Question

What are the three fields E⃗ 1E→1, E⃗ 2E→2, and E⃗ 3E→3 created by the three charges? Write your answer for each as a vector in component form. Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answer 1

Answer:

E₁ = 8500 N/c i − 2800 N/C j

E₂ = 10,000 N/c i

E₃ = 8500 N/c i + 2800 N/C j

Explanation:

The magnitude of electric field from a point charge Q at a distance r is:

E = kQ / r²

where k is Coulomb's constant (9×10⁹ Nm²/C²).

The direction of the electric field is along the radial line outwards from the point.

Use Pythagorean theorem to find the distance from q₁.

r² = (-0.01 m)² + (0.03 m)²

r² = 0.001 m²

r = 0.0316 m

Plugging in values:

E₁ = (9×10⁹ Nm²/C²) (1×10⁻⁹ C) / (0.001 m²)

E₁ = 9000 N/C

Using ratios to find the components:

E₁ₓ = (x/r) E₁

E₁ₓ = (0.03 m / 0.0316 m) (9000 N/C)

E₁ₓ = 8500 N/C

E₁ᵧ = (y/r) E₁

E₁ᵧ = (-0.01 m / 0.0316 m) (9000 N/C)

E₁ᵧ = -2800 N/C

Therefore:

E₁ = 8500 N/c i − 2800 N/C j

Repeat for E₂:

r = 0.03 m

E₂ = (9×10⁹ Nm²/C²) (1×10⁻⁹ C) / (0.03 m)²

E₂ = 10,000 N/C

E₂ₓ = (x/r) E₂

E₂ₓ = (0.03 m / 0.03 m) (10,000 N/C)

E₂ₓ = 10,000 N/C

E₂ᵧ = (y/r) E₂

E₂ᵧ = (0 m / 0.0 m) (10,000 N/C)

E₂ᵧ = 0 N/C

Therefore:

E₂ = 10,000 N/c i

And of course, E₃ is the same as E₁, except the y component is positive.

E₃ = 8500 N/c i + 2800 N/C j