Class 12 Physics · Chapter 2
Short answer:
Is chapter me total 18 numericals/conceptual questions cover kiye gaye hain — point charge/dipole/system of charges ka potential, equipotential surfaces, potential energy calculations, conductors ka electrostatic behaviour, parallel plate capacitor (with/without dielectric), series-parallel combination, aur capacitor me stored energy. Chapter numerical-heavy hai — formula yaad hone ke saath unit aur sign ka dhyan rakhna sabse zaroori hai.
Electrostatic potential ek scalar quantity hai jo batati hai ki unit positive charge ko infinity se kisi point tak laane me kitna kaam karna padega. Pichle chapter me humne electric field (vector) padha tha — is chapter me hum usi charge distribution ka scalar version, potential (V), padhte hain, jisse calculation aasan ho jaati hai. Phir hum capacitors padhte hain — do conductors jo charge store karte hain — jinka use mobile charger se lekar defibrillator tak har jagah hota hai.
Chapter 2 Summary — 5 Minute Revision
1. Electrostatic Potential
Kisi point par potential V, unit positive test charge ko infinity se us point tak laane me kiya gaya kaam hai (bina kinetic energy diye, quasi-statically):
V = W∞→P / q0
SI unit: volt (V) = joule/coulomb. Potential scalar hai — direction nahi hoti, sirf sign (+/-) hota hai.
2. Point Charge ka Potential
V(r) = kq/r, k = 1/(4πε₀) = 9×10⁹ Nm²C⁻²
Positive charge ka potential har jagah positive, negative charge ka har jagah negative — field ke ulat, sign directly formula me aata hai (magnitude nahi lete).
3. Electric Dipole ka Potential
Dipole moment p = q×2a wale dipole ke liye, r>>a (far point) par:
V = kp cosθ / r²
Axial line (θ=0°): V = kp/r² (maximum). Equatorial line (θ=90°): V = 0 — is baat ko exam me bahut poocha jaata hai. Point charge ka V ∝ 1/r hota hai, dipole ka V ∝ 1/r² — dipole ka potential jaldi girta hai.
4. System of Charges ka Potential — Superposition
Multiple charges ke case me total potential un sab ke potentials ka algebraic sum hota hai (vector addition nahi, kyunki scalar hai):
V = k(q₁/r₁ + q₂/r₂ + q₃/r₃ + ...)
5. Equipotential Surfaces
- Wo surface jispar har point ka potential same ho.
- Point charge ke liye equipotential surfaces concentric spheres hoti hain.
- Uniform field me equipotential surfaces flat parallel planes hoti hain.
- Electric field hamesha equipotential surface ke perpendicular hota hai — agar field surface ke along ho to surface ke saath potential badalna chahiye, jo definition se galat hai.
- Do equipotential surfaces kabhi ek dusre ko intersect nahi karti — agar karein to us point ke do potential ho jaayenge, jo impossible hai.
- Closely spaced equipotential surfaces = strong field; widely spaced = weak field.
6. Work Done aur Potential Energy
Charge q ko point A se B tak le jaane me kaam:
W = q(VB − VA) = qΔV
Electrostatic force conservative hai, isliye W path-independent hai — sirf initial aur final position matter karti hai. Do point charges ki system ki potential energy:
U = kq₁q₂/r₁₂
Teen ya zyada charges ke system me sab pairs ki U add hoti hai:
U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃)
7. Conductors aur Electrostatics
- Electrostatic equilibrium me conductor ke andar electric field zero hota hai.
- Conductor ki surface par field surface ke perpendicular hota hai.
- Poore conductor (surface + andar) ka potential same (constant) hota hai — isse hi conductor "equipotential volume" kehlaata hai.
- Charge sirf conductor ki outer surface par rehta hai, andar nahi.
- Cavity ke andar agar koi charge na ho to cavity me field zero hota hai (electrostatic shielding — Faraday cage ka principle).
8. Dielectrics aur Polarisation
Dielectric ek insulator hai jisme free charges nahi hote, lekin external field lagane par molecules polarise ho jaate hain (dipole moment induce hota hai). Isse dielectric ke andar ek induced field ulti direction me banta hai jo net field ko kam karta hai — yahi wajah hai ki dielectric daalne se capacitance badhta hai.
9. Capacitor aur Capacitance
Capacitor do conductors ka system hai jo charge store karta hai. Capacitance:
C = Q/V
SI unit: farad (F) = coulomb/volt. 1F bahut bada unit hai, practical me μF, nF, pF use hote hain.
