Punjab Board Class 12 Physics — Flashcards for Quick Revision

A: Step 1: When spheres touch, charge distributes equally. Total charge = +10 + 20 = +30 μC Each sphere gets = 30/2 = +15 μC Step 2: After separation, both have +15 μC r = 5 cm = 0.05 m Step 3: Apply C

A: When you pull off a woolen sweater, friction occurs between the wool and synthetic cloth. This friction causes electrons to transfer from one material to another, creating a charge separation. The ch

A: Coulomb's Law (Vector Form): 𝐅₂₁ = (1/4πε₀) × (q₁q₂/r₂₁²) × 𝐫̂₂₁ Where: - 𝐅₂₁ = Force on charge q₂ due to charge q₁ (in Newtons) - q₁, q₂ = Point charges (in Coulombs) - r₂₁ = Distance between charg

A: Electrostatic potential (V) at a point is the work done per unit positive charge by an external force in bringing a test charge from infinity to that point. Only potential difference is physically sig

A: Given: Q = 5 × 10⁻⁶ C, r = 20 cm = 0.20 m, k = 9 × 10⁹ N·m²/C² Formula: V = kQ/r Step 1: Substitute values V = (9 × 10⁹ N·m²/C²) × (5 × 10⁻⁶ C) / (0.20 m) Step 2: Calculate numerator Numerator = 9

A: Inside a conductor, free charge carriers (electrons in metals) continuously move until the internal electric field becomes zero. If any field existed, charges would experience a force and continue mov

A: Electric current (I) is defined as the amount of charge flowing through a cross-section of a conductor per unit time. Mathematically: I = Q/t, where Q is charge in coulombs and t is time in seconds. S

A: Given: n = 8.5 × 10²⁸ m⁻³, I = 2.0 A, A = 2.0 × 10⁻⁷ m², e = 1.6 × 10⁻¹⁹ C Formula: I = neAv_d, so v_d = I/(neA) Step 1: Calculate denominator = 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁷ = 8.5 × 1.6 × 2.0

A: Ohm's Law: V = IR, where V is potential difference in volts, I is current in amperes, and R is resistance in ohms. This states that current is directly proportional to voltage if resistance remains co