Conservation of Electric Energy
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AP Physics C: Electricity and Magnetism › Conservation of Electric Energy
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.
The battery supplies $5.0\times10^{-6},\text{J}$ because $E=QV$ with $Q=1.0,\mu\text{C}$.
No energy dissipates because an ideal resistor cannot convert electrical energy.
The battery supplies $1.25\times10^{-5},\text{J}$ because $E=\tfrac12CV^2$ always.
The battery supplies $2.5\times10^{-5},\text{J}$; half stores on $C$ and half dissipates in $R$.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy during capacitor charging through a resistor. Electric energy conservation during charging requires that energy supplied by the battery splits between stored capacitor energy and heat dissipated in the resistor. In this scenario, the battery moves charge Q=CV through potential V, supplying energy E_bat=QV=CV²=2.5×10⁻⁵ J, while the capacitor stores only U_C=½CV²=1.25×10⁻⁵ J. Choice A is correct because it recognizes that exactly half the supplied energy (1.25×10⁻⁵ J) dissipates as heat in the resistor during the charging process. Choice B is incorrect because it considers only the final stored energy, ignoring the energy dissipated during charging. To help students: Emphasize that charging always involves energy loss to heat, regardless of resistance value. Demonstrate that this 50-50 split is universal for RC charging circuits.
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.
The charge gains $4.0\times10^{-7},\text{J}$ of kinetic energy moving to lower potential.
Energy is not conserved because the field removes energy from the system.
The charge loses kinetic energy because $\Delta V<0$ for a positive charge.
The charge gains $4.0\times10^{-7},\text{J}$ of potential energy because $V$ decreases.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy when charges move through potential differences. Electric energy conservation dictates that changes in electric potential energy must be balanced by changes in kinetic energy when no other forces do work. In this scenario, a positive charge moves from higher to lower potential, causing its electric potential energy to decrease by ΔU=qΔV=(2.0×10⁻⁹)(-200)=-4.0×10⁻⁷ J. Choice A is correct because this decrease in potential energy becomes an equal increase in kinetic energy, ΔK=-ΔU=4.0×10⁻⁷ J, as the electric field accelerates the charge. Choice C is incorrect because it confuses the sign convention - positive charges lose potential energy when moving to lower potential, not gain it. To help students: Emphasize that positive charges naturally move toward lower potential, losing potential energy. Practice using energy conservation equations ΔK+ΔU=0 for various charge motions.
A parallel-plate capacitor with capacitance $C=1.0,\mu\text{F}$ is charged by an ideal source to $V=100,\text{V}$ and then disconnected so the capacitor plus surrounding field form a closed system. The stored energy in the electric field is $U=\tfrac12 C V^2=\tfrac12(1.0\times10^{-6})(100^2)=5.0\times10^{-3},\text{J}$. The plates are then connected by a $R=1.0,\text{k}\Omega$ resistor, allowing charge to move due to the potential difference. As the capacitor discharges, electric potential energy decreases, carriers gain kinetic energy momentarily, and the resistor’s lattice gains thermal energy; the integral $\int I^2R,dt$ equals the initial $U$ if no energy leaves the system. Based on the scenario, explain how energy is conserved in the given circuit.
Energy is conserved because the capacitor’s voltage stays at $100,\text{V}$ while heat is produced.
Energy is conserved because $5.0\times10^{-3},\text{J}$ stored in the field becomes exactly thermal energy in $R$.
Energy is not conserved because the electric field energy vanishes without appearing in any other form.
Energy is conserved only if the resistor returns energy to the capacitor so $U$ stays constant.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy when a capacitor discharges through a resistor. Electric energy conservation involves understanding how energy stored in the electric field transforms completely into thermal energy through the discharge process. In this scenario, the capacitor stores 5.0×10⁻³ J in its electric field, and when connected to the resistor, this energy is converted to thermal energy as charge flows due to the potential difference. Choice A is correct because it identifies that the 5.0×10⁻³ J stored in the electric field becomes exactly that amount of thermal energy in the resistor, conserving total energy. Choice C is incorrect because it claims energy vanishes - the electric field energy doesn't disappear but transforms into thermal energy through Joule heating. To help students: Use the integral ∫I²R dt to show mathematically that total thermal energy equals initial stored energy. Emphasize that the electric field is the energy storage mechanism in capacitors, and this field energy is what gets converted.
An inductor $L=0.50,\text{H}$ and resistor $R=5.0,\Omega$ form a closed series loop with an ideal switch; a current $I_0=2.0,\text{A}$ is established and then the source is removed, leaving the $RL$ loop to decay. The magnetic energy initially stored is $U_{B0}=\tfrac12 L I_0^2=\tfrac12(0.50)(2.0^2)=1.0,\text{J}$. As current decays as $I(t)=I_0 e^{-tR/L}$, the inductor’s stored energy becomes $U_B(t)=\tfrac12 L I(t)^2=U_{B0} e^{-2tR/L}$. The induced emf drives charge through the resistor’s potential drop, so energy transfers from magnetic field energy to carrier kinetic energy briefly and then to thermal energy at rate $P_R=I^2R$. Energy conservation predicts $\int_0^\infty I^2R,dt=U_{B0}=1.0\text{ J}$. Based on the scenario, calculate the total energy change in the described system.
