Class 12 Physics · Chapter 1
Short answer:
Ye chapter Class 12 Physics ka foundation chapter hai — poori electrostatics (potential, capacitance, current) isi ke concepts par tiki hai. NCERT me is chapter ke exercise me currently 20 questions hain (numerical + conceptual mix), jinme Coulomb's law force calculations, superposition principle se multiple charges ka net field/force, electric dipole (axial + equatorial field, torque), aur Gauss's law ki applications (line charge, plane sheet, spherical shell) cover hoti hain. Board exam me isse direct numerical aata hai, aur JEE/NEET me Gauss's law applications + dipole ke questions bahut frequent hain.
Electric charge matter ki ek fundamental property hai jiski wajah se woh electric aur magnetic forces experience karta hai — is chapter me hum charge ke basic rules (conservation, quantization), Coulomb's law, electric field ka concept, dipole, aur Gauss's law seekhte hain. Board exam ke liye ye ek high-weightage, numerical-heavy chapter hai, aur JEE Main/Advanced dono me Gauss's law + dipole ke questions almost har saal poochhe jaate hain. Is chapter ki understanding NCERT Class 12 Physics ke agle teen chapters (Electrostatic Potential and Capacitance, Current Electricity, Moving Charges and Magnetism) ki neev hai — isliye concepts clear hona zaroori hai, sirf formula rat lena kaafi nahi.
Chapter 1 Summary — 5 Minute Revision
1. Electric Charge — Basic Properties
Charge ek scalar quantity hai jo do types me hoti hai — positive aur negative. Do fundamental rules hamesha yaad rakho:
- Conservation of charge: Kisi bhi isolated system ka total charge constant rehta hai — charge na banaya ja sakta hai na destroy kiya ja sakta hai, sirf transfer hota hai (rubbing se electrons ek body se dusri body me shift hote hain, naya charge create nahi hota).
- Quantization of charge: Koi bhi charge q hamesha ek integer multiple hota hai elementary charge e ka: q = ne, jahan e = 1.6×10-19 C aur n = ±1, ±2, ... Free state me isse chhota charge exist nahi karta (quarks bound state me hote hain, isliye unka fractional charge is rule ko violate nahi karta).
2. Coulomb's Law
Do point charges ke beech electrostatic force unke charges ke product ke directly proportional aur unke beech ki distance ke square ke inversely proportional hota hai:
F = k q1q2 / r², jahan k = 1/4πε0 ≈ 9×109 N·m²/C²
Force hamesha dono charges ko joinne wali line ke along hota hai (central force) — same sign ke charges repel karte hain, opposite sign attract karte hain. Yaad rakho ki ye vector equation hai, sirf magnitude nahi.
3. Superposition Principle
Jab multiple charges present hon, to kisi ek charge par net force uski wajah se har dusre charge se lagne wale individual forces ka vector sum hota hai — ek-ek pair ka Coulomb force nikaalo, phir vector addition karo. Yehi principle electric field ke liye bhi apply hota hai.
4. Electric Field
Kisi point par electric field us jagah rakhe gaye ek chhote positive test charge q0 par force per unit charge hai:
E = F/q0 = kQ/r² (point charge Q ke liye)
Units N/C (ya equivalently V/m). Field lines conventionally positive charge se nikalti hain aur negative charge me enter karti hain, kabhi cross nahi hoti, aur density field ki strength batati hai.
5. Electric Dipole
Do equal aur opposite charges (+q, -q) jo ek chhoti distance 2a se separated hon, unhe electric dipole kehte hain. Dipole moment:
p = q × 2a (direction: -q se +q ki taraf)
Dipole ka electric field (short dipole approximation, r >> a):
- Axial line (end-on) par: Eaxial = 2kp/r³ (direction p ke parallel)
- Equatorial line (broadside-on) par: Eeq = kp/r³ (direction p ke antiparallel)
Axial field, equatorial field se double hota hai same distance par — ye ek common conceptual/numerical trap hai.
