Question

two pith balls of mass m and charge Q

Repeat with the other producer. Are the results similar? Two pith balls of mass m and charge Q are suspended by threads of length L and are separated by total angle 2theta. Draw a free-body diagram for each pith ball. Using Coulomb's Law for the electrical force of repulsion and the weight of each of the pith balls, show that the charge on each ball is Q = plusminus 2L sin theta squareroot mg tan theta/k. You work should proceed mathematically from each step to the next.

Answer 1

a)

b)The Coulomb's law describes the interaction between point charges.

$F_{e}=\frac{kq_{1}q_{2}}{r^{2}}$

When the pith balls are charged with equal amount of charge, q, they repel each other to a distance r (measured from the centers of the balls), due to a Coulomb force

$F_{e}=\frac{kq^{2}}{r^{2}}$

This Coulomb force Fe is balanced by two other forces, the tension force T and the weight on the ball mg. From the free-body diagram, the relationship between Fe and mg can be determined by the angle θ or the ball-to-ball distance r without knowing T

$\frac{F_{e}}{mg}=\tan\theta=\frac{r/2}{\sqrt(L^{2}-(\frac{r}{2})^{2})}$

This relationship allows us to estimate the amount of charge on the pith balls through measuring the angle θ or the ball-to-ball distance r.

$q=2L\sin\theta\sqrt\frac{mg\tan\theta}{k}=\sqrt\frac{mgr^{3}}{2k(L^{2}-(r/2)^{2})^{1/2}}$