Two point charges q1 and q2, of magnitude +10

1.	Two point charges q1 and q2, of magnitude +10
Answer» 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....

Answer»

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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