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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 .

Vikas TU
14149 Points
one year ago
Hiii
Electric field intensity at A will be E1+E2
and electric field intensity at B will be : E1 – E2
Electric field intensity at A+C will be
sqrt (E12 + E22 + 2*E1E2 Cos 1200 )
Hope this helps.
Khimraj
3007 Points
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/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