# Charge problem: two pith balls

1. Jan 11, 2009

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

Please check if i did this problem right..

two charged pith balls are suspended on very light 12. cm strings. Each has a mass of 10. . In equilibrium the total angle between the string is 50 degrees, Assuming that the pith balls are equally charged, what is the charg on each ball

2. Relevant equations

Fe=kelq1l lq2l $$/$$ r2

the ratio of Fg/Fe= tan $$\theta$$

Fg = mg

3. The attempt at a solution

This is what I tried
I made a diagram in showing all forces in which the graciataional force is downward, the electric force is pointing outward since the two balls repel each other and then their is tension form the string which is upward and inward 25 degree angle since both balls make a 50 degree angle.

Fg= mg = .010 kg x 9.8 m/s2 = .098 N

Using the ratio
Fe = Fg/tan $$\theta$$
= .098 N / tan (25)
= .21 N

Fe=kelq1l lq2l $$/$$ r2

Arranging this equation and solving for q1

r is the distance the distance between the two balls which can be calculated
sin 50 = x / 12

x = 9.2 cm = .092 m

q = $$\sqrt{}$$((Fe X r2)/ k )
= $$\sqrt{}$$( .21N X 5.12 / 9.0X109
= 4.44 X 10 -7 Coulombs

is it right???

2. Jan 11, 2009

### LowlyPion

I think it should be Fe is balanced with mg*Tan25.

The vertical weight component of the ball means that the tension in the string is mg/cos25 and the horizontal force will be sin25 of this in magnitude ... hence mg*tan25.

3. Jan 11, 2009

### LowlyPion

I also think the distance between the balls is given by twice the length of the cord times Sin25 - which is the displacement from the vertical of 1 ball. Hence the need to double it.

4. Jan 12, 2009

Ok so in this case then Fe= mg * tan 25= .098 N * tan 25= .046 N

and then for the distance between you are right
so the distance will be

sin 25 = x / 12

x = 5.07 cm X 2 = 10 cm = .1m

so plugging this new numbers the answer should be

7.15 X 10-7 coulumbs