Answer
$\sigma = 5.0\times 10^{-9}~C/m^2$
Work Step by Step
The ball's weight is equal in magnitude to the vertical component of tension in the thread.
We can find an expression for the tension $F_T$:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
The electrostatic force between the sheet and the ball is equal in magnitude to the horizontal component of tension in the thread.
We can find the surface charge density of the sheet:
$F = F_T~sin~\theta$
$E~q = \frac{mg~sin~\theta}{cos~\theta}$
$\frac{\sigma~q}{2\epsilon_0} = mg~tan~\theta$
$\sigma = \frac{2\epsilon_0~mg~tan~\theta}{q}$
$\sigma = \frac{(2)(8.854\times 10^{-12}~F/m)~(1.0\times 10^{-6}~kg)(9.8~m/s^2)~tan~30^{\circ}}{2.0\times 10^{-8}~C}$
$\sigma = 5.0\times 10^{-9}~C/m^2$
