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

Let $\theta$ be the angle made by the string with the vertical. The distance between the balls is $x = 5.0 \text{ cm} = 0.05 \text{ m}$, and the length of the string is $L = 50 \text{ cm} = 0.5 \text{ m}$.

From the geometry of the suspension:
$\sin\theta = \dfrac{x/2}{L} = \dfrac{0.025}{0.5} = 0.05$
$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (0.05)^2} = \sqrt{0.9975} \approx 0.99875$

The electrostatic force of repulsion between the balls is:
$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{x^2} = \dfrac{9 \times 10^9 \times (2.00 \times 10^{-7})^2}{(0.05)^2}$
$F_e = \dfrac{9 \times 10^9 \times 4 \times 10^{-14}}{0.0025} = 0.144 \text{ N}$

The weight of each ball is:
$mg = 0.1 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.98 \text{ N}$

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

Balancing the forces along the string:
$T - mg \cos\theta - F_e \sin\theta = 0$
$T = mg \cos\theta + F_e \sin\theta$

Substituting the values:
$T = (0.98 \times 0.99875) + (0.144 \times 0.05)$
$T = 0.978775 + 0.0072 = 0.985975 \text{ N}$

Rounding off to three significant figures, we get $T \approx 0.986 \text{ N}$.