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Answer:
A)Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
Data
q₁ = -4 nC = -4*10⁻⁹ C
q₂ = +6nC =+ 6*10⁻⁹ C
k = 9*10⁹ N*m²/C²
d₂ = 0.6 m
Graphic attached
The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:
E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.
E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.
Calculation of the electric field at the origin of x-y coordinates due to the charge q₁
E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C
Components (x-y) of the field due to q1:
E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C
E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C
Calculation of the electric field at the origin of x-y coordinates due to the charge q₂
E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C
Calculation of the electric field components at the origin of x-y coordinates
E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)
E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C
E₀y= E₁y = 28.8 N/C
A )Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
