# Electric Field

1. Jan 5, 2016

### WeiLoong

1. The problem statement, all variables and given/known data

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

2. Relevant equations
Electric Field

3. 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?

2. Jan 5, 2016

### Staff: Mentor

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.

3. Jan 5, 2016

### 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
q1=8q
Did i make some mistake here?

4. Jan 5, 2016

### Staff: Mentor

Looks right.

You also need q2.

5. Jan 5, 2016

### WeiLoong

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

6. Jan 5, 2016

Sure.

7. Jan 5, 2016

### WeiLoong

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

8. Jan 5, 2016

### Staff: Mentor

I get the same result.

9. Jan 5, 2016

thanks!!