A charge q1 = 5 ìC is at the origin. A second charge q2 = -3 ìC is on the x-axis, 0.8 m from the origin. The electric field at a point on the y-axis, 0.5 m from the origin is:

Question

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Answer 1

q₁ = 5 μC =5•10⁻⁶ C,
q₂ = - 3 μC = - 3•10⁻⁶ C,
r₁ =0.5 m,
r₂ = 0.8 m.
k =9•10⁹ N•m²/C²
E₁=k•q₁/ r₁², (directed ↑)
E₂ = k•q₂/ r₂², (directed ↑)
vector E= vector E₁+vectorE₂
The magnitude of vector E is directed ↗ and is calculated using the cosine law.
tanα =0.8/0.5=1.6
α =58⁰
E=sqrt{E₁²+E₂²-2E₁E₂cosα} = …