Calculate the magnitude of the charge on each sphere

[Bradsteeves]

Homework Statement

Two identical small spheres of mass 2.0g are fastened to the ends of an insulating thread of length 0.60m. The spheres are suspended by a hook in the ceiling from the centre of the thread. The spheres are given identical electric and hang in static equillibrium, with an angle of 30 degrees between the string halves. Calculate the magnitude of the charge on each sphere.

Basically looks like a triangle with the top corner angle 30 degrees, and both sides 0.3m each with the bottom side unknown which is r.

-Consider the 3 forces acting on the sphere
-Resolve the tension in the x and y directions
-Find FE
-Find r
-Apply Coulomb's Law

Given: m = 0.002 kg
r = ?
theta = 30 degrees
s1 = 0.3m
s2 = 0.3m
q = ?
FE = ?

Homework Equations

Sin (theta) = Opp/Hyp
Cos(theta)= Adj/Hyp
Sum of all forces in the x and y directions = ma
FE = kq1q2/r^2

The Attempt at a Solution

Sin theta = Opp/Hyp
0 = sin(theta) X Hyp
= sin 15 X .3m
=0.0776 X 2
= 1.55m

Fe = Kq1q2/r^2

Thats how far I got, I know I am supposed to find the force of tension some how but not sure

 

[alphysicist]

Hi Bradsteeves,

The Attempt at a Solution

Sin theta = Opp/Hyp
0 = sin(theta) X Hyp
= sin 15 X .3m
=0.0776 X 2
= 1.55m

I think you have the decimal in the wrong place here; it should be r=0.155 meters.

You can find the tension by looking at the forces in the vertical (y) direction. You know gravity is pulling down (the weight), and you can write the component of the tension that's pulling up. Since it's in equilibrium, these have to cancel. What do you get for the tension?

 

[Bradsteeves]

i got 0.0196 for the tension?

 

[alphysicist]

i got 0.0196 for the tension?

The weight is equal to 0.0196. And so that is equal to the vertical component of tension:

<br /> T_y = m g<br />

But what is T_y? It will be T times a trig function; once you figure out the trig function you can plug it into the above equation and solve for T (since you already know the angle is 15 degrees).