Question

Two point charges q1 and q2 of magnitude +10-8 C and -10-8 C respectively are placed 0.1 m apart. Calculate the electric field at point A.B & C.
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Answer 1

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 = $\bold{\sqrt{E_1^2+E_2^2+2E_1.E_2cos\theta}}$
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 = $\bold{\sqrt{9000^2+9000^2+2.9000.9000.cos120}}$
= 9000 N/C

Answer 2

Answer:

Electricity field intensity at A is E = 72000 N/C

Electricity field intensity at B is E = 32000 N/C

Electricity field intensity at C is E = 9000 N/C

Explanation:

Given,

Charge q₁ = 10⁻⁸ C

Charge q₂ = -10⁻⁸ C

Since one charge (q₁ ) is positive and other charge (q₂ ) is negative

Therefore electric field due to q1 and electric field due to q2 are in same direction.

Now, Electric field intensity at A:

Electric field at A = electric field due to q₁ (at A) + electric field due to q₂ (at A)

E = Kq₁/(0.05)² + kq₂/(0.05)²

E = 9 × 10⁹ × 10⁻⁸/(0.05)² + 9 × 10⁹× 10⁻⁸/(0.05)²

E = 2 × 9 × 10/(1/20)²

E = 180 × 400 N/C

E = 72000 N/C

Now, Electric field intensity at B:

Electric field at B = electric field due to q₁(at B) - electric field due to q₂(at B)

E = kq₁/(0.05)² - kq₂/(3 × 0.05)²

E = 9 × 10⁹ × 10⁻⁸/(1/20)² - 9 × 10⁹ × 10⁻⁸/9 × (1/20)²

E = 36000 - 4000

E = 32000 N/C

Now electric field intensity at C:

Electric field intensity at C =  

Here, Angle = 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