What is the Charge of Two Suspended Pith Balls at Equilibrium?

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SUMMARY

The discussion focuses on calculating the electric force (Fe) and charge of two suspended pith balls, each with a mass of 4.9 g and equal charge, positioned 3.5 cm apart. The suspended ball forms a 30.0° angle with the vertical, indicating equilibrium among the electric force (Fe), gravitational force (Fg), and tension (FT). The key equation used is Fe = (K(qa)(qb))/r², where K is Coulomb's constant. The solution involves determining the tension and electric force to solve for the charge, |q|, without knowing its sign.

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Homework Statement


The two pith balls below each have a mass of 4.9 g and equal charge. One pith ball is suspended by an insulating thread. The other is brought to x = 3.5 cm from the suspended ball. The suspended ball is now hanging with the thread forming an angle of 30.0° with the vertical. The ball is in equilibrium with FE, Fg, and FT. Calculate each of the following.

Fg (got that)

Fe (Havent gotten that)

Charge (Have no real clue how to get that)

Homework Equations


Fe=(K(qa)(qb))/r^2


The Attempt at a Solution


I'm assuming I'd be able to use Fe=(K(qa)(qb))/r^2 except I don't know how to find the charges.

Also, as an unrelated question, how do I find Voltage with the electric field and distance in a parallel plate?
 
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Note that the charges are equal. Though there is no indication that they are + or -.

To determine the charge you need to determine the |Fe| from your force diagram. You know the weight and the angle the string is at, so you can figure the Tension and the Fe.

Then solve for |q|, because you don't know the sign of the charge - only that they repel.
 

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