Two identical balls, each possessing a charge of 2.00 × 10^ -7 C...

The forces acting on each ball are the tension $T$ in the string, the gravitational force $mg$, and the electrostatic repulsion $F_e$.

Just after release, the velocity of the balls is zero, which means the centripetal acceleration is zero. Therefore, the net force along the string is zero.

The electrostatic force between the balls is:

$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2} = \dfrac{9 \times 10^9 \times (2.00 \times 10^{-7})^2}{(0.05)^2} = 0.144 \text{ N}$

Let $\theta$ be the angle the string makes with the vertical. The distance from the vertical axis to each ball is $\dfrac{d}{2} = 2.5 \text{ cm}$.

$\sin\theta = \dfrac{2.5}{50} = 0.05$

For small angles, $\cos\theta \approx 1$.

The component of gravity perpendicular to the string (directed towards the other charge) is:

$mg \sin\theta = 0.1 \times 9.8 \times 0.05 = 0.049 \text{ N}$

The component of the electrostatic force perpendicular to the string (directed away from the other charge) is:

$F_e \cos\theta = 0.144 \times 1 = 0.144 \text{ N}$

The net force perpendicular to the string is:

$F_{\perp} = F_e \cos\theta - mg \sin\theta = 0.144 - 0.049 = 0.095 \text{ N}$

Since $F_e \cos\theta > mg \sin\theta$, the resultant force is directed away from the other charge.