Two charged particles, each carrying a charge of 2.0 × 10^ -5 C ,...

The initial electric potential energy of the system when the charges are at infinity is $U_i = 0$.

The final electric potential energy when the charges are separated by a distance $r$ is given by:

$U_f = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{r}$

Given $q_1 = q_2 = 2.0 \times 10^{-5} \text{ C}$ and $r = 10 \text{ cm} = 0.1 \text{ m}$.

Substituting the values:

$U_f = (9 \times 10^9) \dfrac{(2.0 \times 10^{-5}) \times (2.0 \times 10^{-5})}{0.1}$

$U_f = 9 \times 10^9 \times \dfrac{4.0 \times 10^{-10}}{0.1}$

$U_f = \dfrac{3.6}{0.1} = 36 \text{ J}$

The increase in electric potential energy is $\Delta U = U_f - U_i = 36 \text{ J}$.