Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

Answer 1

Here are the individual electric fields in vector notation:

||E(1)|| = kq/r^2 = (8.99e9 N*m^2/C^2)(10e-9 C)/(3.0e-2 m)^2 = 1.0e5 N/C
E(1) = <0, 1.0e5>

||E(2)|| = (8.99e9 N*m^2/C^2)(5.0e-9 C)/(5.0e-2 m)^2 = 1.8e4 N/C
E(2) = <-1.8e4, 0>

r = sqr(3^2 + 5^2) = 5.83 cm
||E(3)|| = (8.99e9 N*m^2/C^2)(10e-9 C)/(5.83e-2 m)^2 = 2.6e4 N/C
θ = arctan(3/5) = 30.96° (measured counterclockwise from +x axis)
E(3) = <-2.6e4cos 30.96°, -2.6e4sin 30.96°> = <-2.3e4, -1.4e4>

Answer 2

In the triangle use Pythagoras theorem to find the distance of charge q2 from the point. Then use the formula for Electric field to find E1 and E2 (their respective directions are marked by the arrow heads), then add the fields using parallelogram law of vector addition for the resultant field.
Angle between the vector fields has been solved for,\[\theta=tan^{-1}2\]
PS: Remember to convert the given distances into meters before using them for the field calculation.