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CBSE Class 12 Physics 2026: Electrostatics + Current Electricity (Top 20 Most Expected Q &A with Solutions)

 CBSE Class 12 Physics के Electrostatics और Current Electricity के सबसे महत्वपूर्ण 20 सवाल–जवाब, formulas, derivations और numericals के साथ। Q1) Coulomb’s Law लिखिए और SI unit बताइए। Solution: F = 1 4 π ε 0 q 1 q 2 r 2 F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} F = 4 π ε 0 ​ 1 ​ r 2 q 1 ​ q 2 ​ ​ F F F की unit: Newton (N) ε 0 \varepsilon_0 ε 0 ​ : permittivity of free space Force line joining charges के along होता है। Q2) Point charge के कारण Electric Field का सूत्र (vector form)। Solution: E ⃗ = 1 4 π ε 0 q r 2 r ^ \vec E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat r E = 4 π ε 0 ​ 1 ​ r 2 q ​ r ^ दिशा: + q +q + q से बाहर, − q -q − q की तरफ। Q3) Electric Flux की परिभाषा और unit। Solution: Φ E = E ⃗ ⋅ A ⃗ = E A cos ⁡ θ \Phi_E=\vec E\cdot \vec A = EA\cos\theta Φ E ​ = E ⋅ A = E A cos θ Unit: N   m 2 / C N\,m^2/C N m 2 / C Q4) Gauss Law लिखिए। Solution: ∮ E ⃗ ⋅ d A ⃗ = Q e n c ε 0 \oint \vec E\cdot d\vec A=\frac{Q_{enc}}{\varepsilon_0} ∮ E ⋅ d A...

CBSE Class 12 Physics 2026: Top 100 High-Yield Questions (Full Solutions/Steps) – Board Score Booster Set

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 UNIT-1: Electrostatics (Q1–Q10) 1) Coulomb’s Law likho aur vector form बताओ। Force line joining charges ke along hota hai; like charges repel, unlike attract. 2) Electric field due to point charge निकालो। 3) Dipole ka electric field on axial line? Solution (main steps): Dipole moment p = q ( 2 a ) p=q(2a) p = q ( 2 a ) . Axial point distance r ≫ a r\gg a r ≫ a : Direction: dipole moment ke along (from − to +). 4) Dipole ka field on equatorial line? Direction: p ⃗ \vec p p ​ ke opposite. 5) Gauss law statement + formula. Solution: Closed surface se total flux: 6) Infinite plane sheet (surface charge density σ \sigma σ ) ka E-field? Solution: Using Gauss (pillbox): Two sides par equal magnitude. 7) Infinite line charge (linear density λ \lambda λ ) ka E-field at distance r? Solution: Cylindrical Gaussian surface: E ( 2 π r L ) = λ L ε 0 ⇒ E = λ 2 π ε 0 r E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}\Rightarrow E=\frac{\lambda}{2\pi\varepsilon_0 r} E ( 2 π r L ) = ε 0 ​ λ L ​ ...

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