10. Parallel Plate Capacitor
Do parallel plates, area A, separation d, beech me air/vacuum:
C = ε₀A/d
Dielectric (dielectric constant K) daalne par:
C = Kε₀A/d
Capacitance area ke directly proportional aur separation ke inversely proportional hai — plates paas laane se ya dielectric daalne se capacitance badhta hai.
11. Capacitors ka Combination
| Combination | Formula | Kya same rehta hai |
|---|---|---|
| Series | 1/Cs = 1/C₁ + 1/C₂ + ... | Charge Q same, V baatti hai |
| Parallel | Cp = C₁ + C₂ + ... | Voltage V same, Q baatta hai |
↔ Table ko side me swipe karein
Ye resistors ke bilkul ulta hai — resistors series me badhte hain, capacitors series me ghat jaate hain (net capacitance sabse chhote capacitor se bhi kam ho jaata hai). Capacitors parallel me lagane par badh jaate hain, jaise resistors parallel me ghat jaate hain. Yahi sabse zyada confusion create karta hai exams me.
12. Capacitor me Stored Energy
U = ½CV² = ½QV = Q²/2C
Charging process me battery jitna kaam karti hai (W=QV), usme se sirf aadha hi capacitor me energy ban kar store hota hai — baaki aadha resistance/wire me heat ke roop me loss ho jaata hai (chahe resistance kitni bhi chhoti ho). Ye ek common misconception hai ki poora kaam store ho jaata 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–Q18)
Q1. Do point charges 5 μC aur −3 μC ek dusre se 30 cm door (air me) rakhe hain. Inki electrostatic potential energy nikalo.
U = kq₁q₂/r
U = (9×10⁹ × 5×10⁻⁶ × (−3×10⁻⁶)) / 0.3
U = (9×10⁹ × (−15×10⁻¹²)) / 0.3 = −135×10⁻³ / 0.3
U = −0.45 J (negative — charges opposite sign hain, isliye system stable/attractive hai)
Q2. Ek point charge 2×10⁻⁷ C se 9 cm door potential kitna hoga?
V = kq/r
V = (9×10⁹ × 2×10⁻⁷) / 0.09 = 1800 / 0.09
V = 2×10⁴ V = 20,000 V
Q3. Origin par 8 mC ka charge rakha hai. Ek chhota charge −2×10⁻⁹ C ko point P(0,0,3cm) se point Q(0,4cm,0) tak kisi bhi path se le jaane me kitna kaam karna padega?
Electrostatic force conservative hai, isliye sirf VP aur VQ chahiye — path matter nahi karta.
r_P = 0.03 m, r_Q = 0.04 m
V_P = (9×10⁹ × 8×10⁻³)/0.03 = 2.4×10⁹ V
V_Q = (9×10⁹ × 8×10⁻³)/0.04 = 1.8×10⁹ V
W = q(V_Q − V_P) = (−2×10⁻⁹)(1.8×10⁹ − 2.4×10⁹) = (−2×10⁻⁹)(−0.6×10⁹)
W = 1.2 J
Q4. Do charges 2 μC (at x=+6cm) aur −2 μC (at x=−6cm) x-axis par rakhe hain. (a) Origin par potential kya hoga? (b) Point (20cm, 0, 0) par potential kya hoga?
(a) Origin par: dono charge origin se equal distance (6 cm) par hain lekin opposite sign ke — dono potential cancel ho jaate hain.
V_origin = kq/0.06 + k(−q)/0.06 = 0 V
(b) Point (20cm,0,0) par: distance from +2μC = 20−6 = 14 cm; distance from −2μC = 20−(−6) = 26 cm
V = (9×10⁹×2×10⁻⁶)/0.14 + (9×10⁹×(−2×10⁻⁶))/0.26
V = 128571.4 − 69230.8
V ≈ 5.93×10⁴ V (≈ 59.3 kV)
Q5. Ek electric dipole (p = 4×10⁻⁹ C·m) uniform field E = 5×10⁴ N/C me is tarah rakha hai ki field ke saath 30° ka angle banta hai. Dipole par torque nikalo.
τ = pE sinθ
τ = 4×10⁻⁹ × 5×10⁴ × sin30° = 2×10⁻⁴ × 0.5
τ = 1×10⁻⁴ N·m
Q6. Ek parallel plate capacitor ki plate area 6×10⁻³ m² hai aur plates ke beech separation 3 mm hai (air medium). Capacitance nikalo. Agar 100 V lagaya jaaye to plates par kitna charge aayega?
