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

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

Since the balls are released from rest, the initial velocity is zero. Thus, the radial acceleration is zero, and the net acceleration is purely tangential (perpendicular to the string).

The electrostatic force between the balls is given by Coulomb's law:

$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2}$

$F_e = \dfrac{9 \times 10^9 \times (2.00 \times 10^{-7})^2}{(0.05)^2} = 0.144 \text{ N}$

The angle $\theta$ made by each string with the vertical is:

$\sin \theta = \dfrac{d/2}{L} = \dfrac{2.5 \text{ cm}}{50 \text{ cm}} = 0.05$

For this small angle, $\cos \theta \approx 1$.

The tangential component of the net force is:

$F_t = F_e \cos \theta - mg \sin \theta$

$F_t = 0.144 \times 1 - (0.1 \times 9.8 \times 0.05) = 0.144 - 0.049 = 0.095 \text{ N}$

The tangential acceleration of the ball is:

$a = \dfrac{F_t}{m} = \dfrac{0.095}{0.1} = 0.95 \text{ m s}^{-2}$

Since $F_e \cos \theta > mg \sin \theta$, the net force and acceleration are directed perpendicular to the string, away from the other charge.