Conservation of Electric Energy Practice Test

A $5.0,\text{V}$ battery charges a capacitor $C=1.0,\mu\text{F}$ through a resistor $R=1.0,\text{k}\Omega$; after a long time the capacitor reaches $V=5.0,\text{V}$. The energy stored on the capacitor is $U_C=\tfrac12CV^2=\tfrac12(1.0\times10^{-6})(5.0)^2=1.25\times10^{-5},\text{J}$. The battery moves total charge $Q=CV=(1.0\times10^{-6})(5.0)=5.0\times10^{-6},\text{C}$ through a potential difference of $5.0,\text{V}$, so the chemical energy converted to electrical energy is $E_{\text{bat}}=QV=CV^2=2.5\times10^{-5},\text{J}$. The remaining $E_{\text{bat}}-U_C=1.25\times10^{-5},\text{J}$ becomes thermal energy in the resistor during the transient current as charges move and collide. In this classical circuit, energy is conserved: chemical energy decreases in the battery and appears as stored electric energy plus heat.

A point charge $q=+2.0,\text{nC}$ is released from rest in a region where the electric potential decreases from $V_A=300,\text{V}$ at point A to $V_B=100,\text{V}$ at point B along its path. The potential difference causes charge motion, and the electric field does work on the charge. The change in electric potential energy is $\Delta U = q\Delta V = q(V_B-V_A) = (2.0\times10^{-9})(100-300)=-4.0\times10^{-7},\text{J}$. In a closed system with negligible nonconservative forces, the decrease in electric potential energy becomes an increase in kinetic energy: $\Delta K = -\Delta U$. Because the charge starts from rest, $K_A=0$ and $K_B=\Delta K=4.0\times10^{-7},\text{J}$. This illustrates conservation of energy: the electric field’s work converts stored electric potential energy associated with position in the potential into kinetic energy of the moving charge, without creating or destroying energy.