This problem tests conservation of electric energy. Initially, the charged 2.0μF capacitor stores U₁ = ½(2.0μF)(1.5V)² = 2.25μJ. When connected to an identical uncharged capacitor, charge redistributes until both reach the same voltage. The total charge Q = (2.0μF)(1.5V) = 3.0μC is conserved, giving final voltage Vf = Q/Ctotal = 3.0μC/4.0μF = 0.75V. The final total energy is Uf = ½(4.0μF)(0.75V)² = 1.125μJ, which is half the initial energy. Choice A incorrectly assumes charge conservation implies energy conservation. Choice C wrongly thinks parallel connection increases energy, missing that energy is lost during charge redistribution. The strategy is to recognize that connecting capacitors at different voltages always dissipates energy, even though charge is conserved.

This problem tests conservation of electric energy. When an isolated capacitor's plate separation decreases by factor of 3, its capacitance increases by factor of 3 (since C ∝ 1/d). With charge Q held constant, the stored energy U = Q²/2C decreases by factor of 3. If initially U₁ = Q²/2C, then finally U₂ = Q²/(2×3C) = U₁/3. Choice A incorrectly claims energy increases, confusing the inverse relationship between C and U at constant charge. Choice C wrongly assumes energy can't change without a battery, missing that the external agent does negative work by allowing attractive force between plates to pull them together. The strategy is to use U = Q²/2C for constant charge situations and remember that increasing capacitance decreases stored energy when charge is fixed.

The work-energy theorem states that the net work done on an object equals its change in kinetic energy. Since the charge moves at a constant speed, its kinetic energy does not change, so the net work is zero. The net work is the sum of the work done by the external agent ($$W_{ext}$$) and the work done by the electric field ($$W_E$$). So, $$W_{net} = W_{ext} + W_E = 0$$. This means $$W_E = -W_{ext} = -10 \text{ J}$$. The change in potential energy is related to the work done by the conservative electric field by $$\Delta U_E = -W_E$$. Therefore, $$\Delta U_E = -(-10 \text{ J}) = +10 \text{ J}$$.

This question assesses the skill of conservation of electric energy. With the battery disconnected, charge Q is conserved, and increasing plate separation decreases C, so stored energy U = Q²/(2C) increases. This increase comes from mechanical work done pulling the plates apart, redistributing energy into the electric field. Energy is conserved as the external work adds to the system's stored energy without dissipation. Choice A wrongly says it decreases due to weakened field, missing the misconception that energy depends only on field strength without considering capacitance changes. Track where energy goes, not just where charge ends up.

To escape, the particle's total energy must be at least zero. The total energy is the sum of kinetic and potential energy: $$E = K + U_E$$. To just escape ($$v_f=0$$ at $$r=\infty$$), the final total energy is zero. By conservation of energy, the initial total energy must also be zero. The initial potential energy is $$U_{E,i} = k\frac{(+Q)(-q)}{R} = -k\frac{Qq}{R}$$. The initial kinetic energy is $$K_i = \frac{1}{2}mv_{esc}^2$$. Setting $$E_i=0$$ gives $$\frac{1}{2}mv_{esc}^2 - k\frac{Qq}{R} = 0$$. Solving for $$v_{esc}$$ yields $$v_{esc} = \sqrt{\frac{2kQq}{mR}}$$. Note that the initial speed for a circular orbit is less than this value.

To bring the particle to rest, the change in its kinetic energy must be $$\Delta K = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}mv^2$$. By the work-energy theorem for conservative forces, $$\Delta K = -\Delta U_E$$. The change in potential energy is $$\Delta U_E = q \Delta V$$. Therefore, $$-\frac{1}{2}mv^2 = -q \Delta V$$. Solving for the magnitude of the potential difference $$|\Delta V|$$ gives $$|\Delta V| = \frac{mv^2}{2q}$$

Let the potential difference between the plates be $$V$$. The potential at the midpoint is $$V/2$$ relative to the negative plate. The proton (charge $$+e$$) moves through a potential difference of $$V/2$$ to reach the negative plate. The electron (charge $$-e$$) moves through a potential difference of $$V/2$$ to reach the positive plate. The magnitude of the change in potential energy for both is $$|\Delta U_E| = |q \Delta V| = e(V/2)$$. By conservation of energy, the kinetic energy gained by each is equal to this value. Thus, they strike the plates with the same kinetic energy.

The electrostatic force is a conservative force. This means the work done by the force, and therefore the change in potential energy, depends only on the initial and final positions of the particle, not on the path taken between them. Since both paths start at A and end at B, the change in electric potential energy is the same for both paths: $$\Delta U_1 = \Delta U_2$$.

This problem tests conservation of electric energy. In parallel, both capacitors have the same voltage V = 1.5 V but different charges proportional to their capacitances. The energy stored in each is U = ½CV². Since C₂ = 6.0 μF = 3×C₁, we have U₂ = 3U₁. Specifically, U₁ = ½(2.0×10⁻⁶)(1.5)² = 2.25 μJ and U₂ = ½(6.0×10⁻⁶)(1.5)² = 6.75 μJ. Choice B incorrectly relates energy to charge amount, missing that for parallel capacitors at the same voltage, energy is directly proportional to capacitance. The strategy is to remember that for parallel capacitors, U ∝ C when V is constant.

The gain in kinetic energy is equal to the magnitude of the loss in potential energy: $$K_f = |\Delta U_E| = |q \Delta V|$$. The charge of an alpha particle is $$q = +2e$$. The potential difference is $$500 \text{ V}$$. Thus, the kinetic energy gained is $$K_f = |(+2e)(-500 \text{ V})| = 1000 \text{ eV}$$. One electron volt (eV) is the energy gained by a charge of $$e$$ moving through a potential difference of $$1 \text{ V}$$.