Two charges of magnitude + - 2 into 10 raise to power minus 8 column respectively are placed at 0 10 cm and 0 - 10 cm find electric field strength at 200
Answers
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
Electric field intensity at C =
Here ,Θ = 120° [ see attachment for understanding ]
E₁ = kq₁/(0.1)² = 9 × 10⁹ × 10⁻⁸/(0.1)² = 9 × 10³ = 9000 N/C
E₂ = 9000 N/C [ because magnitude of charges and separation are same ]
Now, Electric field intensity at C =
= 9000 N/C
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