Question

Two point charges, Q1=-1.0 uC and Q2= +4.0 uC are placed as shown .The y-component of the electric field, at the origin 0 is ?

Answer 1

The point charges, Q1=-1.0 uC and Q2= +4.0 uC are placed asshown in the below figure:
The electric field due to the charge Q₁ at the originisE₁= (1/4πε_o) *(Q₁/r²)Here,(1/4πε_o) = 9 * 10⁹Nm²/C² and r = 2.5 mor E₁= 9 * 10⁹ * (-1.0 *10^-6/(2.5)²)or E₁= -1440 N/CThe electric field due to the charge Q₂ at theorigin isE₂= (1/4πε_o) *(Q₂/r₁²)Here,r₁²= ((2.5)² +(1.4)²) = 8.21 m²Substituting the values in the above equation,we getE₂= 9 * 10⁹ * (4.0 *10^-6/(8.21))or E₂= 4384.9 N/CThe net electric field at the origin isE = (E_x² +E_y²)^1/2Here,E_x= E_1x + E_2x----------------(1)and E_y= E_1y + E_2y-----------------(2)Here,E_1x= E₁* cosθ andE_2x= E₂* cosθSimilarly,E_1y= E₁* sinθ andE_2y= E₂* sinθSubstituting the above values in equations (1) and (2),wegetE_x= E₁* cosθ + E₂*cosθ = (E₁+ E₂) * cosθ----------------(3)and E_y= E₁* sinθ + E₂*sinθ = (E₁+ E₂) * sinθ-----------------(4)Here,θ = 45^oFrom equations (1) and (2),we getE_x= (-1440 + 4384.9) * cos(45^o) =2082.67 N/Cand E_y= (-1440 + 4384.9) * sin(45^o) =2082.67 N/CTherefore,the y-component of the electric field at the originis E_y= 2082.67 N/C.