# Two spheres hanging, find charge.

1. Jan 30, 2012

### DrKrunklehorn

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

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.

2. Relevant equations

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

3. 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

2. Jan 30, 2012

### 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

3. Jan 30, 2012

### DrKrunklehorn

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.

4. Jan 30, 2012

### nure

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

Sx=S*sin15

and

S=Sy/cos15

Therefore

q=1.19*10-7

since

Fe=m*g*tan15

Last edited: Jan 30, 2012
5. Jan 30, 2012

### DrKrunklehorn

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?

6. Jan 30, 2012

### nure

The equation

Fe=(k*q2)/(r2)

can(as you know) be rewritten as

q=r*Sqrt[Fe/k]

Sum of forces in x-direction equals zero:

Fe-Sx=0

therefore

Fe=Sx

Similarly in the y-direction:

Sy-w=0

therefore

Sy=w=m*g

Now, from simple trigonometry you see that

Sy=S*cos15

and

Sx=S*sin15

This means that you can write Sx as

Sx=(Sy/cos15)*sin15=m*g*tan15

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

Fe=m*g*tan15

Now you have Fe. 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

Last edited: Jan 30, 2012
7. Jan 30, 2012

### DrKrunklehorn

Thanks for the help! I understand completely now.

8. Jan 30, 2012

### technician

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