# Calculating electric field at a point due to 2 charges

1. ### mr_coffee

Hello everyone, I'm just starting this problem and i'm already confused. I'm suppose to calculat the electric feild at the point P(x =3, y = 3) due to two charges, q1 = 4.0x10^-6C at the origin, and q2 = -3.0x10^-6C at x = 3.0m.

I'll try and draw the diagram.

....................................^ E1
.................................../
................................../
................................./
...............................(P)
................................|
................................|
................................v E2

(q1).........................(q2)

How did they know E2 will be point downard and E1 would be diagnoal like that? Thanks

2. ### dextercioby

12,324
Decompose each vector in the two perpendicular components and add those 2 vectors using components.At the end,you'll have to use Pythagora's theorem to find the modulus,knowing the components.

Daniel.

### Staff: Mentor

The charge q2 is negative and directly below the point P; so the field E2 points straight down (towards q2).

Similarly, the field (E1) from q1, a positive charge, points away from q1 towards P, thus is at a diagonal. (What angle does it make with the x-axis?)

4. ### El Hombre Invisible

I don't know, but you can just add the two electric fields together separately. The electric field at distance r is:

E(r) = Q/(4*pi*e0*r^2), where e0 is the permittivity of free space (8.854*10^-12).

At P, r from the charge q1 is 4.24 (root of 2*3^2), while r at P from q2 is 3. Calculate these two fields, add 'em together. Remember that the two fields aren't pointing in the same direction though - the y part of the field due to q1 can be added to the field due to q2, but not the x part.

5. ### mr_coffee

ahh i remember now! Sorry its been a year since i took this course and i forgot literally everything. It will make a 45 degree angle. I ended up getting the answer! thanks!