10.Iuutun4. Two point charges q, and q, of 2 x 10-C and- 2 x 10 °C re

1.	10.Iuutun4. Two point charges q, and q, of 2 x 10-C and- 2 x 10 °C respectively are placed 0.4 m apart.Calculate the electric field at the centre of the line[CBSE F 94C]joining the two charges.(Ans. 900 NC-1, towards the -ve charge)
Answer» Given, q₁ = 10⁻⁸ C q₂ = -10⁻⁸ C ∵ one charge is positive and other charge is negative ∴ electric field due to q1 and electric field due to q2 are in same direction. ∴ Electric field at A = electric field due to q₁ + electric field due to q₂ = Kq₁/(0.05)² + kq₂/(0.05)² = 9 × 10⁹ × 10⁻⁸/(0.05)² + 9 × 10⁹× 10⁻⁸/(0.05)² = 2 × 9 × 10/(1/20)²= 180 × 400 N/C = 72000 N/C Electric field at B = electric field due to q₁ - electric field due to q₂ = kq₁/(0.05)² - kq₂/(3 × 0.05)² = 9 × 10⁹ × 10⁻⁸/(1/20)² - 9 × 10⁹ × 10⁻⁸/9 × (1/20)² = 36000 - 4000 = 32000 N/C

10.Iuutun4. Two point charges q, and q, of 2 x 10-C and- 2 x 10 °C respectively are placed 0.4 m apart.Calculate the electric field at the centre of the line[CBSE F 94C]joining the two charges.(Ans. 900 NC-1, towards the -ve charge)

Answer»

Given, q₁ = 10⁻⁸ C q₂ = -10⁻⁸ C ∵ one charge is positive and other charge is negative ∴ electric field due to q1 and electric field due to q2 are in same direction. ∴ Electric field at A = electric field due to q₁ + electric field due to q₂ = Kq₁/(0.05)² + kq₂/(0.05)² = 9 × 10⁹ × 10⁻⁸/(0.05)² + 9 × 10⁹× 10⁻⁸/(0.05)² = 2 × 9 × 10/(1/20)²= 180 × 400 N/C = 72000 N/C

Electric field at B = electric field due to q₁ - electric field due to q₂ = kq₁/(0.05)² - kq₂/(3 × 0.05)² = 9 × 10⁹ × 10⁻⁸/(1/20)² - 9 × 10⁹ × 10⁻⁸/9 × (1/20)² = 36000 - 4000 = 32000 N/C

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