवैद्युत क्षेत्र रेखाएँ बंद लूप नहीं बनातीं (क्योंकि स्थिरवैद्युत क्षेत्र संरक्षी है — अध्याय 2)।Electrostatic field lines closed loops nahi banati (kyunki electrostatic field conservative hai — Chapter 2).Electrostatic field lines do not form closed loops (they never come back to where they started). This is because the electrostatic field is conservative — you will learn this in Chapter 2.
Drag the charges and watch the field lines redraw instantly. Add + or − charges, try a dipole or two like charges, and tap anywhere to measure the field there.आवेशों को खिसकाएँ और क्षेत्र रेखाएँ तुरंत बदलती देखें। + या − आवेश जोड़ें, द्विध्रुव या दो सजातीय आवेश आज़माएँ, और कहीं भी टैप करके वहाँ का क्षेत्र मापें।Charges ko drag kijiye aur field lines turant badalti dekhiye. + ya − charge jodiye, dipole ya do like charges try kijiye, aur kahin bhi tap karke wahan ka field naapo.
Field lines really fill space in all directions. Spin this real 3D model of a point charge, a dipole and two like charges.क्षेत्र रेखाएँ वास्तव में हर दिशा में फैली होती हैं। बिंदु आवेश, द्विध्रुव और दो सजातीय आवेशों के इस असली 3D मॉडल को घुमाएँ।Field lines asal mein har direction mein phaili hoti hain. Point charge, dipole aur do like charges ke is real 3D model ko ghumaao.
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1.9Electric Fluxविद्युत फ्लक्सElectric Flux
Think of water flowing through a ring: how much passes depends on the flow speed, the ring’s area, and how the ring is tilted. Electric flux is the same idea for field lines — roughly, the number of field lines crossing a surface. For a small flat area ΔS in a field E, with the area’s normal n̂ at angle θ to E:किसी छल्ले से बहते पानी के बारे में सोचें: कितना पानी गुज़रेगा, यह बहाव की चाल, छल्ले के क्षेत्रफल और छल्ले के झुकाव पर निर्भर है। विद्युत फ्लक्स क्षेत्र रेखाओं के लिए यही विचार है — मोटे तौर पर, किसी पृष्ठ से गुज़रने वाली क्षेत्र रेखाओं की संख्या। क्षेत्र E में छोटे समतल क्षेत्रफल ΔS के लिए, जिसका अभिलंब n̂, E से θ कोण पर हो:Ek ring se behte paani ke baare mein sochiye: kitna paani guzrega ye flow speed, ring ke area aur ring ke tilt par depend karta hai. Electric flux field lines ke liye yahi idea hai — roughly, kisi surface se guzarne wali field lines ki ginti. Field E mein chhote flat area ΔS ke liye, jiska normal n̂, E se θ angle par ho:Imagine water flowing through a ring. How much water passes depends on three things: how fast the water flows, how big the ring is, and how the ring is tilted. Electric flux is the same idea for field lines. Roughly, it is the number of field lines that pass through a surface. Take a small flat area ΔS in a field E. Let its normal n̂ (a line standing straight out of the surface) make an angle θ with E. Then:
Flux is maximum (EΔS) when the area faces the field (θ = 0), zero when the area is parallel to the field (θ = 90°), and negative when lines enter a closed surface (θ > 90°). For a big surface we add up Δφ over all small pieces: φ = Σ E·ΔS.फ्लक्स अधिकतम (EΔS) तब है जब क्षेत्रफल क्षेत्र के सामने हो (θ = 0), शून्य जब क्षेत्रफल क्षेत्र के समांतर हो (θ = 90°), और ऋणात्मक जब रेखाएँ बंद पृष्ठ में प्रवेश करें (θ > 90°)। बड़े पृष्ठ के लिए सभी छोटे टुकड़ों का Δφ जोड़ते हैं: φ = Σ E·ΔS।Flux maximum (EΔS) tab hai jab area field ke saamne ho (θ = 0), zero jab area field ke parallel ho (θ = 90°), aur negative jab lines closed surface mein enter karein (θ > 90°). Badi surface ke liye saare chhote tukdon ka Δφ jodte hain: φ = Σ E·ΔS.The flux is largest (EΔS) when the area faces the field straight on (θ = 0). It is zero when the area lies along the field (θ = 90°). It is negative where lines go into a closed surface (θ > 90°). For a big surface, we break it into small pieces and add Δφ for all of them: φ = Σ E·ΔS.
Tilt the area and change E. Count the lines that pass through, and see φ = EA cos θ change live.क्षेत्रफल झुकाएँ और E बदलें। गुज़रने वाली रेखाएँ गिनें और φ = EA cos θ को बदलता देखें।Area ko tilt kijiye aur E badliye. Guzarne wali lines gino aur φ = EA cos θ ko live badalta dekhiye.
