Calculate the magnitude and direction of the resultant electric field at the origin due to these charges

Question

A point charge 𝑞1 = +6.00 𝑥 10−⁹𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0.600 𝑚 and a second point charge 𝑞2 = −2.00 𝑥 10−⁹𝐶 is at the point 𝑥 = 0.800 𝑚, 𝑦 = 0 𝑚. Calculate the magnitude and direction of the resultant electric field at the origin due to these charges.

Answer 1

The electric field from a point charge will have a magnitude |k_eQ/d²| in a direction towards the charge if Q<0 and away from the charge if Q>0

The directions are facilitated as we are looking at vectors that aim towards and away from the origin (coordinates give components)

Easiest is the charge at .8,0 which creates a field E₁ = k_e(2 x 10^-9 C)/(.8 m)² in the + x direction

The other charge at (.8,.6) creates a field E₂ = k_e(6 x 10^-9 C)/(1m)² in the (-.8,-.6) direction (the vector is already a unit vector) E_2x = -.8E₂ and E_2y = -.6E₂

Now just add the x components and y components separately (To find E_Total,x and E_Total,y and combine them using the usual transform:

|E_T| = sqrt(E_T,x² + E_Ty²) and θ = arctan(E_Ty/E_Tx) (+/- 180 if E_Tx <0)