Answer
$x = 0.0315 m = 3.15 cm$
Work Step by Step
The equilibrium separation is when $\frac{|q|}{2}$
$x = (\frac{(\frac{q}{2})^2L}{2\pi\epsilon_omg})^{1/3}$
$x = (\frac{(\frac{q^2}{4})2L}{2\pi\epsilon_omg})^{1/3}$
Taking from question (42.b)
$|q| = 2.38 \times 10^{-8} C, L = 1.2 m, m = 0.01kg $
$x = (\frac{(\frac{1}{4})( 2.38 \times 10^{-8} C)^2( 1.2 m)}{2\pi(8.85 x 10^{-12} F/m)(0.01kg)(9.8 m/s^2)})^{1/3}$
$x = 0.0315 m = 3.15 cm$