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1.10Electric Dipoleविद्युत द्विध्रुवElectric Dipole
An electric dipole is a pair of equal and opposite charges, q and −q, separated by a distance 2a. The line from −q to +q is the dipole’s axis. Its strength is measured by the dipole moment:विद्युत द्विध्रुव बराबर और विपरीत आवेशों q और −q का युग्म है जो 2a दूरी पर हों। −q से +q की ओर की रेखा द्विध्रुव का अक्ष है। इसकी प्रबलता द्विध्रुव आघूर्ण से मापी जाती है:Electric dipole barabar aur opposite charges q aur −q ka joda hai jo 2a doori par hon. −q se +q ki taraf ki line dipole ka axis hai. Iski strength dipole moment se naapi jaati hai:An electric dipole is a pair of equal and opposite charges, q and −q, kept a distance 2a apart. The line going from −q to +q is called the axis of the dipole. How strong a dipole is, is measured by its dipole moment:
1.10.1The field of an electric dipoleविद्युत द्विध्रुव का क्षेत्रElectric Dipole ka Field
The two charges are close together, so far away their fields almost cancel — but not completely. Let r be the distance of the point from the centre O. (a) On the axis (point P), the nearer +q wins, so the net field points along p. (b) On the equatorial plane (point Q), the two fields are equal in size; their sideways parts cancel and the rest adds up, giving a field opposite to p (see the arrows).दोनों आवेश पास-पास हैं, इसलिए दूर उनके क्षेत्र लगभग निरस्त होते हैं — पर पूरी तरह नहीं। मान लें बिंदु की केंद्र O से दूरी r है। (a) अक्ष पर (बिंदु P) पास वाला +q भारी पड़ता है, इसलिए नेट क्षेत्र p की दिशा में है। (b) विषुवतीय तल पर (बिंदु Q) दोनों क्षेत्र बराबर हैं; उनके बगल वाले भाग निरस्त होकर बाकी जुड़ते हैं, और क्षेत्र p के विपरीत मिलता है (तीर देखें)।Dono charges paas-paas hain, isliye door unke fields lagbhag cancel hote hain — par poori tarah nahi. Maano point ki centre O se doori r hai. (a) Axis par (point P) paas wala +q bhaari padta hai, isliye net field p ki direction mein hai. (b) Equatorial plane par (point Q) dono fields barabar hain; unke side wale parts cancel hokar baaki jud jaate hain, aur field p ke opposite milta hai (arrows dekhiye).The two charges are close to each other. So far away, their fields almost cancel — but not fully. Let r be the distance of the point from the centre O. (a) On the axis (point P): the +q is nearer, so its field wins. The net field points along p. (b) On the equatorial plane (point Q, on the line cutting the dipole in the middle at right angles): the two fields are equal in size. Their sideways parts cancel, and the other parts add up. So the net field points opposite to p. 👉 (see the arrows)
A dipole’s field falls as 1/r³, faster than a point charge’s 1/r², because the two opposite charges nearly cancel each other from far away. At large distances the field depends only on the product p = q × 2a, not on q and a separately.द्विध्रुव का क्षेत्र 1/r³ से घटता है, जो बिंदु आवेश के 1/r² से तेज़ है, क्योंकि दूर से दोनों विपरीत आवेश लगभग एक-दूसरे को निरस्त कर देते हैं। अधिक दूरी पर क्षेत्र केवल गुणनफल p = q × 2a पर निर्भर करता है, q और a पर अलग-अलग नहीं।Dipole ka field 1/r³ se ghatata hai, jo point charge ke 1/r² se tez hai, kyunki door se dono opposite charges lagbhag ek-doosre ko cancel kar dete hain. Zyada doori par field sirf product p = q × 2a par depend karta hai, q aur a par alag-alag nahi.The field of a dipole falls as 1/r³. This is faster than a point charge, whose field falls as 1/r². The reason: from far away, the two opposite charges almost cancel each other. At large distances, the field depends only on the product p = q × 2a, not on q and a separately.
1.10.2Physical significance of dipolesद्विध्रुवों का भौतिक महत्वDipoles ka Physical Meaning
In most molecules the centres of positive and negative charge are at the same place, so they have no dipole moment — for example CO₂ and CH₄. They can, however, develop a dipole moment when an electric field is applied. In some molecules the two centres do not coincide, so they have a permanent dipole moment even without a field — for example water (H₂O). Such molecules are called polar molecules. Dipoles explain how many materials behave in electric fields (Chapter 2).अधिकांश अणुओं में धन और ऋण आवेश के केंद्र एक ही स्थान पर होते हैं, इसलिए उनका द्विध्रुव आघूर्ण शून्य होता है — जैसे CO₂ और CH₄। पर विद्युत क्षेत्र लगाने पर उनमें द्विध्रुव आघूर्ण उत्पन्न हो सकता है। कुछ अणुओं में दोनों केंद्र एक स्थान पर नहीं होते, इसलिए बिना क्षेत्र के भी उनका स्थायी द्विध्रुव आघूर्ण होता है — जैसे जल (H₂O)। ऐसे अणु ध्रुवीय अणु कहलाते हैं। द्विध्रुव समझाते हैं कि बहुत से पदार्थ विद्युत क्षेत्र में कैसे व्यवहार करते हैं (अध्याय 2)।Zyadatar molecules mein positive aur negative charge ke centres ek hi jagah hote hain, isliye unka dipole moment zero hota hai — jaise CO₂ aur CH₄. Par electric field lagaane par unmein dipole moment ban sakta hai. Kuch molecules mein dono centres ek jagah nahi hote, isliye bina field ke bhi unka permanent dipole moment hota hai — jaise paani (H₂O). Aise molecules polar molecules kehlaate hain. Dipoles samjhaate hain ki bahut se materials electric field mein kaise behave karte hain (Chapter 2).In most molecules, the centre of positive charge and the centre of negative charge are at the same place. So they have no dipole moment — for example CO₂ and CH₄. But they can get a dipole moment when we apply an electric field. In some molecules, the two centres are at different places. So they have a permanent dipole moment even without any field — for example water (H₂O). Such molecules are called polar molecules. Dipoles help explain how many materials behave in electric fields (Chapter 2).
Two charges ±10 μC are 5.0 mm apart. Find the field (a) on the axis 15 cm from the centre, on the side of the positive charge, and (b) on the equatorial line 15 cm from the centre.±10 μC के दो आवेश 5.0 mm दूर हैं। (a) अक्ष पर केंद्र से 15 cm दूर (धनावेश की ओर) और (b) विषुवतीय रेखा पर केंद्र से 15 cm दूर क्षेत्र ज्ञात करें।±10 μC ke do charges 5.0 mm door hain. (a) Axis par centre se 15 cm door (positive charge ki taraf) aur (b) equatorial line par centre se 15 cm door field nikaalo.Two charges of +10 μC and −10 μC are kept 5.0 mm apart. Find the electric field (a) on the axis, 15 cm from the centre, on the side of the positive charge, and (b) on the equatorial line, 15 cm from the centre.
