Question

two points charges q_1 and q_2 of 10^-8 and -10^-8 respectively are placed 0.1m apart . Calculate electric field at points a,b,c as shown in the fig​

Answer 1

GiVEN

• q₁ = 10⁻⁸ C
• q₂ = -10⁻⁸

• ∵ 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

Electric field intensity at C = E1² +E2² +2E1 . E2 cosθ

Here ,

• Θ = 120°
• E₁ = kq₁/(0.1)² = 9 × 10⁹ × 10⁻⁸/(0.1)² = 9 × 10³ = 9000 N/C
• E₂ = 9000 N/C

Now, Electric field intensity at C

= 9000² +9000² +2.9000.9000.cos120

= 9000 N/C