Jab dipole ko uniform external field E me rakha jaata hai, to net force zero hota hai (equal opposite forces on +q aur -q), lekin ek torque lagta hai jo dipole ko field ke along align karne ki koshish karta hai:
τ = p × E = pE sinθ
Torque maximum hota hai θ=90° par aur zero hota hai θ=0° ya 180° par (dono stable/unstable equilibrium positions). Dipole ko ek orientation se dusri orientation tak rotate karne me kiya gaya work: W = pE(cosθ1 - cosθ2).
6. Continuous Charge Distribution
Bahut saare closely-spaced charges ko ek continuous charge distribution maan sakte hain, described by:
- Linear charge density λ = charge/length (C/m) — wire jaisi lambi cheez ke liye
- Surface charge density σ = charge/area (C/m²) — sheet/plate ke liye
- Volume charge density ρ = charge/volume (C/m³) — solid body ke liye
7. Electric Flux aur Gauss's Law
Electric flux ek surface se guzarne wali field lines ki measure hai: Δφ = E·ΔA = EΔA cosθ.
Gauss's law kehta hai ki kisi bhi closed surface (Gaussian surface) se guzarne wala total electric flux sirf usme enclosed charge par depend karta hai — surface ka shape ya size matter nahi karta:
∮ E·dA = Qenclosed / ε0
Surface ke bahar rakhe charges net flux me contribute nahi karte (unki field lines surface me enter karke wapas exit ho jaati hain). Gauss's law Coulomb's law ka hi ek different (aur zyada powerful) form hai — high symmetry problems me field nikalna bahut easy ho jaata hai.
8. Gauss's Law ki Standard Applications
- Infinite line charge (linear density λ), cylindrical Gaussian surface use karke: E = λ/(2πε0r), radially outward, distance r ke inversely proportional (1/r, na ki 1/r²).
- Infinite plane sheet of charge (surface density σ), pillbox Gaussian surface use karke: E = σ/(2ε0), distance se independent (uniform field), dono taraf directed perpendicular to sheet.
- Charged conductor ki surface ke paas (charge sirf ek side available): E = σ/ε0 — sheet wale case se factor of 2 zyada, kyunki conductor ke andar field zero hoti hai.
- Uniformly charged thin spherical shell (total charge Q, radius R), spherical
Gaussian surface use karke:
- Andar (r < R): E = 0
- Surface par (r = R): E = kQ/R²
- Bahar (r > R): E = kQ/r² (bilkul point charge jaisa behave karta hai)
Poore Class 12 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q19)
Q1. Do point charges q1 = 2×10-7 C aur q2 = 3×10-7 C ek dusre se 30 cm door hain. Unke beech electrostatic force ka magnitude nikaalo.
Coulomb's law se force nikalte hain.
Given: q1 = 2×10-7 C, q2 = 3×10-7 C, r = 0.30 m
F = kq1q2/r² = (9×109 × 2×10-7 × 3×10-7) / (0.30)²
F = (9×109 × 6×10-14) / 0.09 = 5.4×10-4 / 0.09
F = 6×10-3 N = 6 mN (repulsive, kyunki dono charges same sign hain)
Q2. Hydrogen atom me proton aur electron ke beech electrostatic force ki gravitational force se tulna karo (Felectric/Fgravity ka ratio nikaalo).
Dono forces me distance r same hoti hai, isliye ratio nikaalte waqt r² cancel ho jaata hai.
Fe/Fg = ke² / (Gmemp)
Given: k = 9×109, e = 1.6×10-19 C, G = 6.67×10-11, me = 9.1×10-31 kg, mp = 1.67×10-27 kg
Numerator: ke² = 9×109 × (1.6×10-19)² = 2.30×10-28
Denominator: Gmemp = 6.67×10-11 × 9.1×10-31 × 1.67×10-27 = 1.01×10-67
Fe/Fg ≈ 2.27×1039 — electric force gravitational force se lagbhag 1039 guna strong hai, isliye atomic/molecular scale par gravity completely negligible hoti hai.
