Electric field due to three point charges

[emmanual]

Homework Statement

Three point charges q1(+3C), q2(+4C), q3(+3C) are located in a square-shaped form with a distance of 2cm with a 90° cone in every edge(picture in the attached file 39png). Calculate the resultant intensity of point P.

Relevant Equations

Coulomb's law :
F=q1q2/4πεr2
Parallelogram law:
R=√(P²+ Q²+ 2PQcosα)

I've calculated the intensity for every point charge which are
EA = 6.741 x 10¹³ NC¯¹
EB = 4.494 x 10¹¹ NC¯¹
EC = 6.741 x 10¹³ NC¯¹
and I am pretty sure about this far but I am struggling to calculate the X-axis intensity and Y-axis intensity to find the entire approximate intensity with the parallelogram law.
PLEASE HELPPP!

 

[berkeman]

Welcome to PF.

It looks like you have a good start, but those electric field values should be vectors so that you can do the vector sum at P. Can you express all 3 of them as vectors in rectangular coordinates at P so that you can add them up component-wise?

 

[emmanual]

Welcome to PF.

It looks like you have a good start, but those electric field values should be vectors so that you can do the vector sum at P. Can you express all 3 of them as vectors in rectangular coordinates at P so that you can add them up component-wise?

I've already expressed all 3 in vectors in the first file w the arrows on p and identified them with the EA, EB and EC and which I used to express the resultant intensity too and Idk how I'm supposed to add it up that's where I'm stuck like I know it's supposed to be 2 values, not 3 and IDK how to reduce it into 2 values which will be valid mathematically

 

[kuruman]

Are you saying YDK how to decompose a vector into its components? The magnitude of the vector is represented as the hypotenuse of a right triangle with the x and y components as the two right sides. Have you not seen this before?

 

[salmanshu322]

to keep things simple, find separate electrics field of all the charges and in the end add them to get a net electric field due to all the charges, if you want to do other way around its up to you.