Finding the point where the net electric field is 0

[012983]

Homework Statement

Three charges of equal magnitude q reside at the corners of an equilateral triangle of side length a. The topmost charge is positive while the charges at the bottom left and bottom right of the triangle are negative. If point P is midway between the negative charges, at what distance above point P along the +y axis must a -8q charge must be placed so that any charge located at point P experiences no net electric force? The distance from point P to the positive charge is 9 meters.

Homework Equations

E = (kq)/r^2

The Attempt at a Solution

No clue. I think you would find the the electrical field for one of the points and then, using that, find the x and y components of the magnitude. Then solve for 0 = sqrt{(Ex)^2 - (Ey)^2}... but the "any charge located at point P" is confusing me. Or else I'm not thinking correctly.

 

[ideasrule]

It just means that the electric field at P has to be 0. So assume that the third charge is a distance "a" from P, calculate the electric field at P, and set the result to 0.

 

[012983]

It just means that the electric field at P has to be 0. So assume that the third charge is a distance "a" from P, calculate the electric field at P, and set the result to 0.

If the electric field at P is equal to (-4*ke*q)/(3a^2), where ke = 8.99 * 10^9, do I proceed by plugging in -8q for q and setting equal to 0?

If so, how would I solve for a? a could be any number so long as the numerator is 0.

 

[ideasrule]

If the electric field at P is equal to (-4*ke*q)/(3a^2), where ke = 8.99 * 10^9, do I proceed by plugging in -8q for q and setting equal to 0?

That's not the total electric field. There are 3 charges, not just one, and the electric field has both an x and a y component.