Show solutionहल देखेंSolution dekhiye
p = q × 2a = 10⁻⁵ C × 5 × 10⁻³ m = 5 × 10⁻⁸ C m. Here r = 15 cm ≫ a = 0.25 cm, so use the short formulas. (a) E = 2kp/r³ = 2 × 9 × 10⁹ × 5 × 10⁻⁸/(0.15)³ ≈ 2.7 × 10⁵ N C⁻¹, along p (adding the two charges’ fields exactly, as NCERT does, also gives 2.7 × 10⁵). (b) E = kp/r³ ≈ 1.33 × 10⁵ N C⁻¹, opposite to p — half the axial value.p = q × 2a = 10⁻⁵ C × 5 × 10⁻³ m = 5 × 10⁻⁸ C m। यहाँ r = 15 cm ≫ a = 0.25 cm, इसलिए छोटे सूत्र लगाएँ। (a) E = 2kp/r³ = 2 × 9 × 10⁹ × 5 × 10⁻⁸/(0.15)³ ≈ 2.7 × 10⁵ N C⁻¹, p की दिशा में (NCERT की तरह दोनों आवेशों के क्षेत्र सटीक जोड़ने पर भी 2.7 × 10⁵ आता है)। (b) E = kp/r³ ≈ 1.33 × 10⁵ N C⁻¹, p के विपरीत — अक्षीय मान का आधा।p = q × 2a = 10⁻⁵ C × 5 × 10⁻³ m = 5 × 10⁻⁸ C m. Yahan r = 15 cm ≫ a = 0.25 cm, isliye short formulas lagaiye. (a) E = 2kp/r³ = 2 × 9 × 10⁹ × 5 × 10⁻⁸/(0.15)³ ≈ 2.7 × 10⁵ N C⁻¹, p ki direction mein (NCERT ki tarah dono charges ke fields exactly jodne par bhi 2.7 × 10⁵ aata hai). (b) E = kp/r³ ≈ 1.33 × 10⁵ N C⁻¹, p ke opposite — axial value ka aadha.First find the dipole moment: p = q × 2a = 10⁻⁵ C × 5 × 10⁻³ m = 5 × 10⁻⁸ C m. Here r = 15 cm is much bigger than a = 0.25 cm, so we can use the short formulas. (a) On the axis: E = 2kp/r³ = 2 × 9 × 10⁹ × 5 × 10⁻⁸/(0.15)³ ≈ 2.7 × 10⁵ N C⁻¹, along p. (NCERT adds the fields of the two charges exactly, and it also gets 2.7 × 10⁵.) (b) On the equatorial line: E = kp/r³ ≈ 1.33 × 10⁵ N C⁻¹, opposite to p. This is half the value on the axis.
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1.11Dipole in a Uniform External Fieldएकसमान बाह्य क्षेत्र में द्विध्रुवUniform External Field mein Dipole
Put a dipole in a uniform field E, its moment p making angle θ with E. The force on +q is qE (along E) and on −q is qE (opposite to E). They are equal and opposite, so the net force is zero. But they act at different points, so they form a couple that turns the dipole (see the torque).द्विध्रुव को एकसमान क्षेत्र E में रखें, उसका आघूर्ण p, E से θ कोण बनाए। +q पर बल qE (E की दिशा में) और −q पर qE (E के विपरीत) है। ये बराबर और विपरीत हैं, इसलिए नेट बल शून्य है। पर ये अलग बिंदुओं पर लगते हैं, इसलिए एक बलयुग्म बनाते हैं जो द्विध्रुव को घुमाता है (बल आघूर्ण देखें)।Dipole ko uniform field E mein rakhiye, uska moment p, E se θ angle banaye. +q par force qE (E ki direction mein) aur −q par qE (E ke opposite). Ye barabar aur opposite hain, isliye net force zero hai. Par ye alag points par lagte hain, isliye ek couple banaate hain jo dipole ko ghumaata hai (torque dekhiye).In a uniform field a dipole does not move as a whole; it only turns until it lines up with the field.
Set the angle, p and E, read τ = pE sin θ, then let go and watch the dipole swing into line with the field.कोण, p और E चुनें, τ = pE sin θ पढ़ें, फिर छोड़ दें और द्विध्रुव को क्षेत्र की सीध में झूलते देखें।Angle, p aur E set kijiye, τ = pE sin θ padhiye, phir chhod do aur dipole ko field ki seedh mein jhoolte dekhiye.
Non-uniform field: now the net force is generally not zero. If p is parallel to E, the dipole is pulled towards the stronger field; if antiparallel, towards the weaker field. This explains why a comb charged by combing dry hair attracts bits of paper: the comb’s field polarises the paper (induces a dipole along the field), and since the comb’s field is non-uniform, the paper is pulled towards the comb.असमान क्षेत्र: अब नेट बल सामान्यतः शून्य नहीं होता। यदि p, E के समांतर है तो द्विध्रुव प्रबल क्षेत्र की ओर खिंचता है; प्रतिसमांतर हो तो दुर्बल क्षेत्र की ओर। इससे समझ आता है कि सूखे बालों में फेरी गई कंघी काग़ज़ के टुकड़ों को क्यों खींचती है: कंघी का क्षेत्र काग़ज़ को ध्रुवित करता है (क्षेत्र की दिशा में द्विध्रुव प्रेरित करता है), और कंघी का क्षेत्र असमान है, इसलिए काग़ज़ कंघी की ओर खिंचता है।Non-uniform field: ab net force aam taur par zero nahi hota. Agar p, E ke parallel hai to dipole strong field ki taraf kheenchta hai; antiparallel ho to weak field ki taraf. Isse samajh aata hai ki sookhe baalon mein ghumaayi kanghi kaagaz ke tukdon ko kyun kheenchti hai: kanghi ka field kaagaz ko polarise karta hai (field ki direction mein dipole induce karta hai), aur kanghi ka field non-uniform hai, isliye kaagaz kanghi ki taraf kheenchta hai.Non-uniform field (a field that is not the same everywhere): now the net force on the dipole is usually not zero. If p points along E, the dipole is pulled towards the place where the field is stronger. If p points opposite to E, it is pulled towards the weaker field. This explains a common thing: a comb charged by combing dry hair pulls small bits of paper. The comb’s field polarises the paper (makes a dipole in it, along the field). The comb’s field is non-uniform, so the paper is pulled towards the comb.
