The work done by an external agent is equal to the change in potential energy, which is $$W_{ext} = q_{new}V_{final}$$, where $$V_{final}$$ is the potential at the destination point due to the existing charges. The potential at $$x=2d$$ due to the charges at $$x=0$$ and $$x=d$$ is $$V_{final} = k \frac{+q}{2d} + k \frac{-2q}{d} = k\left(\frac{q}{2d} - \frac{4q}{2d}\right) = -k\frac{3q}{2d}$$. The work done is $$W_{ext} = (+3q) V_{final} = (+3q) \left(-k\frac{3q}{2d}\right) = -k\frac{9q^2}{2d}$$.

Let the elementary charge be $$e$$. The potential energies are: $$U_A = k \frac{(+e)(+e)}{d} = +k \frac{e^2}{d}$$. $$U_B = k \frac{(+e)(-e)}{d} = -k \frac{e^2}{d}$$. $$U_C = k \frac{(-e)(-e)}{d} = +k \frac{e^2}{d}$$. Since $$k, e, d$$ are positive, $$U_A$$ and $$U_C$$ are equal and positive, while $$U_B$$ is negative. Therefore, the correct ranking from most positive to most negative is $$U_A = U_C > U_B$$.

The work done by the conservative electric field is the negative of the change in potential energy: $$W_E = -\Delta U = -(U_f - U_i)$$. The initial potential energy is $$U_i = kQq/R$$. The final potential energy is $$U_f = kQq/(3R)$$. The change in potential energy is $$\Delta U = U_f - U_i = \frac{kQq}{3R} - \frac{kQq}{R} = \frac{kQq - 3kQq}{3R} = -\frac{2kQq}{3R}$$. The work done by the field is $$W_E = -(-\frac{2kQq}{3R}) = +\frac{2kQq}{3R}$$.

According to the principle of conservation of energy for the proton-field system, the total energy remains constant. Since only the conservative electric force does work, the sum of the changes in kinetic energy ($$\Delta K$$) and electric potential energy ($$\Delta U_E$$) must be zero. $$\Delta K + \Delta U_E = 0$$. Given that the kinetic energy increases by $$5.0 \times 10^{-16} \text{ J}$$, we have $$\Delta K = +5.0 \times 10^{-16} \text{ J}$$. Therefore, $$\Delta U_E = -\Delta K = -5.0 \times 10^{-16} \text{ J}$$.

For the particle to just escape, its total mechanical energy must be zero. The total energy at infinity ($$r=\infty$$) is zero ($$K_f = 0, U_f = 0$$). By conservation of energy, the initial total energy must also be zero. $$E_i = K_i + U_i = 0$$. The initial kinetic energy is $$K_i = \frac{1}{2}mv_0^2$$. The initial potential energy is $$U_i = k \frac{(+Q)(-q)}{r_0} = -\frac{kQq}{r_0}$$. Setting the sum to zero: $$\frac{1}{2}mv_0^2 - \frac{kQq}{r_0} = 0$$. Solving for $$v_0$$ gives $$v_0 = \sqrt{\frac{2kQq}{mr_0}}$$.

The electric potential energy of two point charges is $$U_E = k \frac{q_1 q_2}{r}$$. For a proton ($$q_1 = +e$$) and an electron ($$q_2 = -e$$), the product $$q_1 q_2$$ is negative, so the potential energy is negative. A negative potential energy for a bound system implies that energy must be added (positive work must be done by an external force) to overcome the attractive force and separate the components to a state of zero potential energy (infinite separation).

The total potential energy is the sum of the potential energies of all six unique pairs. There are four adjacent pairs (distance $$s$$) and two diagonal pairs (distance $$s\sqrt{2}$$). Each adjacent pair consists of opposite charges ($$+q, -q$$), so there are four such terms: $$4 \times k\frac{(+q)(-q)}{s} = -4k\frac{q^2}{s}$$. The two diagonal pairs consist of like charges ($$+q, +q$$ and $$-q, -q$$): $$k\frac{(+q)(+q)}{s\sqrt{2}} + k\frac{(-q)(-q)}{s\sqrt{2}} = \frac{2kq^2}{s\sqrt{2}} = \frac{\sqrt{2}kq^2}{s}$$. The total energy is $$U_{total} = -4k\frac{q^2}{s} + \sqrt{2}k\frac{q^2}{s} = k \frac{q^2}{s}(\sqrt{2} - 4)$$.

During acceleration from rest, the electron's kinetic energy increases. By conservation of energy, its potential energy must decrease. So, $$\Delta U_1 < 0$$. The change is $$\Delta U_1 = (-e)V$$. During deceleration to rest, its kinetic energy decreases back to zero. Thus, its potential energy must increase, so $$\Delta U_2 > 0$$. By the work-energy theorem, the total change in kinetic energy is zero, so the net work done by the electric field is zero. This means the total change in potential energy is zero: $$\Delta U_1 + \Delta U_2 = 0$$, which implies $$\Delta U_2 = -\Delta U_1$$. Therefore, their magnitudes are equal.

This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the charge moved 0.50 m across a 200 V/m field, experiencing a potential difference of ΔV = -Ed = -200 × 0.50 = -100 V (negative because moving along field direction). The correct answer, B, is derived by applying ΔU = qΔV = (1.0×10⁻⁶ C)(-100 V) = -1.0×10⁻⁴ J. A common distractor, A, arises from sign confusion about field direction. To assist students, emphasize that "across" typically means along the field direction (high to low potential), and practice visualizing field lines and equipotential surfaces.

This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge moved through a battery from 0 V to 12 V experiences a potential difference of ΔV = 12 - 0 = 12 V. The correct answer, A, is derived by applying ΔU = qΔV = (1.5×10⁻⁶ C)(12 V) = 1.8×10⁻⁵ J. A common distractor, B, arises from incorrect unit conversion, treating microCoulombs as Coulombs. To assist students, emphasize careful unit conversion (1 μC = 10⁻⁶ C) and practice dimensional analysis to ensure energy units (Joules) result from charge (Coulombs) times potential (Volts).