Point charges acting on a point

[fal01]

Homework Statement

Three point charges are located at the vertices of an equilateral triangle. The charge at the top vertex of the triangle is -4.4 µC. The two charges q at the bottom vertices of the triangle are equal. A fourth charge 2 µC is placed below the triangle on its symmetryaxis, and experiences a zero net force from the other three charges, as shown in the figure below. Find q.
One side of the triangle is 7.3m. The distance from 2 µC to the triangle is 4.8m.

-4.4 µC {1}
2 µC {2}

Homework Equations

F=kQ1Q2/r^2
E=kQ/r^2

The Attempt at a Solution

F (1 on q)= k (-4.4*10^-6)q/7.3^2

(7.3/2)^2+(4.8)^2=b^2
=6.03

F(2 on q)=k(2*10^-6)q/6.03^2

 

[vela]

Homework Statement

Three point charges are located at the vertices of an equilateral triangle. The charge at the top vertex of the triangle is -4.4 µC. The two charges q at the bottom vertices of the triangle are equal. A fourth charge 2 µC is placed below the triangle on its symmetryaxis, and experiences a zero net force from the other three charges, as shown in the figure below. Find q.
One side of the triangle is 7.3m. The distance from 2 µC to the triangle is 4.8m.

-4.4 µC {1}
2 µC {2}

Homework Equations

F=kQ1Q2/r^2
E=kQ/r^2

The Attempt at a Solution

F (1 on q)= k (-4.4*10^-6)q/7.3^2

(7.3/2)^2+(4.8)^2=b^2
=6.03

F(2 on q)=k(2*10^-6)q/6.03^2

You actually want to calculate the forces on the 2 uC charge, not on the q's, since the problem tells you the net force on that charge is 0.

 

[fal01]

So something like this?

F (1 on 2)= k (4.4*10^-)(2*10^-6)/(11.12)^2

F(q on 2)= k*Q*(2*10^-6)/(6.03)^2

 

[vela]

Yes, and just remember you need to sum them as vectors.

 

[fal01]

So F(1 on 2)x=0
F(1 on 2) y=6.305*10^-4

but how do I apply this to F(q on 2)

F(q on 2)x= F(q on 2) Cos 60
F(q on 2)y= F(q on 2) Sin 60

but F(q on 2)= k*Q*(2*10^-6)/(6.03)^2

and Q is unknown...

also would 2F(q on 2)+F(1 on 2)=0

 

[vela]

So F(1 on 2)x=0
F(1 on 2) y=6.305*10^-4

but how do I apply this to F(q on 2)

F(q on 2)x= F(q on 2) Cos 60
F(q on 2)y= F(q on 2) Sin 60

but F(q on 2)= k*Q*(2*10^-6)/(6.03)^2

and Q is unknown...

also would 2F(q on 2)+F(1 on 2)=0

Pretty close. I think you have the right idea.

For simplicity, let's assume q is positive. The force of charge 1 on charge 2 will point up. The force of a q-charge on charge 2 will point away from the q. So in component form, you have
\begin{align*}
\sum F_x &= F_{q~\textrm{on}~2} \cos 60 - F_{q~\textrm{on}~2} \cos 60 = 0 \\
\sum F_y &= F_{1~\textrm{on}~2} - 2F_{q~\textrm{on}~2} \sin 60 = 0
\end{align*}
The first equation simply tells you the x-components cancel out, regardless of what q is, which you could have deduced beforehand based on symmetry. The second equation is the one that will let you solve for q.