Answer
$q = \pm 2.38 \times 10^{-8} C$
Work Step by Step
Taking the equation from (a),
$x = (\frac{q^2L}{2\pi\epsilon_omg})^{1/3}$
$x^3 = \frac{q^2L}{2\pi\epsilon_omg}$
$(\frac{x^32\pi\epsilon_omg}{L})^{1/2} = q $
From the question, $ L = 1.2 m, m = 0.01kg, x = 0.05 m $
$q = (\frac{x^32\pi\epsilon_omg}{L})^{1/2} $
$q = (\frac{( 0.05 m)^32\pi(8.85 x 10^{-12} F/m)(0.01kg)(9.8 m/s^2)}{1.2 m})^{1/2} $
$q = \pm 2.38 \times 10^{-8} C$
