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## Homework Statement

When the pith balls are suspended, all forces are balanced. Assuming a mass of 0.060g for each pith ball, calculate the electric force which is repelling the pith balls. Hint: draw a free body diagram and apply Newton's 2nd Law seperately in the x and y directions. Assuming an equal charge on each pith ball, calculate the charge in Coulombs of each pith ball.

The pith balls form an isosceles triangle with the stand.

## Homework Equations

Fg = mg

Fe = [tex] \frac{kq1q2}{r^{2}} [\tex]

Fe = mg

## The Attempt at a Solution

Angle of pith balls = [tex] tan^{-1} (0.10/0.0125) [/tex] = 82.87

Because gravity is acting at an angle, so the acceleration due to gravity wouldn't be 9.81m/s^2 right?

Fg = (6.0*10^-5) * (Tan 82.87) * (9.81) = 4.705 * 10^-3

Fg = Fe

4.705 * 10^-3 = k q^2 / r^2

r^2 = 6.25*10^-4

q = sqrt[ (Fg)(r)^2) / k ]

q = sqrt[ (4.705 * 10^-3) (6.25*10^-4) / (8.99*10^9) ]

q = 7.2*10^-7 C

I'm not sure if my use of 'angles' was correct. If anyone can double check my work I would really appreciate it. Thanks in advance.