Q3. Ek equilateral triangle ke teeno corners par equal charge q = 2×10-6 C rakhe hain, side length 10 cm hai. Kisi ek charge par lagne wala net force nikaalo.
Pehle ek pair ke beech force nikaalte hain, phir vector addition karte hain.
Each pair force: F = kq²/r² = 9×109 × (2×10-6)² / (0.10)² = 9×109 × 4×10-12 / 0.01 = 3.6 N
Kisi ek charge par baaki do charges se do forces lagte hain, dono magnitude 3.6 N ke, aur unke beech angle = triangle ka internal angle = 60°
Resultant = √(F² + F² + 2F²cos60°) = F√3 = 3.6 × 1.732
Fnet ≈ 6.24 N, direction — triangle ke centroid se us charge ki taraf jaane wali line ke along, bahar ki taraf (sab charges same sign hain, isliye repulsion)
Q4. Ek point charge q = 5×10-9 C se 20 cm door electric field ka magnitude nikaalo.
E = kq/r² = (9×109 × 5×10-9) / (0.20)² = 45 / 0.04
E = 1125 N/C, direction charge se radially outward (kyunki charge positive hai)
Q5. Ek electric dipole me charges ±3×10-6 C hain, jo 4 cm (2a) se separated hain. Dipole moment nikaalo.
Given: q = 3×10-6 C, 2a = 0.04 m
p = q × 2a = 3×10-6 × 0.04
p = 1.2×10-7 C·m, direction -q se +q ki taraf
Q6. Q5 wale dipole ka electric field uski axial line par, center se 10 cm door, nikaalo.
Given: p = 1.2×10-7 C·m, r = 0.10 m
Eaxial = 2kp/r³ = (2 × 9×109 × 1.2×10-7) / (0.10)³
Eaxial = 2160 / 0.001
Eaxial = 2.16×106 N/C, dipole moment p ke parallel direction me
Q7. Wahi dipole (Q5), same 10 cm distance par lekin equatorial line par electric field nikaalo. Axial field se compare karo.
Eeq = kp/r³ = (9×109 × 1.2×10-7) / (0.10)³ = 1080/0.001
Eeq = 1.08×106 N/C, dipole moment p ke antiparallel direction me
Comparison: Eaxial = 2.16×106 N/C = 2 × Eeq — same distance par axial field hamesha equatorial field ka double hota hai.
Q8. Wahi dipole (p = 1.2×10-7 C·m) ek uniform electric field E = 1×105 N/C me field ke saath 30° angle par rakha hai. Torque nikaalo.
τ = pE sinθ = 1.2×10-7 × 1×105 × sin30°
τ = 1.2×10-7 × 1×105 × 0.5 = 1.2×10-2 × 0.5
τ = 6×10-3 N·m, direction — dipole ko field ke saath align karne ki taraf
Q9. Ek dipole (p = 2×10-8 C·m) ek uniform field E = 5×104 N/C me θ=0° se θ=180° tak rotate kiya jaata hai. Kiya gaya work nikaalo.
W = pE(cosθ1 - cosθ2) = pE(cos0° - cos180°) = pE(1-(-1)) = 2pE
W = 2 × 2×10-8 × 5×104 = 2 × 1×10-3
W = 2×10-3 J = 2 mJ — ye maximum possible work hai kyunki hum stable equilibrium (0°) se unstable equilibrium (180°) tak rotate kar rahe hain
Q10. Ek point charge q = 1 μC ek cube ke bilkul center par rakha hai. Cube se guzarne wala total electric flux aur ek face se guzarne wala flux nikaalo.
Gauss's law shape-independent hai, isliye cube ke liye bhi wahi formula chalega jo sphere ke liye chalta hai.
φtotal = q/ε0 = 1×10-6 / 8.854×10-12
φtotal ≈ 1.13×105 N·m²/C
Cube ke 6 identical faces hain aur charge exact center par hai, isliye symmetry se: φone face = φtotal/6
φone face ≈ 1.88×104 N·m²/C
Q11. Ek infinite line charge ka linear charge density λ = 2×10-6 C/m hai. Isse 5 cm perpendicular distance par electric field nikaalo.
