Tension on a string and charge -

[thercias]

Homework Statement

Object A has mass 90.0 g and hangs from an insulated thread. When object B, which has a charge of +130 nC, is held nearby, A is attracted to it. In equilibrium, A hangs at an angle θ = 7.20° with respect to the vertical and is 5.00 cm to the left of B.
(a) What is the charge on A?
(b) What is the tension in the thread?

http://i694.photobucket.com/albums/vv301/redraven45/physics1_zps6759b55e.png

Homework Equations

F = ma
F = kq1q2/r^2

The Attempt at a Solution

Im pretty clueless on this one. I know it's some type of equilibrium question, and it looks like the forces on the x direction cancel out. To find the charge on A, I need the force between the two charges but I don't have that and I'm not sure if I'm doing this right.

The only force on the string in the y direction is from gravity so

Fy = 0.09*9.8
=0.882N
The only force pulling the string in the x direction is the force between charge A and charge B
tantheta = Fx/Fy
Fx = tantheta*Fy
= tan7.2*0.882
= 1.15N

This equals to the force between the two charges, so

1.15 = k(qA)(qB)/(r^2)
(qA) = 1.15 *0.05^2/ (8.85e-12)(130e-9)
(qA) = 2.49e15

this value doesn't make much sense. not sure what to do.

 

[szynkasz]

The angle is in degrees not in radians:

$\tan 7.2^o=0.126\Rightarrow F_x=0.111\,N$

and $k=9\cdot 10^9\,N\cdot m^2/C^2$

 

[thercias]

The angle is in degrees not in radians:

$\tan 7.2^o=0.126\Rightarrow F_x=0.111\,N$

and $k=9\cdot 10^9\,N\cdot m^2/C^2$

thanks for the correction, my final answer for the charge is now 237 nC. That seems a lot better

Also, the force of tension I got was sqrt((0.0882^2)+(0.111^2))= 0.122N

Can you confirm that these are correct?

 

[szynkasz]

The charge is ok, but $F_y=0.882\,N$ not $0.0882$

 

[thercias]

Thanks man, just got to pay more attention to these clumsy errors I guess.