Answer:
|E(t)| = 1258,46 [N/C]
α = 67,5⁰ (angle with respect x-axis)
Explanation:
E(t) Electric Field is a vector, so we need to determine module and direction
E(t) = E(q₁) + E (q₂) Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂ respectively.
E(q₁) = K * q₁/ (d₁)² K = 9 *10⁹ [N*m²/C²] d₁ = 0,350 m
E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225 [N*m²/C²] *C/m²
E(q₁) = 484,9 [N/C]
E(q₂) = 9 *10⁹ * 3,1*10⁻⁹ / 0,024025
E(q₂) = 1161,29
Then
|E(t)| = √ |Eq₁|² + |Eq₂|²
|E(t)| = √ ( 484,9)² +( 1161,29)²
|E(t)| = √ 235128 + 1348594,46
|E(t)| = 1258,46 [N/C]
And tanα = 1161,29/484,9 tanα = 2,395 α = 67,5⁰
The angle of the vector electric field with the x-axis
