How Do Charge Magnitudes and Signs Affect Electric Field Directions?

[azila]

Homework Statement

A negative point charge q1 = -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q2 is on the x-axis at x=-1.20 m.

What must the sign and magnitude of q2 be for the net electric field at the origin to be 50.0 N/C in the +x direction be?

What must the sign and magnitude of q2 be for the net electric field at the origin to be 50.0 N/C in the -x direction be?

Homework Equations

Coulomb's Law
E = q1 + q2

The Attempt at a Solution

Ok, i got the part when it is in the + x direction but how would i do it for the - x direction? I was doing 50 = 100-q2, but i don't think it will work?? Any help would be appreciated. Thanks in advance.

 

[Kurdt]

The electric field is a vector and isn't given by the sum of the charges.

Electric Field: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elefie.html#c1

Electric field of multiple point charges: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/mulpoi.html#c2

 

[azila]

Thanks for the sites, which I understand. But I know how to get the electric force in the +x direction but not in the -x direction. Could I do that to make it go in the -x direction, the charge of q1 could be +4.00 nC since electric field lines go away from positive charges? Any help would be appreciated. Thanks.

 

[HallsofIvy]

Since q1 have negative charge, it will attract a positive charge and repel a positive charge. Since q1 is to the right of q2 ( 0.6> -1.2) an attraction, toward q1, will be "in the positive direction" (from -1.2 toward +0.6) while a repulsion, away from q1, will be "in the negative direction" (from 0.6 toward -1.2).