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1.12Continuous Charge Distributionसंतत आवेश वितरणContinuous Charge Distribution
Charged rods, plates and spheres carry so many charges that it is easier to treat the charge as spread continuously. We describe it by a charge density (see all three):आवेशित छड़ों, प्लेटों और गोलों पर इतने अधिक आवेश होते हैं कि आवेश को संतत फैला हुआ मानना आसान है। इसे आवेश घनत्व से बताते हैं (तीनों देखें):Charged rods, plates aur spheres par itne zyada charges hote hain ki charge ko continuously phaila hua maanna aasaan hai. Ise charge density se batate hain (teeno dekhiye):Charged rods, plates and spheres have a huge number of charges. So it is easier to think of the charge as spread out smoothly (continuously). We describe this using a charge density (how much charge there is in a given length, area or volume): 👉 (see all three)
- Linear charge density λ = Δq/Δl — charge per unit length (C m⁻¹), for a wire or thin rod.रैखिक आवेश घनत्व λ = Δq/Δl — प्रति एकांक लंबाई आवेश (C m⁻¹), तार या पतली छड़ के लिए।Linear charge density λ = Δq/Δl — per unit length charge (C m⁻¹), wire ya patli rod ke liye.Linear charge density λ = Δq/Δl — charge on each unit of length (C m⁻¹). We use it for a wire or a thin rod.
- Surface charge density σ = ΔQ/ΔS — charge per unit area (C m⁻²), for a sheet or the surface of a conductor.पृष्ठ आवेश घनत्व σ = ΔQ/ΔS — प्रति एकांक क्षेत्रफल आवेश (C m⁻²), चादर या चालक के पृष्ठ के लिए।Surface charge density σ = ΔQ/ΔS — per unit area charge (C m⁻²), sheet ya conductor ki surface ke liye.Surface charge density σ = ΔQ/ΔS — charge on each unit of area (C m⁻²). We use it for a sheet, or for the surface of a conductor.
- Volume charge density ρ = ΔQ/ΔV — charge per unit volume (C m⁻³), for charge filling a region.आयतन आवेश घनत्व ρ = ΔQ/ΔV — प्रति एकांक आयतन आवेश (C m⁻³), किसी क्षेत्र में भरे आवेश के लिए।Volume charge density ρ = ΔQ/ΔV — per unit volume charge (C m⁻³), kisi region mein bhare charge ke liye.Volume charge density ρ = ΔQ/ΔV — charge in each unit of volume (C m⁻³). We use it when charge fills up a region of space.
Here ΔS (or Δl, ΔV) is small on the everyday scale but still contains a very large number of electrons — so the density is a smooth (macroscopic) quantity and we can ignore quantisation. The field of the whole body is found by superposition: break it into small elements, take each element’s field k(ρΔV)/r² and add them as vectors (in the limit, an integral).यहाँ ΔS (या Δl, ΔV) रोज़मर्रा के पैमाने पर छोटा है पर उसमें भी बहुत अधिक इलेक्ट्रॉन हैं — इसलिए घनत्व एक चिकनी (स्थूल) राशि है और क्वांटमीकरण की उपेक्षा कर सकते हैं। पूरी वस्तु का क्षेत्र अध्यारोपण से निकलता है: उसे छोटे अवयवों में बाँटें, हर अवयव का क्षेत्र k(ρΔV)/r² लें और सदिश रूप से जोड़ें (सीमा में, समाकलन)।Yahan ΔS (ya Δl, ΔV) roz ke scale par chhota hai par usmein bhi bahut zyada electrons hain — isliye density ek smooth (macroscopic) quantity hai aur quantisation ignore kar sakte hain. Poori body ka field superposition se nikalta hai: use chhote elements mein todo, har element ka field k(ρΔV)/r² lo aur vector ki tarah jodiye (limit mein, integral).Here ΔS (or Δl, or ΔV) is small for everyday purposes, but it still holds a very large number of electrons. So the density is a smooth (macroscopic) quantity, and we can ignore quantisation. To find the field of the whole body, we use superposition. We break the body into small pieces. We find the field of each piece, k(ρΔV)/r². Then we add all these fields as vectors (in the limit, this sum becomes an integral).
