Solving Electric Field with 3 Point Charges: Find q1 & q2

[WeiLoong]

Homework Statement

IF the resultant electric field at D due to the three point charges is zero, find the magnitudes of q1 q2 in terms of q
[/B]

https://scontent-sin1-1.xx.fbcdn.net/hphotos-xft1/v/l/t34.0-12/12498556_10205376500763571_410670163_n.jpg?oh=f586685c3d06d630fad4469546c534dc&oe=568DB0E8

Homework Equations

Electric Field

The Attempt at a Solution

Ex = E(q1) + E(q2)x = k q1 / AB^2 + k q2 cosθ / BD^2
Ey = E(q2)y + E(q) = k q2 sinθ / BD^2 + k q / CD^2
must be
k q2 cosθ / BD^2 = - k q1 / AB^2
k q2 sinθ / BD^2 = - k q / CD^2
then
q1 = q AB^2 / (tanθ CD^2)
AB = CD
tanθ = AB / BC
then
q1 = q BC / AB = 2 q
and
q2 = - q1 BD^2 / (cosθ AB^2)
cosθ= BC / BD
then
q2 = - q1 BD^3 / (BC AB^2) = - q (BD / AB)^3 = -11,2 q

Did i make some mistake here?

 

[mfb]

Ex = E(q1) + E(q2)x = k q1 / AB^2 + k q2 cosθ / BD^2

I don't think this should be AB here.

The rest looks fine, but you should define the coordinate axes and the angle in some way.

 

[WeiLoong]

Oopss!

Correction
Ex = E(q1) + E(q2)x = k q1 / AD^2 + k q2 cosθ / BD^2
Ey = E(q2)y + E(q) = k q2 sinθ / BD^2 + k q / CD^2
must be
k q2 cosθ / BD^2 = - k q1 / AD^2 ...(1)
k q2 sinθ / BD^2 = - k q / CD^2 ...(2)
(2)/(1) I got
tanθ = (q/CD^2)(AD^2/q1)
q1 = q. (AD^2/tanθ.CD^2)
q1=8q
Did i make some mistake here?

 

[mfb]

Looks right.

You also need q2.

 

[WeiLoong]

Yea i am stucked here T_T
Can i just sub 8q into eq(1)?

 

[mfb]

Sure.

 

[WeiLoong]

So i got q2 = (-8q/AD^2)(BD^2/cos)
q2=-11.18q am i doing right?

 

[mfb]

I get the same result.

 

[WeiLoong]

thanks!