Two identical pith balls, each with a charge q , are suspended from a common point using two strings of equal length l . Determine the mass of each ball given that the angle between the strings is 2θ in equilibrium.

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Correct answer: C. (q^2 θ)/(16π _0 g l^2 ^2θ)

The angle made by each string with the vertical is

\theta

.

Distance between the two balls,

r = 2l \sin\theta

.

Electrostatic force of repulsion between the balls,

F_e = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q^2}{r^2} = \dfrac{q^2}{16\pi\varepsilon_0 l^2 \sin^2\theta}

.

For the equilibrium of each ball, balancing the forces in the horizontal and vertical directions:

T \sin\theta = F_e

T \cos\theta = mg

Dividing the two equations:

\tan\theta = \dfrac{F_e}{mg}

m = \dfrac{F_e}{g \tan\theta} = \dfrac{F_e \cot\theta}{g}

Substituting the value of

F_e

:

m = \dfrac{q^2 \cot\theta}{16\pi\varepsilon_0 g l^2 \sin^2\theta}

Two identical pith balls, each with a charge q , are suspended from...

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