Two spheres hanging, find charge.

[DrKrunklehorn]

Homework Statement

Two identical, small spheres of mass 2.0g are fastened to the ends of a 0.6m long light, flexible, insulating fishline. The fishline is suspended by a hook in the ceiling at it's exact center. The spheres are each given the same electric charge. They are in static equilibrium, creating an angle of 30° between the halves of the string. Calculate the magnitude of the charge on each sphere.

Here's a sketch of the picture given on the worksheet:

Red lines are what I drew over the diagram to make a right triangle.

This being a bonus type question, I was given three possible answers and had to find out which one was the real one:

q =
3.7 x 10^-7 C or
1.2 x 10^3 C or
1.2 x 10^-3 C.

Homework Equations

Coulombs Law: F = k \ \frac{q_1 q_2}{r^2}

The Attempt at a Solution

First I converted the mass into kilograms: m = 0.002kg

Then I split the angle into two right triangles to get the distance between the spheres.

r/2 = Sin15 x 0.3 m
r = approx. 0.16m

Then I figured since the spheres are at rest, then a = 0 and Fg = Fe...so I tried coulombs law to solve for q:

F_{g} = F_{e}
mg = k \ \frac{q_1 q_2}{r^2}
\frac{mgr^2}{k} = q_1 q_2

I ended up with 2.6 x 10^-11...obviously way off any of the true answers.

Thanks for taking the time to read over all this. =D

Cheers - Krunklehorn

 

[technician]

You need to look at the forces acting on the balls.
There is a vertical force = weight of a ball = 2x10^-3x9.81 = 0.0196N
There is a sideways force, F, due to the repulsion.
Can you see that F/0.0196 = Tan15
This will give you F to use in the repulsion equation.
I got Q = 1.2x10^-7C, on your list there is 1.2x10^-3C
I use k = 8.98x10^9 in the repulsion equation

 

[DrKrunklehorn]

You need to look at the forces acting on the balls.
There is a vertical force = weight of a ball = 2x10^-3x9.81 = 0.0196N
There is a sideways force, F, due to the repulsion.
Can you see that F/0.0196 = Tan15
This will give you F to use in the repulsion equation.
I got Q = 1.2x10^-7C, on your list there is 1.2x10^-3C
I use k = 8.98x10^9 in the repulsion equation

Oh I see! Fe pushes the spheres to the side while Fg pushes them down to the ground.

Tan = Opp / Adj so my equation should be:

Tan15 Fg = Fe

\frac{Tan15mgr^2}{2k} = q

Awww too bad it's actually wrong. I am getting 7.03x10^-15 as an answer.

 

[nure]

Agree with technician. The force on one of the spheres is equal to the x-component of the fishline force(S):

S_x=S*sin15

and

S=S_y/cos15

Therefore

q=1.19*10^-7

since

F_e=m*g*tan15

 

[DrKrunklehorn]

since

F_e=m*g*tan15

Oh I completely agree with him and appreciate the help immensely...I just don't see why I'm getting such a mangled answer.

Did I rearrange the equation wrong?

 

[nure]

The equation

F_e=(k*q²)/(r²)

can(as you know) be rewritten as

q=r*Sqrt[F_e/k]

Sum of forces in x-direction equals zero:

F_e-S_x=0

therefore

F_e=S_x

Similarly in the y-direction:

S_y-w=0

therefore

S_y=w=m*g

Now, from simple trigonometry you see that

S_y=S*cos15

and

S_x=S*sin15

This means that you can write S_x as

S_x=(S_y/cos15)*sin15=m*g*tan15

And from the equilibrium equation in the x-direction you see that

F_e=m*g*tan15

Now you have F_e. You then need the distance r between the spheres:

r=2*0.3*sin15

Filling all this information into the expression for q then gives you

q=1.187*10^-7 C

 

[DrKrunklehorn]

Thanks for the help! I understand completely now.

 

[technician]

Great...well done...don't always believe published answers