A charge q1 = 5 µC, is at the origin, and a second charge, q2 = -3 µC, is on the x-axis 1 m from the origin. F

A charge q1 = 5 µC, is at the origin, and a second charge, q2 = -3 µC, is on the x-axis 1 m from the origin. Find the electric field at a point on the y-axis 0.5 m from the origin.
Magnitude of Field:
Direction of Field [ angle between 0 and 359 °]

draw charge 1 and charge 2 on the x-y axis... from this, you can get get the vector length between charge 2 on the x-axis and the point 0.5 on the y-axis, as well as the angle.

sqrt(0.5²+1²) = 1.12 m ---> vector length from charge 2 to desired point
theta = atan(0.5/1) = 26.6º ---> angle above x axis for that vector

the electric field is a vector quantity, so E1 + E2 = E
E = F / q , where F = k q1 q2 / r²

for E1 (all in the j direction)
E1 = k q1 q2 / (q2 r²) = kq1 / r² j = (8.99E9)(5E-6)/(0.5²) = 179800 N/C j

for E2 (i and j directions)
E2 = k q1 q2 / (q1 r²) = kq2 / r² (cos θ i + sin θ j)
= [(8.99E9)(-3E-6) / (1.12²)] (cos 26.6º i + sin 26.6º j)
= - 19225 N/C i - 9627 N/C j

Now add E1 + E2
E = -19255 i + ( 179800 - 9627) j = -19255 i + 170173 j
E = sqrt(19255² + 170173² )
E = 171259 N/C <----


direction
tan θ = Ey / Ex = 170173 / 19255
θ = 83.5º

charge q1 = 5 µC, at the origin,
charge, q2 = -3 µC, on x-axis 1 m from origin.
electric field on y-axis 0.5 m from origin = ?

E = 5 x 10^-6 / 0.5 - 3 x 10^-6 / sqrt(1+0.5^2)

E = (10^-6)[10 - 2.4) = 7.6 µ units >============< ANSWER

Angle = arctan(-2.4/10) = 166.5 ° >============< ANSWER