C = ε₀A/d
C = (8.854×10⁻¹² × 6×10⁻³) / (3×10⁻³) = 8.854×10⁻¹² × 2
C ≈ 17.7 pF
Q = CV = 17.7×10⁻¹² × 100
Q ≈ 1.77 nC
Q7. Ek parallel plate capacitor ki plate separation aadhi kar di jaaye aur dielectric constant teen guna kar diya jaaye, to capacitance kitna badhega?
C = Kε₀A/d, jismein d → d/2 aur K → 3K
C_new = 3K·ε₀A / (d/2) = 6 × (Kε₀A/d) = 6 × C_original
Capacitance 6 guna ho jaayega.
Q8. Teen capacitors 2 pF, 3 pF, aur 4 pF ko parallel me jodkar 100 V ki supply di gayi hai. Equivalent capacitance aur har capacitor par charge nikalo.
Parallel: Cp = C₁+C₂+C₃
C_p = 2+3+4 = 9 pF
Parallel me voltage sab par same (100 V) rehta hai:
Q₁ = 2pF×100V = 200 pC; Q₂ = 3pF×100V = 300 pC; Q₃ = 4pF×100V = 400 pC
C_p = 9 pF, total Q = 900 pC
Q9. Wahi teen capacitors 2 pF, 3 pF, 4 pF ab series me jodkar 100 V diya gaya hai. Equivalent capacitance aur har capacitor ke aar-paar voltage nikalo.
Series: 1/Cs = 1/C₁+1/C₂+1/C₃
1/C_s = 1/2 + 1/3 + 1/4 = (6+4+3)/12 = 13/12
C_s = 12/13 ≈ 0.923 pF
Series me charge sab par same rehta hai:
Q = C_s×V = 0.923×100 ≈ 92.3 pC
V₁ = 92.3/2 = 46.15 V; V₂ = 92.3/3 = 30.77 V; V₃ = 92.3/4 = 23.08 V
Check: 46.15+30.77+23.08 ≈ 100 V ✓
C_s ≈ 0.923 pF
Q10. Ek air-filled parallel plate capacitor ka capacitance 6 pF hai. Plates ke beech ki doori double kar di jaati hai aur beech me wax bhar diya jaata hai, jisse naya capacitance 12 pF ho jaata hai. Wax ka dielectric constant nikalo.
Air wale case me: C₁ = ε₀A/d = 6 pF, isliye ε₀A = 6d
Wax wale case me d'=2d:
C₂ = Kε₀A/(2d) = K(6d)/(2d) = 3K
3K = 12 pF
K = 4
Q11. Ek 8 pF ka capacitor 50 V ki battery se jodkar charge kiya jaata hai. Isme kitni electrostatic energy store hogi?
U = ½CV²
U = 0.5 × 8×10⁻¹² × (50)² = 0.5 × 8×10⁻¹² × 2500
U = 1×10⁻⁸ J = 10 nJ
Q12. Ek 12 pF capacitor ko 50 V battery se charge kiya gaya, phir battery hata di gayi aur ise ek uncharged 6 pF capacitor se jod diya gaya. Electrostatic energy me kitni kami aayi?
Initial energy:
U₁ = ½C₁V² = 0.5 × 12×10⁻¹² × 2500 = 1.5×10⁻⁸ J
Charge conserve rehta hai:
Q = C₁V = 12×10⁻¹² × 50 = 6×10⁻¹⁰ C
Dono jud jaane par common voltage:
V' = Q/(C₁+C₂) = 6×10⁻¹⁰ / 18×10⁻¹² = 33.33 V
Final energy:
U₂ = ½(C₁+C₂)V'² = 0.5 × 18×10⁻¹² × (33.33)² ≈ 1×10⁻⁸ J
Loss = U₁ − U₂ = 1.5×10⁻⁸ − 1×10⁻⁸
Energy loss ≈ 5×10⁻⁹ J (5 nJ) — ye energy sharing ke waqt heat/radiation ke roop me nikal jaati hai.
Q13. Ek dipole ka moment 2×10⁻⁸ C·m hai. Iski axial line par 1 m door potential nikalo.
Axial line par θ=0°, cosθ=1, r>>a maankar:
V = kp cosθ / r² = (9×10⁹ × 2×10⁻⁸ × 1) / 1²
V = 180 V
Q14. Kya do equipotential surfaces ek dusre ko kabhi intersect kar sakti hain? Reason do.
