(adsbygoogle = window.adsbygoogle || []).push({}); 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

F_{e}=k_{e}lq_{1}l lq_{2}l [tex]/[/tex] r^{2}

the ratio of F_{g}/F_{e}= tan [tex]\theta[/tex]

F_{g}= 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.

F_{g}= mg = .010 kg x 9.8 m/s^{2}= .098 N

Using the ratio

F_{e}= F_{g}/tan [tex]\theta[/tex]

= .098 N / tan (25)

= .21 N

F_{e}=k_{e}lq_{1}l lq_{2}l [tex]/[/tex] r^{2}

Arranging this equation and solving for q_{1}

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 = [tex]\sqrt{}[/tex]((F_{e}X r^{2})/ k )

= [tex]\sqrt{}[/tex]( .21N X 5.1^{2}/ 9.0X10^{9}

= 4.44 X 10^{-7}Coulombs

is it right???

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# Homework Help: Charge problem: two pith balls

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