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1.13Gauss’s Lawगाउस नियमGauss’s Law
Take a point charge q and imagine a sphere of radius r around it, with q at the centre (the Gaussian sphere). On the sphere, E = q/(4πε₀r²) everywhere and it points along the outward normal. So the total flux is φ = E × (area) = [q/(4πε₀r²)] × 4πr² = q/ε₀. The radius cancels! Gauss showed this is true for any closed surface of any shape:एक बिंदु आवेश q लें और उसके चारों ओर r त्रिज्या का गोला सोचें, q केंद्र पर (गाउसीय गोला)। गोले पर हर जगह E = q/(4πε₀r²) है और बाहरी अभिलंब के अनुदिश है। अतः कुल फ्लक्स φ = E × (क्षेत्रफल) = [q/(4πε₀r²)] × 4πr² = q/ε₀। त्रिज्या कट गई! गाउस ने दिखाया कि यह किसी भी आकार के हर बंद पृष्ठ के लिए सत्य है:Ek point charge q lo aur uske chaaron taraf r radius ka sphere sochiye, q centre par (Gaussian sphere). Sphere par har jagah E = q/(4πε₀r²) hai aur outward normal ke along hai. Isliye total flux φ = E × (area) = [q/(4πε₀r²)] × 4πr² = q/ε₀. Radius cancel ho gaya! Gauss ne dikhaya ki ye kisi bhi shape ki har closed surface ke liye sach hai:Take a point charge q. Imagine a sphere of radius r around it, with q at the centre (this is called a Gaussian sphere). Everywhere on this sphere, E = q/(4πε₀r²), and it points straight outward. So the total flux is φ = E × (area) = [q/(4πε₀r²)] × 4πr² = q/ε₀. The radius r cancels out! Gauss showed that this result is true for any closed surface, of any shape: 👉 (the Gaussian sphere)
If a closed surface encloses no charge, the total flux through it is zero: every line that enters also leaves. For example, a closed cylinder placed in a uniform field: flux −EA through one end, +EA through the other, and zero through the curved side — total zero (see the cylinder).यदि किसी बंद पृष्ठ के अंदर कोई आवेश नहीं है, तो उससे कुल फ्लक्स शून्य है: जो रेखा अंदर आती है, वह बाहर भी जाती है। उदाहरण: एकसमान क्षेत्र में रखा बंद बेलन: एक सिरे से फ्लक्स −EA, दूसरे से +EA, और वक्र पृष्ठ से शून्य — कुल शून्य (बेलन देखें)।Agar kisi closed surface ke andar koi charge nahi, to usse total flux zero hai: jo line andar aati hai wo bahar bhi jaati hai. Example: uniform field mein rakha closed cylinder: ek end se flux −EA, doosre se +EA, aur curved side se zero — total zero (cylinder dekhiye).If a closed surface has no charge inside it, the total flux through it is zero. Every line that goes in also comes out. For example, put a closed cylinder in a uniform field. The flux is −EA through one flat end, +EA through the other flat end, and zero through the curved side. So the total is zero. 👉 (see the cylinder)
- Gauss’s law holds for any closed surface, whatever its shape or size.गाउस नियम किसी भी बंद पृष्ठ के लिए सत्य है, आकार या माप कुछ भी हो।Gauss’s law kisi bhi closed surface ke liye sach hai, shape ya size kuch bhi ho.Gauss’s law is true for any closed surface, whatever its shape or size.
- q on the right side is the sum of all charges enclosed, wherever they sit inside. Charges outside give zero net flux.दाईं ओर q अंदर के सभी आवेशों का योग है, वे अंदर कहीं भी हों। बाहर के आवेशों से नेट फ्लक्स शून्य।Right side ka q andar ke saare charges ka sum hai, wo andar kahin bhi hon. Bahar ke charges se net flux zero.The q on the right side is the total of all the charges inside the surface, wherever they are placed inside. Charges outside the surface give zero net flux.
- But the E in the flux is the total field at the surface — due to charges inside and outside.पर फ्लक्स वाला E पृष्ठ पर कुल क्षेत्र है — अंदर और बाहर दोनों आवेशों के कारण।Par flux wala E surface par total field hai — andar aur bahar dono charges ki wajah se.But be careful: the E used in the flux is the total field at the surface. It comes from the charges inside and the charges outside.
- The chosen surface is called the Gaussian surface. It should not pass through a discrete point charge (the field there is not defined); it may pass through a continuous charge distribution.चुना गया पृष्ठ गाउसीय पृष्ठ कहलाता है। यह किसी बिंदु आवेश से होकर नहीं जाना चाहिए (वहाँ क्षेत्र परिभाषित नहीं); संतत वितरण से होकर जा सकता है।Chuni gayi surface Gaussian surface kehlaati hai. Ye kisi point charge se hokar nahi jaani chahiye (wahan field defined nahi); continuous distribution se hokar ja sakti hai.The closed surface we choose is called the Gaussian surface. It should not pass through a single point charge, because the field at that point is not defined. But it may pass through charge that is spread out continuously.
- Gauss’s law makes field calculations easy when there is symmetry (spherical, cylindrical, plane). It is based on the inverse-square law: it would not hold for a different power of r.गाउस नियम सममिति (गोलीय, बेलनाकार, समतल) होने पर क्षेत्र की गणना आसान कर देता है। यह व्युत्क्रम-वर्ग नियम पर आधारित है: r की किसी और घात के लिए सत्य न होता।Gauss’s law symmetry (spherical, cylindrical, plane) hone par field ki calculation aasaan kar deta hai. Ye inverse-square law par based hai: r ki kisi aur power ke liye sach na hota.Gauss’s law makes finding the field easy when the charge has symmetry — spherical (like a ball), cylindrical (like a pipe) or plane (like a flat sheet). The law comes from the inverse-square law. If force depended on some other power of r, Gauss’s law would not be true.
Drag and resize the Gaussian surface around the charges. The flux depends only on the charge inside — check it with charges inside and outside.गाउसीय पृष्ठ को आवेशों के चारों ओर खिसकाएँ और छोटा-बड़ा करें। फ्लक्स केवल अंदर के आवेश पर निर्भर है — अंदर और बाहर आवेश रखकर जाँचें।Gaussian surface ko charges ke aas-paas drag aur resize kijiye. Flux sirf andar ke charge par depend karta hai — andar aur bahar charges rakh kar check kijiye.
A field is E = +200 î N/C for x > 0 and −200 î N/C for x < 0. A cylinder of length 20 cm and radius 5 cm has its centre at the origin and axis along x. Find (a) the flux through each flat face, (b) through the side, (c) the net flux, (d) the charge inside.क्षेत्र x > 0 के लिए E = +200 î N/C और x < 0 के लिए −200 î N/C है। 20 cm लंबा, 5 cm त्रिज्या का बेलन मूल बिंदु पर केंद्रित है, अक्ष x के अनुदिश। ज्ञात करें (a) हर चपटे फलक से फ्लक्स, (b) पार्श्व से, (c) कुल फ्लक्स, (d) अंदर का आवेश।Field x > 0 ke liye E = +200 î N/C aur x < 0 ke liye −200 î N/C hai. 20 cm lamba, 5 cm radius ka cylinder origin par centred hai, axis x ke along. Nikaalo (a) har flat face se flux, (b) side se, (c) net flux, (d) andar ka charge.An electric field is E = +200 î N/C when x > 0, and −200 î N/C when x < 0. A cylinder of length 20 cm and radius 5 cm has its centre at the origin, and its axis lies along x. Find (a) the flux through each flat end, (b) the flux through the curved side, (c) the net flux, and (d) the charge inside the cylinder.
