# Electric Field due to two points

1. Sep 19, 2009

### Digdug12

1. The problem statement, all variables and given/known data
A point charge q_1 = -4.00 {\rm nC} is at the point x = 0.60 {\rm m}, y = 0.80 {\rm m} , and a second point charge q_2 = +6.00 {\rm nC} is at the point x = 0.60 {\rm m} , y = 0
Calculate the magnitude of the net electric field at the origin due to these two point charges.
Calculate the direction of the net electric field at the origin due to these two point charges.

2. Relevant equations
E=F/Q
F=k*q*q/r^2
E=(KQ1/r^2) + (KQ2/r^2)

3. The attempt at a solution
I searched the threads and I came up with one very similar to this problem, however i tried their solution and still comes out wrong. I am summing of the E fields produced by both charges, it looks like this

E=(8.99x10^9)(-4x10^-6)/1 + ((8.99x10^9)(6x10^-6)/.6^2
I get an answer around 113842 N/C, i tried different specific digits of K to get a more specific answer however they were all wrong, so i changed the sign of -4 to 4, just because, and i got 185793 N/C which is STILL wrong. I have no idea what I'm doing wrong?
and to solve for the angle, do i just use their vector components?

2. Sep 19, 2009

### rl.bhat

Electric field is a vector quantity. You cannot add them directly. Find the direction of electric fields. One field due to 6.0 nC is along the -x-axis. Find the direction of the other field. Resolve it into x and y components and then find the net x and y components and find the resultant.