E = λ/(2πε0r) = 2kλ/r (kyunki 1/2πε0 = 2k)
E = (2 × 9×109 × 2×10-6) / 0.05 = 3.6×104 / 0.05
E = 7.2×105 N/C, radially outward line se perpendicular direction me
Q12. Ek badi infinite plane sheet ka surface charge density σ = 8.85×10-12 C/m² hai. Sheet ke paas electric field nikaalo.
E = σ/(2ε0) = 8.85×10-12 / (2 × 8.854×10-12)
E ≈ 0.5 N/C, sheet se perpendicular direction me, dono taraf
Note: agar ye ek charged conducting surface hoti (charge sirf ek side available), to field double hoti: E = σ/ε0 ≈ 1 N/C
Q13. Ek uniformly charged thin spherical shell ka radius R = 10 cm aur total charge Q = 2×10-6 C hai. Electric field nikaalo: (a) shell ke andar r = 5 cm par, (b) shell ki surface par, (c) shell ke bahar r = 20 cm par.
Gauss's law ki spherical shell application — teeno regions ke liye alag formula.
(a) Andar (r = 5 cm < R): Gaussian sphere ke andar koi charge enclosed nahi hai, isliye E = 0
(b) Surface par (r = R = 0.10 m): E = kQ/R² = (9×109 × 2×10-6) / (0.10)² = 1.8×104/0.01
Esurface = 1.8×106 N/C
(c) Bahar (r = 20 cm = 0.20 m): E = kQ/r² = (9×109 × 2×10-6) / (0.20)² = 1.8×104/0.04
Eoutside = 4.5×105 N/C (bilkul waisa jaise pura charge Q center par point charge ho)
Q14. Ek closed surface ke andar koi charge nahi hai, lekin surface ke bahar kai charges rakhe hain. Is closed surface se guzarne wala net electric flux kya hoga? Reasoning do.
Ye ek conceptual application hai.
Gauss's law: φ = Qenclosed/ε0. Yahan Qenclosed = 0 (koi charge andar nahi hai)
φnet = 0
Reasoning: bahar ke charges ki field lines surface me kisi jagah se enter karke doosri jagah se exit ho jaati hain — net contribution zero rehta hai, chahe local field E kisi bhi point par zero na ho.
Q15. Kisi body me -1 C charge produce karne ke liye kitne electrons transfer karne honge?
q = ne ⇒ n = q/e
n = 1 / (1.6×10-19)
n = 6.25×1018 electrons (body me extra electrons transfer honge, taaki net charge negative ho)
Q16. Do identical charges q ek dusre se distance r par F force se repel karte hain. Agar separation double kar di jaaye, to naya force kya hoga?
Coulomb's law inverse-square law hai (F ∝ 1/r²), isliye ratio approach use karte hain.
Fnew/F = (r/2r)² = 1/4
Fnew = F/4 — distance double karne se force chauthai (1/4) ho jaata hai
Q17. Do point charges q1 = +4 μC aur q2 = -2 μC ek dusre se 12 cm door rakhe hain. Woh point dhundo jahan net electric field zero ho.
Charges opposite sign ke hain, isliye null point segment ke bahar hoga, chhote-magnitude charge (q2) ki taraf, us charge se aage.
Maan lo q2 se x distance par null point hai (q1 se door jaate hue), to q1 se distance = (0.12 + x)
Field magnitudes barabar: k(4μC)/(0.12+x)² = k(2μC)/x²
4x² = 2(0.12+x)² ⇒ 2x² = (0.12+x)² ⇒ √2 x = 0.12 + x
x(√2 - 1) = 0.12 ⇒ x = 0.12/0.4142
x ≈ 0.29 m (29 cm) q2 se aage, yani q1 se lagbhag 41 cm door, seedhi line ke us paar jahan q2 hai
Q18. Do charges +5μC aur +5μC 20 cm door rakhe hain. Unke beech midpoint par net electric field kitna hoga?
