Answer:
E = 8.85 N/C
Explanation:
Look at the attached graphic:
Because all the charges are positive the electric fields leave the charge towards the point P:
The electric field due to each of the charges is calculated by Coulomb's law:
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)
Problem development
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
E₁, E₂, E₃: Electric field at point P due to charge q₁, q₂ and q₃ respectively
q = q₁ = q₂ = q₃ = 2.8 µC = 2.8*10⁻⁶ C
K*q = 8.98755 × 10⁹ N*m²/C² * 2.8*10⁻⁶ C = 25.16*10³ N*m²/ C
E₁= k*q/d₁² = (25.16*10³)/(68)² = 5.44 N/C
E₂ = k*q/d₂² = (25.16*10³)/(71)² = 4.99 N/C
[tex]d_3 = \sqrt{51^2+49^2} = 70.72 m[/tex]
E₃ = k*q/d₃² = (25.16*10³)/(70.72)² = 5.03 N/C
Calculation of the electric field components at the origin of x-y coordinates
Eₓ = E₁ₓ+E₂ₓ+ E₃ₓ
E₁ₓ = 0
E₂x = -5 N/C
β=tan⁻¹(49/51) = 43.85°
E₃ₓ = -5.03*cos(43.85°) = -3.63 N/C
Eₓ= 0 - 5 N/C - 3.63 N/C = -8.63 N/C
Ey = E₁y + E₂y + E₃y
E₁y= -5.44 N/C
E₂y = 0
E₃y = 5.03 sin(43.85°)= 3.48 N/C
Ey = -5.44 N/C + 0 + 3.48 N/C = -1.96 N/C
Magnitude of the net electric field at the origin due q₁ ,q₂ and q₃
[tex]E = \sqrt{E_x^2+E_y^2}[/tex]
[tex]E = \sqrt{(-8.63)^2 +(-1.96)^2}[/tex]
E = 8.85 N/C
