Electric Charges and Fields

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Given: q₁ = +3 μC = 3×10⁻⁶ C, q₂ = -5 μC = -5×10⁻⁶ C, r = 30 cm = 0.3 m Formula: F = k|q₁q₂|/r² Step 1: F = (9×10⁹ × 3×10⁻⁶ × 5×10⁻⁶)/(0.3)² Step 2: F = (135×10⁻³)/(0.09) = 1.5 N Since charges are opp

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This happens due to electrostatic charging by friction. When synthetic clothes rub against skin, electrons transfer between surfaces, creating opposite charges. The attractive force between these oppo

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Coulomb's Law: The electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them. Mathematical form

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Given: q = 8 μC = 8×10⁻⁶ C, P(3,4) m Step 1: Distance r = √(3² + 4²) = 5 m Step 2: Electric field magnitude E = kq/r² = (9×10⁹ × 8×10⁻⁶)/25 = 2880 N/C Step 3: Unit vector r̂ = (3î + 4ĵ)/5 = 0.6î + 0.8

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Electric field (E) is the region around a charge where another charge experiences force. It's defined as force per unit positive test charge: E = F/q₀. Differences: • Force depends on both source and

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Given: q₁ = +2μC, q₂ = -3μC, q₃ = +4μC, side = 10cm = 0.1m Step 1: F₁₃ = kq₁q₃/r² = (9×10⁹×2×4×10⁻¹²)/(0.1)² = 7.2 N (repulsive) Step 2: F₂₃ = kq₂q₃/r² = (9×10⁹×3×4×10⁻¹²)/(0.1)² = 10.8 N (attractive)

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Electric field lines never intersect because if they did, it would mean there are two different directions of electric field at the intersection point. This is impossible since electric field has a un

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Given: E = 5×10³ N/C, q = 2μC = 2×10⁻⁶ C, d = 3cm = 0.03m, motion opposite to field Formula: W = qEd cosθ Step 1: Since motion is opposite to field, θ = 180°, cos180° = -1 Step 2: W = qEd cos180° = 2×