A silk thread 10 cm in length has one end attached to the large...

Let $T$ be the tension in the thread, $E$ be the electric field due to the plate, $m$ be the mass of the ball, and $q$ be its charge.

In equilibrium, the forces acting on the ball in the horizontal and vertical directions balance each other:

$T \sin\theta = qE$

$T \cos\theta = mg$

Dividing the two equations, we get:

$\tan\theta = \dfrac{qE}{mg}$

The electric field produced by a large nonconducting charged plate is given by $E = \dfrac{\sigma}{2\epsilon_0}$.

Substituting this into the equation for $\tan\theta$:

$\tan\theta = \dfrac{q\sigma}{2\epsilon_0 mg}$

Rearranging for $\sigma$:

$\sigma = \dfrac{2\epsilon_0 mg \tan\theta}{q}$

Given values are $m = 10 \text{ g} = 10^{-2} \text{ kg}$, $q = 4.0 \times 10^{-6} \text{ C}$, \theta = 60^{}^{{}^{\circ}}, $g = 9.8 \text{ m/s}^2$, and $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2$.

$\sigma = \dfrac{2 \times 8.85 \times 10^{-12} \times 10^{-2} \times 9.8 \times \sqrt{3}}{4.0 \times 10^{-6}}$

$\sigma = \dfrac{17.7 \times 10^{-12} \times 0.098 \times 1.732}{4.0 \times 10^{-6}}$

$\sigma \approx 7.5 \times 10^{-7} \text{ C/m}^2$