Nahi. Agar do equipotential surfaces kisi point par intersect karein, to us intersection point par ek saath do alag potential values honi chahiye (har surface apna potential leke), jo ek point ke liye impossible hai kyunki potential single-valued quantity hai. Isliye equipotential surfaces kabhi cross nahi karti.
Q15. Ek 4 μF capacitor ko 200 V tak charge karke battery se disconnect kar diya jaata hai, phir ek uncharged 2 μF capacitor se jod diya jaata hai. Final common voltage, dono capacitors par final charge, aur energy loss nikalo.
Total charge conserve:
Q = C₁V = 4×10⁻⁶ × 200 = 8×10⁻⁴ C
Common voltage:
V' = Q/(C₁+C₂) = 8×10⁻⁴ / 6×10⁻⁶ = 133.33 V
Q₁' = 4×10⁻⁶ × 133.33 = 5.33×10⁻⁴ C; Q₂' = 2×10⁻⁶ × 133.33 = 2.67×10⁻⁴ C
Energy:
U_i = ½×4×10⁻⁶×200² = 0.08 J
U_f = ½×6×10⁻⁶×133.33² ≈ 0.0533 J
Energy loss ≈ 0.08 − 0.0533 = 0.0267 J (≈2.67×10⁻² J)
Q16. Teen charges q₁=1 μC (origin par), q₂=−2 μC (3cm, 0 par), q₃=3 μC (0, 4cm par) rakhe hain. System ki total electrostatic potential energy nikalo.
r₁₂ = 3 cm = 0.03 m, r₁₃ = 4 cm = 0.04 m, r₂₃ = √(3²+4²) = 5 cm = 0.05 m
U₁₂ = k q₁q₂/r₁₂ = 9×10⁹×(1×10⁻⁶)(−2×10⁻⁶)/0.03 = −0.6 J
U₁₃ = k q₁q₃/r₁₃ = 9×10⁹×(1×10⁻⁶)(3×10⁻⁶)/0.04 = 0.675 J
U₂₃ = k q₂q₃/r₂₃ = 9×10⁹×(−2×10⁻⁶)(3×10⁻⁶)/0.05 = −1.08 J
U_total = −0.6 + 0.675 − 1.08
U_total = −1.005 J
Q17. Ek spherical conductor ki radius 12 cm hai aur usme 1.6×10⁻⁷ C charge hai. (a) Surface par potential, (b) 18 cm door (centre se, bahar) potential, (c) centre par potential nikalo.
(a) Surface par (r=0.12m):
V = kq/r = (9×10⁹×1.6×10⁻⁷)/0.12 = 1440/0.12
V = 1.2×10⁴ V
(b) 18 cm par (r=0.18m):
V = 1440/0.18 = 8000 V = 8×10³ V
(c) Centre par: conductor ke andar field zero hota hai, isliye poora conductor (andar bhi) surface jitna hi potential rakhta hai.
V_centre = 1.2×10⁴ V (surface ke barabar)
Q18. Ek 2 pF ka parallel plate capacitor 12 V se charge karke battery hata di jaati hai. Ab beech me K=6 wala dielectric slab pura bhar diya jaata hai. Naya capacitance, naya voltage, aur energy me farak nikalo.
Battery disconnected hai isliye charge Q constant rahega:
Q = C₀V = 2×10⁻¹² × 12 = 24×10⁻¹² C
Dielectric daalne ke baad:
C' = KC₀ = 6×2 = 12 pF
V' = Q/C' = 24×10⁻¹²/12×10⁻¹² = 2 V
Energy:
U_i = ½QV = 0.5×24×10⁻¹²×12 = 144×10⁻¹² J
U_f = ½QV' = 0.5×24×10⁻¹²×2 = 24×10⁻¹² J
Energy 144pJ se ghatkar 24pJ ho gayi — kami is wajah se hai ki dielectric slab ko capacitor andar khinchta hai aur uspar kaam hota hai (battery disconnected hone se energy conserve nahi, balance dielectric ko andar kheenchne me lagta hai).
