Answer:
Ex = k q / b2 [- ¾] , Ey = k q / b2 3
Explanation:
For this exercise we calculate the electric field created by each load on a test charge located at the origin
Field created by load q = + q
E₁ = k q / r²
r = b-0
E₁ = k q / b²
in the negative direction of the x axis
Field created by load q = -2q
E₂ = k (2q) / (2b-0)₂
E₂ = K q / 2b₂
The field is in the negative direction of the x-axis
Field created by charge q = + q / 3
this charge creates a field that has components on the x and y axes
X axis
E3x = K (q / 3) / (2b / 3) 2
E3x = K q 3 / 4b²
as the charge is on the negative side of the x axis. The field goes to the bright side
Axis y
E3y = k (q / 3) / (b / 3) 2
E3y = k q / b²
directed up
therefore the electric field is the sum of the field created by each charge
X axis
Ex = -E1 + E2 + E3x
Ex = -k q / b2 - k q / 2b2 + k q 3 / 4b2
Ex = k q / b2 [-1 -1/2 + ¾]
Ex = k q / b2 [- ¾]
Axis y
Ey = k q / b2 3