Show solutionहल देखेंSolution dekhiye
(a) On each face, E points outward along the face’s outward normal, so each flux = +EA = 200 × π(0.05)² = +1.57 N m² C⁻¹. (b) On the side, E ⊥ normal → 0. (c) Net = 1.57 + 1.57 + 0 = 3.14 N m² C⁻¹. (d) q = ε₀φ = 8.854 × 10⁻¹² × 3.14 ≈ 2.78 × 10⁻¹¹ C (positive — the field points away from the yz-plane on both sides).(a) हर फलक पर E बाहरी अभिलंब की दिशा में है, इसलिए हर फ्लक्स = +EA = 200 × π(0.05)² = +1.57 N m² C⁻¹। (b) पार्श्व पर E ⊥ अभिलंब → 0। (c) कुल = 1.57 + 1.57 + 0 = 3.14 N m² C⁻¹। (d) q = ε₀φ = 8.854 × 10⁻¹² × 3.14 ≈ 2.78 × 10⁻¹¹ C (धनात्मक — दोनों ओर क्षेत्र yz-तल से दूर है)।(a) Har face par E outward normal ki direction mein hai, isliye har flux = +EA = 200 × π(0.05)² = +1.57 N m² C⁻¹. (b) Side par E ⊥ normal → 0. (c) Net = 1.57 + 1.57 + 0 = 3.14 N m² C⁻¹. (d) q = ε₀φ = 8.854 × 10⁻¹² × 3.14 ≈ 2.78 × 10⁻¹¹ C (positive — dono taraf field yz-plane se door hai).(a) On each flat end, E points outward, along the outward normal of that end. So the flux through each end = +EA = 200 × π(0.05)² = +1.57 N m² C⁻¹. (b) On the curved side, E is perpendicular to the normal (at right angles), so the flux is 0. (c) Net flux = 1.57 + 1.57 + 0 = 3.14 N m² C⁻¹. (d) From Gauss’s law, q = ε₀φ = 8.854 × 10⁻¹² × 3.14 ≈ 2.78 × 10⁻¹¹ C. The charge is positive, because the field points away from the yz-plane on both sides.
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1.14Applications of Gauss’s Lawगाउस नियम के अनुप्रयोगGauss’s Law ke Applications
The trick: pick a Gaussian surface on which E is either constant and along the normal or parallel to the surface (zero flux). Then φ is just E × (area), and Gauss’s law gives E in one line.तरकीब: ऐसा गाउसीय पृष्ठ चुनें जिस पर E या तो स्थिर और अभिलंब के अनुदिश हो या पृष्ठ के समांतर (शून्य फ्लक्स)। तब φ सीधे E × (क्षेत्रफल) है, और गाउस नियम एक पंक्ति में E दे देता है।Trick: aisi Gaussian surface chuniye jis par E ya to constant aur normal ke along ho ya surface ke parallel (zero flux). Tab φ seedha E × (area) hai, aur Gauss’s law ek line mein E de deta hai.The trick: choose a Gaussian surface where E is either the same everywhere and along the normal, or along the surface (so the flux is zero). Then the flux φ is simply E × (area), and Gauss’s law gives E in one line.
1.14.1Field due to an infinitely long straight uniformly charged wireअनंत लंबे सीधे एकसमान आवेशित तार के कारण क्षेत्रInfinitely long straight charged wire ka Field
By symmetry the field is radial (perpendicular to the wire) and has the same size at the same distance r. Take a coaxial cylinder of radius r and length l. The flat ends give no flux (E lies along them); the curved surface gives E × 2πrl. Charge inside = λl. So E × 2πrl = λl/ε₀:सममिति से क्षेत्र त्रिज्य (तार के लंबवत) है और समान दूरी r पर समान परिमाण का है। r त्रिज्या और l लंबाई का समाक्ष बेलन लें। चपटे सिरों से फ्लक्स नहीं (E उनके अनुदिश है); वक्र पृष्ठ से E × 2πrl। अंदर का आवेश = λl। अतः E × 2πrl = λl/ε₀:Symmetry se field radial (wire ke perpendicular) hai aur same doori r par same magnitude ka hai. r radius aur l length ka coaxial cylinder lo. Flat ends se flux nahi (E unke along hai); curved surface se E × 2πrl. Andar ka charge = λl. Isliye E × 2πrl = λl/ε₀:Think of a long, straight, charged wire. By symmetry, the field is radial (it points straight out from the wire, at right angles to it), and it has the same size at the same distance r. Take a cylinder of radius r and length l with the wire along its axis. The flat ends give no flux, because E runs along them. The curved surface gives E × 2πrl. The charge inside is λl. So E × 2πrl = λl/ε₀. This gives:
1.14.2Field due to a uniformly charged infinite plane sheetएकसमान आवेशित अनंत समतल चादर के कारण क्षेत्रInfinite charged plane sheet ka Field
By symmetry E is perpendicular to the sheet, pointing away on both sides (σ > 0). Take a cylinder (pillbox) through the sheet with end faces of area A. Only the two ends carry flux: 2EA. Charge inside = σA. So 2EA = σA/ε₀:सममिति से E चादर के लंबवत है, दोनों ओर दूर की ओर (σ > 0)। चादर के आर-पार एक बेलन लें जिसके सिरों का क्षेत्रफल A हो। केवल दोनों सिरों से फ्लक्स: 2EA। अंदर का आवेश = σA। अतः 2EA = σA/ε₀:Symmetry se E sheet ke perpendicular hai, dono taraf door ki taraf (σ > 0). Sheet ke aar-paar ek cylinder lo jiske ends ka area A ho. Sirf dono ends se flux: 2EA. Andar ka charge = σA. Isliye 2EA = σA/ε₀:Think of a flat, charged sheet. By symmetry, E is perpendicular to the sheet (at right angles to it), and points away on both sides (when σ > 0). Take a small cylinder (a pillbox) that passes through the sheet, with end faces of area A. Only the two ends have flux, so the total flux is 2EA. The charge inside is σA. So 2EA = σA/ε₀. This gives:
