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two point charges of 10^-8 and -10^-8C are placed 0.1m apart . calculate electric feild intensity at A,B and C .
two point charges of 10^-8 and -10^-8C are placed 0.1m apart . calculate electric feild intensity at A,B and C .

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one year ago

```							Hiii Electric field intensity at A will be E1+E2and electric field intensity at B will be : E1 – E2Electric field intensity at A+C will be sqrt (E12 + E22 + 2*E1E2 Cos 1200 )Hope this helps.
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one year ago
```							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/CElectric 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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one year ago
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