Important Equations — Ek Nazar Me
| Quantity | Formula |
|---|---|
| Potential (definition) | V = W∞→P/q₀ |
| Point charge ka V | V = kq/r, k = 9×10⁹ Nm²C⁻² |
| Dipole ka V (r>>a) | V = kp cosθ / r² |
| Kaam done (work-energy) | W = q(VB − VA) = qΔV |
| Do charges ki PE | U = kq₁q₂/r |
| Capacitance (definition) | C = Q/V |
| Parallel plate (air) | C = ε₀A/d |
| Parallel plate (dielectric) | C = Kε₀A/d |
| Series combination | 1/Cs = 1/C₁ + 1/C₂ + ... |
| Parallel combination | Cp = C₁ + C₂ + ... |
| Energy stored | U = ½CV² = ½QV = Q²/2C |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Series aur parallel capacitor formula ulti kar dena. Resistors ke habit se students series capacitors me C₁+C₂ likh dete hain aur parallel me 1/C formula — capacitors resistors ke bilkul ulte behave karte hain, ye baar-baar revise karo.
- Energy formula me ½ bhool jaana. U=CV² likh dena galat hai, sahi hai U=½CV² — battery jitna kaam karti hai (QV) usme se sirf aadha capacitor me store hota hai, baaki heat me chala jaata hai.
- Point charge ke potential me magnitude le lena, sign nahi. V=kq/r me charge ka sign (+/−) directly daalna hota hai — potential ke case me sign bahut matter karta hai, field ki tarah magnitude nahi le sakte.
- System of charges ki potential energy me sirf ek pair ka U likh dena. Teen ya zyada charges ho to har unique pair (q₁q₂, q₂q₃, q₁q₃...) ki U alag-alag nikaal kar add karni hoti hai, koi pair chhoot na jaaye.
- Equipotential surface field lines ke parallel maan lena. Electric field hamesha equipotential surface ke perpendicular hota hai — parallel maanna seedha galat hai aur conceptual MCQs me trap hota hai.
- Dielectric daalne par har case me energy badhna maan lena. Agar battery connected rahe to charge badhta hai aur energy badhti hai, lekin agar battery disconnect karke dielectric daala jaaye (charge constant) to energy ghatti hai — dono cases alag result dete hain, question dhyan se padho ki battery connected hai ya nahi.
Board-Style Important Questions
- 1 mark: Do equipotential surfaces kabhi intersect kyun nahi karti?
- 1 mark: Conductor ke andar electrostatic field ka value kya hota hai?
- 2 marks: Electric dipole ke equatorial point par potential zero kyun hota hai, dikhaओ.
- 3 marks: Do charges q₁ aur q₂ diye gaye hain unke beech ki doori r hai — inki electrostatic potential energy ka formula derive karo.
- 3 marks: Series aur parallel me capacitors jodne par equivalent capacitance kaise nikalte hain — dono formulas derive karo.
- 5 marks: Ek parallel plate capacitor me plates ke beech dielectric daalne se capacitance kaise badhta hai, samjhao aur is process me energy conservation discuss karo.
Aksar Poochhe Jaane Wale Sawaal
Capacitance ka unit farad kitna bada hota hai?
1 farad bahut bada unit hai — practical capacitors usually pF (10⁻¹² F) ya μF (10⁻⁶ F) range me hote hain. 1F ka capacitor banane ke liye bahut badi plate area chahiye hogi.
Series me capacitors lagane se capacitance kyun ghat jaata hai?
Series me effective plate separation badh jaata hai (jaise do capacitors ka d add ho raha ho), aur capacitance separation ke inversely proportional hota hai, isliye net capacitance ghat jaata hai — ye resistors ke bilkul ulta hai.
Dielectric daalne se capacitance kyun badhta hai?
External field lagne par dielectric polarise ho jaata hai aur ek induced field ulti direction me banata hai jo net field ko kam karta hai. Kam field ka matlab hai same charge ke liye kam voltage chahiye, aur C=Q/V hone se voltage kam hone par C badh jaata hai.
Conductor ke andar potential surface jitna hi kyun hota hai?
Electrostatic equilibrium me conductor ke andar field zero hota hai. Field zero ka matlab hai potential me koi change nahi ho raha, isliye poora conductor (surface se lekar centre tak) same potential par hota hai.
Capacitor me stored energy poori battery se aayi energy ke barabar kyun nahi hoti?
Battery total kaam W=QV karti hai charging ke dauraan, lekin capacitor me sirf ½QV hi energy ke roop me store hoti hai — baaki aadha wire/resistance me heat ke roop me nikal jaata hai, chahe resistance kitni bhi chhoti kyun na ho.
Dipole ka potential point charge se jaldi kyun kam hota hai?
Point charge ka V ∝ 1/r hota hai jabki dipole ka V ∝ 1/r² hota hai (kyunki dipole ke do opposite charges door se ek dusre ko partially cancel karte hain), isliye dipole ka potential distance ke saath jaldi girta hai.
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