1.14.3Field due to a uniformly charged thin spherical shellएकसमान आवेशित पतले गोलीय कोश के कारण क्षेत्रThin charged spherical shell ka Field
Let a shell of radius R carry charge q = 4πR²σ. Outside (r ≥ R): a concentric Gaussian sphere of radius r encloses q, so E × 4πr² = q/ε₀ and E = q/(4πε₀r²) — exactly as if all the charge were at the centre. Inside (r < R): the Gaussian sphere encloses no charge, so the flux is zero and E = 0 everywhere inside the shell.R त्रिज्या के कोश पर आवेश q = 4πR²σ हो। बाहर (r ≥ R): r त्रिज्या का संकेंद्री गाउसीय गोला q को घेरता है, इसलिए E × 4πr² = q/ε₀ और E = q/(4πε₀r²) — ठीक वैसे जैसे सारा आवेश केंद्र पर हो। अंदर (r < R): गाउसीय गोला कोई आवेश नहीं घेरता, इसलिए फ्लक्स शून्य और कोश के अंदर हर जगह E = 0।R radius ke shell par charge q = 4πR²σ ho. Bahar (r ≥ R): r radius ka concentric Gaussian sphere q ko gherta hai, isliye E × 4πr² = q/ε₀ aur E = q/(4πε₀r²) — bilkul waise jaise saara charge centre par ho. Andar (r < R): Gaussian sphere koi charge nahi gherta, isliye flux zero aur shell ke andar har jagah E = 0.Outside the shell: same as a point charge at the centre. Inside the shell: no field at all.
Slide the distance r and compare: the wire’s field falls as 1/r, the sheet’s stays constant, and the shell’s is zero inside and falls as 1/r² outside.दूरी r खिसकाएँ और तुलना करें: तार का क्षेत्र 1/r से घटता है, चादर का स्थिर रहता है, और कोश का अंदर शून्य व बाहर 1/r² से घटता है।Doori r slide kijiye aur compare kijiye: wire ka field 1/r se ghatata hai, sheet ka constant rehta hai, aur shell ka andar zero aur bahar 1/r² se ghatata hai.
Why is E = 0 inside the shell? Gauss’s law says the flux through the inner sphere is zero, and by symmetry E must have the same size everywhere on it and point radially — so E itself must be zero. (Zero flux alone would not be enough without symmetry.) The same Gauss argument shows a uniformly charged solid sphere also has E = q/(4πε₀r²) outside. Inside a solid sphere, only the charge within radius r counts, which makes E grow in proportion to r.कोश के अंदर E = 0 क्यों? गाउस नियम कहता है कि भीतरी गोले से फ्लक्स शून्य है, और सममिति से E उस पर हर जगह समान परिमाण का और त्रिज्य होना चाहिए — इसलिए E स्वयं शून्य है। (सममिति के बिना केवल शून्य फ्लक्स पर्याप्त नहीं होता।) यही तर्क दिखाता है कि एकसमान आवेशित ठोस गोले का भी बाहर E = q/(4πε₀r²) है। ठोस गोले के अंदर केवल r त्रिज्या के भीतर का आवेश गिना जाता है, जिससे E, r के अनुपात में बढ़ता है।Shell ke andar E = 0 kyun? Gauss’s law kehta hai ki andar wale sphere se flux zero hai, aur symmetry se E us par har jagah same magnitude ka aur radial hona chahiye — isliye E khud zero hai. (Symmetry ke bina sirf zero flux kaafi nahi hota.) Yahi argument dikhata hai ki uniformly charged solid sphere ka bhi bahar E = q/(4πε₀r²) hai. Solid sphere ke andar sirf r radius ke andar ka charge gina jaata hai, jisse E, r ke proportion mein badhta hai.Why is E = 0 inside the shell? Gauss’s law says the flux through a sphere drawn inside the shell is zero. By symmetry, E must have the same size everywhere on that sphere and must point radially (straight out). So E itself must be zero. (Zero flux alone would not prove this; we also need the symmetry.) The same Gauss argument shows that a uniformly charged solid sphere also has E = q/(4πε₀r²) outside it. Inside a solid sphere, only the charge within radius r counts. This makes E grow in proportion to r (E is directly proportional to r).
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★Formula sheetसूत्र सूचीFormula sheet
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- Two kinds of charge; like repel, unlike attract. Glass rubbed with silk: +; plastic rubbed with fur: −. Charging by rubbing = transfer of electrons.दो प्रकार के आवेश; सजातीय प्रतिकर्षित, विजातीय आकर्षित। रेशम से रगड़ा काँच: +; फर से रगड़ा प्लास्टिक: −। रगड़कर आवेशन = इलेक्ट्रॉनों का स्थानांतरण।Do type ke charge; like repel, unlike attract. Silk se ragda glass: +; fur se ragda plastic: −. Ragad kar charging = electrons ka transfer.There are two kinds of charge. Like charges push each other away (repel); unlike charges pull each other (attract). Glass rubbed with silk becomes +. Plastic rubbed with fur becomes −. Charging by rubbing simply means electrons move (are transferred) from one body to the other.