Midpoint par dono charges se distance equal hai, aur dono positive hain — fields opposite directions me point karengi.
Midpoint se har charge ki distance = 0.10 m
Har charge ka field: E = kq/r² = 9×109 × 5×10-6/(0.10)² = 4.5×106 N/C
Dono fields magnitude me equal hain lekin opposite directions me (kyunki dono charges positive hain aur midpoint dono ke beech me hai)
Enet = 0 at the midpoint
Q19. Ek electric dipole ko uniform field ke saath align kiya gaya hai (θ=0°). Explain karo ki ye stable equilibrium kyun hai, aur θ=180° unstable equilibrium kyun hai.
θ=0° par torque τ=pEsin0°=0, aur agar dipole ko thoda displace karo to restoring torque lagta hai jo use wapas θ=0° par le aata hai — stable equilibrium
θ=180° par bhi torque zero hai (τ=pEsin180°=0), lekin agar dipole ko thoda bhi displace karo to torque use aur door le jaata hai, wapas nahi laata — unstable equilibrium
Potential energy approach se: U = -pEcosθ hai; θ=0° par U minimum hota hai (stable), θ=180° par U maximum hota hai (unstable)
Important Equations — Ek Nazar Me
| Quantity | Formula | Notes |
|---|---|---|
| Coulomb's law | F = kq1q2/r² | k = 1/4πε0 ≈ 9×109 N·m²/C² |
| Permittivity of free space | ε0 ≈ 8.854×10-12 C²/N·m² | k = 1/4πε0 |
| Charge quantization | q = ne | e = 1.6×10-19 C, n = integer |
| Electric field (point charge) | E = kQ/r² = F/q0 | Units: N/C |
| Dipole moment | p = q × 2a | Direction: -q se +q ki taraf |
| Dipole field — axial (end-on) | Eaxial = 2kp/r³ | r >> a, direction p ke parallel |
| Dipole field — equatorial (broadside-on) | Eeq = kp/r³ | r >> a, direction p ke antiparallel; axial ka half |
| Torque on dipole | τ = pE sinθ (= p × E) | Max at θ=90°, zero at 0°/180° |
| Work to rotate dipole | W = pE(cosθ1 - cosθ2) | Potential energy: U = -pEcosθ |
| Electric flux | Δφ = E·ΔA = EA cosθ | θ = angle between E aur area normal |
| Gauss's law | ∮ E·dA = Qenclosed/ε0 | Shape/size independent, sirf enclosed charge matter karta hai |
| Infinite line charge | E = λ/(2πε0r) | λ = linear charge density; E ∝ 1/r |
| Infinite plane sheet | E = σ/(2ε0) | σ = surface charge density; distance-independent |
| Charged conductor surface | E = σ/ε0 | Sheet ke case se double (ek hi taraf charge available) |
| Charged spherical shell — andar | E = 0 | r < R |
| Charged spherical shell — surface | E = kQ/R² | r = R |
| Charged spherical shell — bahar | E = kQ/r² | r > R; point charge jaisa behavior |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Spherical shell ke andar field ko zero na maanna. Students galti se kQ/r² formula har jagah laga dete hain — lekin uniformly charged shell ke andar (r < R) electric field hamesha exactly zero hota hai, kyunki us radius ki Gaussian sphere ke andar koi charge enclosed nahi hota.
- Axial aur equatorial dipole field formula ko mix karna. Axial field (2kp/r³) equatorial field (kp/r³) se exactly double hoti hai — exam me galat formula lagane se answer factor-of-2 galat aa jaata hai. Direction bhi yaad rakho: axial field p ke parallel, equatorial field p ke antiparallel hoti hai.
- Vector addition ke bajaye scalar addition karna. Superposition principle me multiple charges ka net force/field nikalte waqt magnitudes ko seedha add nahi karte — pehle har contribution ka direction dekho, phir vector (component-wise ya law of cosines se) addition karo.
