Conservation of Electric Energy Practice Test

A $12,\text{V}$ ideal battery charges a $C=200,\mu\text{F}$ capacitor through a resistor $R=300,\Omega$; after a long time, the capacitor reaches $12,\text{V}$. The capacitor then disconnects from the battery and discharges through the same resistor in a closed loop. During discharge, charge moves through a potential difference, so capacitor potential energy decreases and becomes thermal energy in $R$ (via brief carrier kinetic energy). The initial discharge energy is $U_0=\tfrac12 C V^2=\tfrac12(200\times10^{-6})(12^2)=1.44\times10^{-2},\text{J}$. With $RC=0.060,\text{s}$, after $t=0.030,\text{s}$, $U_C=U_0 e^{-2t/RC}=U_0 e^{-1}\approx5.30\times10^{-3},\text{J}$, so $9.1\times10^{-3},\text{J}$ has been converted to heat while total energy remains conserved. Based on the scenario, how does the energy stored in the capacitor change as it discharges?

A proton ($q=+e$) moves from point A to point B in an electrostatic field inside an insulated chamber, so no external work is done and no energy leaves the system. The potential difference is $\Delta V = V_B - V_A = -500,\text{V}$. The change in electric potential energy is $\Delta U = q\Delta V = (1.60\times10^{-19})(-500)=-8.0\times10^{-17},\text{J}$. If the proton starts with kinetic energy $K_i=2.0\times10^{-17},\text{J}$, then energy conservation gives $K_f = K_i - \Delta U = 1.0\times10^{-16},\text{J}$. The potential energy decrease equals the kinetic energy increase, showing interplay between $U$ and $K$ driven by potential difference. In the context of this scenario, what is the relationship between potential difference and energy conservation in the context?