- Conductors let charge move (metals, body, earth); insulators do not (glass, plastic, wood).चालक आवेश को चलने देते हैं (धातु, शरीर, पृथ्वी); विद्युतरोधी नहीं (काँच, प्लास्टिक, लकड़ी)।Conductors charge ko chalne dete hain (metals, body, earth); insulators nahi (glass, plastic, wood).Conductors let charge move through them (metals, the human body, the earth). Insulators do not (glass, plastic, wood).
- Three properties of charge: additivity (add with signs), conservation (total of an isolated system is constant), quantisation (q = ne).आवेश के तीन गुण: योज्यता (चिह्न सहित जोड़), संरक्षण (विलगित निकाय का कुल स्थिर), क्वांटमीकरण (q = ne)।Charge ke teen properties: additivity (sign ke saath jodiye), conservation (isolated system ka total constant), quantisation (q = ne).Charge has three properties. Additivity: charges add up, keeping their signs. Conservation: the total charge of an isolated system stays the same. Quantisation: charge comes only in whole-number steps, q = ne.
- Coulomb’s law: F ∝ q₁q₂/r², along the joining line; k = 9 × 10⁹ N m² C⁻². Electric force ≫ gravity (≈ 10³⁹ times for e–p).कूलॉम नियम: F ∝ q₁q₂/r², जोड़ने वाली रेखा के अनुदिश; k = 9 × 10⁹ N m² C⁻²। विद्युत बल ≫ गुरुत्व (e–p के लिए ≈ 10³⁹ गुना)।Coulomb’s law: F ∝ q₁q₂/r², joining line ke along; k = 9 × 10⁹ N m² C⁻². Electric force ≫ gravity (e–p ke liye ≈ 10³⁹ guna).Coulomb’s law: F ∝ q₁q₂/r². The force acts along the line joining the two charges. k = 9 × 10⁹ N m² C⁻². Electric force is far bigger than gravity (about 10³⁹ times for an electron and a proton).
- Superposition: net force (or field) = vector sum of individual forces (fields).अध्यारोपण: नेट बल (या क्षेत्र) = अलग-अलग बलों (क्षेत्रों) का सदिश योग।Superposition: net force (ya field) = alag-alag forces (fields) ka vector sum.Superposition: the net force (or field) is the vector sum of the separate forces (or fields) from each charge.
- Field E = F/q; point charge E = kq/r², outward for +, inward for −.क्षेत्र E = F/q; बिंदु आवेश E = kq/r², + के लिए बाहर, − के लिए अंदर।Field E = F/q; point charge E = kq/r², + ke liye bahar, − ke liye andar.Electric field E = F/q (force per unit charge). For a point charge, E = kq/r². It points outward for +, inward for −.
- Field lines: start on +, end on −; never cross; no breaks in charge-free space; no closed loops; crowded = strong.क्षेत्र रेखाएँ: + से शुरू, − पर अंत; कभी नहीं काटतीं; आवेश-रहित स्थान में अटूट; बंद लूप नहीं; घनी = प्रबल।Field lines: + se shuru, − par khatam; kabhi cross nahi; charge-free space mein toot-ti nahi; closed loop nahi; ghani = strong.Field lines: they begin on + charges and end on − charges. They never cross. They have no breaks in a space with no charge. They do not form closed loops. Where lines are crowded close together, the field is strong.
- Flux Δφ = EΔS cos θ; for a closed surface take the outward normal.फ्लक्स Δφ = EΔS cos θ; बंद पृष्ठ के लिए बाहरी अभिलंब लें।Flux Δφ = EΔS cos θ; closed surface ke liye outward normal lo.Flux Δφ = EΔS cos θ. For a closed surface, always take the normal that points outward.
- Dipole: p = q·2a (−q → +q). Far field ∝ 1/r³: axial 2kp/r³ along p, equatorial kp/r³ opposite to p.द्विध्रुव: p = q·2a (−q → +q)। दूर का क्षेत्र ∝ 1/r³: अक्षीय 2kp/r³ p की दिशा में, विषुवतीय kp/r³ p के विपरीत।Dipole: p = q·2a (−q → +q). Door ka field ∝ 1/r³: axial 2kp/r³ p ki direction, equatorial kp/r³ p ke opposite.Dipole: p = q·2a, pointing from −q to +q. Far away, its field ∝ 1/r³. On the axis, the field is 2kp/r³ along p. On the equatorial line, it is kp/r³ opposite to p.
- Dipole in uniform E: net force 0, torque τ = p × E (pE sin θ). Non-uniform E: net force too.एकसमान E में द्विध्रुव: नेट बल 0, बल आघूर्ण τ = p × E (pE sin θ)। असमान E: नेट बल भी।Uniform E mein dipole: net force 0, torque τ = p × E (pE sin θ). Non-uniform E: net force bhi.A dipole in a uniform E: the net force is 0, but there is a torque (turning effect) τ = p × E, of size pE sin θ. In a non-uniform E, there is a net force as well.
- Gauss’s law: φ = q_enclosed/ε₀ for any closed surface. Wire λ/(2πε₀r); sheet σ/(2ε₀); shell: kq/r² outside, 0 inside.गाउस नियम: किसी भी बंद पृष्ठ के लिए φ = q_अंदर/ε₀। तार λ/(2πε₀r); चादर σ/(2ε₀); कोश: बाहर kq/r², अंदर 0।Gauss’s law: kisi bhi closed surface ke liye φ = q_andar/ε₀. Wire λ/(2πε₀r); sheet σ/(2ε₀); shell: bahar kq/r², andar 0.Gauss’s law: φ = q_enclosed/ε₀ (flux = charge inside ÷ ε₀) for any closed surface. Long wire: E = λ/(2πε₀r). Flat sheet: E = σ/(2ε₀). Thin shell: kq/r² outside, and 0 inside.
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These are Simple Learning Hub’s own study notes written to follow the current (rationalised) NCERT Class 12 Physics textbook section by section; read them with the official NCERT chapter. Diagrams and simulations are original SLH work based on the NCERT figures they name. Report a mistake