- k aur ε0 ke beech relation bhool jaana. k = 1/4πε0 hota hai, na ki k = 1/ε0. Gauss's law applications (line charge, sheet) me is confusion se factor of 4π ka error aa jaata hai.
- Charged sheet vs charged conductor surface ki field confuse karna. Isolated thin sheet ke paas E = σ/2ε0 hoti hai (dono taraf charge available), lekin conductor ki surface ke paas E = σ/ε0 hoti hai (sirf ek taraf) — dono alag situations hain, formula swap mat karo.
- Gauss's law ko sirf symmetric cases me hi apply kar paana samajhna galat scenario me try karna. Gauss's law hamesha true hoti hai, lekin field easily nikalne ke liye high symmetry (spherical/cylindrical/planar) chahiye hoti hai — asymmetric charge distribution me flux integral seedha solve nahi ho sakta, wahan is method ko force mat karo.
Board-Style Important Questions
- 1 mark: Electric charge ki do basic properties (conservation aur quantization) likho.
- 2 marks: Coulomb's law ka statement do aur uska vector form likho.
- 2 marks: Electric dipole ki equatorial line par field ka formula derive karo (short dipole approximation).
- 3 marks: Gauss's law use karke ek infinite plane sheet of charge ke paas electric field ka expression derive karo.
- 3 marks: Uniformly charged spherical shell ke liye Gauss's law se electric field ke teeno regions (andar, surface, bahar) ka expression derive karo.
- 5 marks: Ek numerical: do point charges diye gaye hain (magnitude aur separation), unke beech force, aur unke beech ek diye gaye point par net electric field dono nikalo.
Aksar Poochhe Jaane Wale Sawaal
Coulomb's law aur Gauss's law me kya farq hai?
Dono ek hi underlying physics describe karte hain, lekin Coulomb's law point charges ke beech direct force deta hai, jabki Gauss's law flux aur enclosed charge ka relation deta hai. High-symmetry charge distributions (line, sheet, sphere) me Gauss's law se field nikalna bahut aasan hota hai, jahan Coulomb's law se direct integration karna mushkil hota.
Dipole field axial line par equatorial line se double kyun hoti hai?
Axial line par dono charges ke fields same direction me add hoti hain (thoda unequal magnitude ke saath, jo net dipole contribution deta hai), jabki equatorial line par dono charges ke fields ka sirf component subtract hota hai jo p ke opposite direction me hota hai — detailed derivation se ye factor-of-2 relation nikalta hai.
Agar Gaussian surface ke andar charge zero hai, to kya E bhi har jagah zero hoga?
Nahi — flux zero hoga, lekin local electric field E kisi bhi point par non-zero ho sakta hai (bahar ke charges ki wajah se). Bas surface se guzarne wala net flux zero hoga, kyunki jitni field lines enter karti hain utni hi exit ho jaati hain.
Electric field lines kabhi ek dusre ko cross kyun nahi karti?
Agar do field lines kisi point par cross karti, to us point par field ki do alag directions ho jaati — jo possible nahi hai kyunki har point par net electric field ek hi unique direction rakhti hai.
Quarks ka charge e/3 ya 2e/3 hota hai — kya ye quantization rule ko violate karta hai?
Nahi, kyunki quarks free state me exist nahi karte — woh hamesha bound combinations (protons, neutrons, etc.) me hote hain jinka net charge hamesha e ka integer multiple hota hai. Isolated, free charge ke liye quantization rule (q=ne) hamesha valid rehta hai.
Dipole ko uniform field me rakhne par net force zero kyun hota hai jabki torque non-zero hota hai?
+q aur -q par lagne wale forces (qE aur -qE) magnitude me equal aur direction me opposite hote hain, isliye vector sum (net force) zero ho jaata hai. Lekin ye do forces different points par act karte hain (charges ke beech separation ki wajah se), isliye ek net torque produce hota hai jab tak dipole field ke exactly parallel ya antiparallel na ho.
Class 12 Physics — Saare Chapters
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