# Charge on two suspended metal balls

## Homework Statement

Two very small metal coated foam spheres, each of mass 2.80x10^-6kg, are attached to nylon threads 45cm long and hung from a common point. When the spheres are given equal quantities of negative charge, each supporting thread makes an angle of 15 degrees with the vertical. Find the charge on each sphere.

## Homework Equations

F=(kQ1Q2)/r^2 in the form of Q=√((Fr^2)/k) as Q1 and Q2 will be the same.
F=ma
probably Pythagoras and sine/tangent/cosine ratios

## The Attempt at a Solution

ok, so r or the distance between the charges was pretty obvious. If the strings are 0.45m long and the angle at the point where the strings combine is 30° (2x15°), then its going to be an equilateral triangle so r=0.45m (I think). I suppose we are assuming k=9X10^9. Noe for the force. The force of gravity will be 2.80x10^-6x9.8 (F=ma), which equals approximately 2.7x10^-7. I then did (2.7x10^-7)/tan(90-15), which equals about 7.4x10^-8. I thought that equaled half the force of repulsion of the sphere, but I'm not sure. If thats right then the total force of repulsion is 1.5x10^-7. So then
Q=√((1.5X10^-7x0.45^2)/9x10^9). That equaled 1.8X10^-9. But the actual answer is 6.66X10^-9. I know my rounding is having quite and effect on it, so i did it again, using as many decimals as I could but I was still quite a bit off. What am I doing wrong?

Related Introductory Physics Homework Help News on Phys.org
An equilateral triangle has angles of 60° - this one has a top angle of 30°.

oh wow, oopsies!

I don't really understand your solution attempt here.
Start by drawing a free-body diagram for one of the charges and resolve the forces along the x and y directions.
You should be able to eliminate the tension leaving only 1 unknown ( the charge ).

## Homework Statement

... angle at the point where the strings combine is 30° (2x15°), then its going to be an equilateral triangle so r=0.45m
First and foremost, when did you learn that an equilateral triangle has an angle 30°? It's not an equilateral triangle!
This one will be isosceles!
Next why are you taking the weight as it is? Take its horizontal components!
Then I don't get why you are dividing by tan75°. Could you explain that?

Rookie mistake. You are unworthy of this forum. Next time go to yahoo answers...