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Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle 'θ' with the vertical ?

1. x = ((q² l)/(2πε₀ mg))^(1/2)
2. x = ((q² l)/(2πε₀ mg))^(1/3)
3. x = ((q² l²)/(2πε₀ m² g))^(1/3)
4. x = ((q² l²)/(2πε₀ m² g²))^(1/3)

Correct answer: x = ((q² l)/(2πε₀ mg))^(1/3)

Solution

For small theta, separation x = 2*l*sin(theta) ~ 2*l*theta, and tan(theta) ~ x/(2l) equals the Coulomb-to-weight ratio k*q^2/(x^2*m*g). Setting x/(2l) = q^2/(4*pi*e0*x^2*m*g) gives x^3 = q^2*l/(2*pi*e0*m*g), so x = (q^2*l/(2*pi*e0*m*g))^(1/3).

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