The system’s total energy change is $-1.0,\text{J}$ because the inductor loses energy during decay.
The system’s total energy change is $+1.0,\text{J}$ because induced emf creates additional energy.
The system’s total energy change is $+0.50,\text{J}$ because only half the magnetic energy dissipates.
The system’s total energy change is $0,\text{J}$ because magnetic energy converts to thermal energy in $R$.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy in RL circuits and magnetic energy transformation. Electric energy conservation involves understanding how energy stored in magnetic fields transforms into thermal energy when current decays through a resistor. In this scenario, the inductor initially stores 1.0 J of magnetic energy, and as current decays exponentially, this energy is converted to thermal energy in the resistor through the induced emf driving current. Choice B is correct because it recognizes that the total energy change is 0 J - the magnetic energy stored in the inductor's field is completely converted to thermal energy in the resistor, conserving total energy. Choice A is incorrect because it suggests energy is lost from the system rather than transformed within it - the inductor doesn't lose energy to the surroundings but transfers it to the resistor. To help students: Draw analogies between LC and RC circuits to show similar energy transformation patterns. Emphasize that induced emf opposes current change but still allows energy transfer from magnetic field to thermal form.
A $C=50,\mu\text{F}$ capacitor is charged to $V_0=20,\text{V}$ and then discharged through a $R=1.0,\text{k}\Omega$ resistor in an otherwise ideal closed circuit. As charge moves, electric potential energy decreases and becomes kinetic energy of charge carriers that is rapidly randomized into thermal energy in the resistor. The capacitor’s initial energy is $U_0=\tfrac12 C V_0^2=\tfrac12(50\times10^{-6})(20^2)=1.0\times10^{-2},\text{J}$. The discharge obeys $V(t)=V_0 e^{-t/RC}$ and $I(t)=\tfrac{V(t)}{R}$, so $U_C(t)=\tfrac12 C V(t)^2=U_0 e^{-2t/RC}$. At $t=RC=0.050,\text{s}$, $U_C=U_0 e^{-2}\approx1.35\times10^{-3},\text{J}$, meaning $8.65\times10^{-3},\text{J}$ has been converted to thermal energy in $R$ while total energy of the closed system remains constant. Based on the scenario, how does the energy stored in the capacitor change as it discharges?
It remains constant because energy conservation prevents any energy conversion in the circuit.
It increases because decreasing current implies increasing capacitor potential energy during discharge.
It decreases exponentially as $U_C(t)=U_0 e^{-2t/RC}$ while heating increases in the resistor.
It decreases linearly as $U_C(t)=U_0\left(1-\tfrac{t}{RC}\right)$ until it reaches zero.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy and exponential decay in RC circuits. Electric energy conservation involves understanding how stored capacitor energy transforms into thermal energy during discharge while total energy remains constant. In this scenario, the capacitor's stored energy decreases exponentially according to U_C(t) = U₀e^(-2t/RC) as charge flows through the resistor, converting potential energy to thermal energy. Choice A is correct because it accurately describes the exponential decay of capacitor energy and acknowledges that this energy becomes thermal energy in the resistor. Choice B is incorrect because it suggests linear decay, which doesn't match the exponential nature of RC circuit discharge governed by the differential equations. To help students: Use graphs to show the exponential decay of voltage, current, and energy in RC circuits. Emphasize that the factor of 2 in the energy decay exponent comes from squaring the voltage in the energy formula U = ½CV².
In a vacuum region between parallel plates, a particle with charge $q=+2.0,\mu\text{C}$ starts from rest at electric potential $V_A=+300,\text{V}$ and moves along the electric field to a point at $V_B=+100,\text{V}$, with no nonconservative forces. The electric potential energy change is $\Delta U = q\Delta V = q(V_B-V_A)= (2.0\times10^{-6})(100-300)=-4.0\times10^{-4},\text{J}$. Because the field does work, the particle’s kinetic energy increases by $\Delta K = -\Delta U = 4.0\times10^{-4},\text{J}$, so $K_f=4.0\times10^{-4},\text{J}$ and total mechanical energy $K+U$ remains constant for this closed system. The potential difference drives charge motion, converting potential energy into kinetic energy without loss. In the scenario, what happens to the potential energy as the charge moves through the field?
It increases by $4.0\times10^{-4},\text{J}$ because the charge moves to lower potential.
It decreases by $4.0\times10^{-4},\text{J}$ and the kinetic energy increases by the same amount.
It becomes zero because only kinetic energy exists once the charge starts moving.
It decreases, but the lost energy disappears since the electric field does no work.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy and work-energy theorem for charged particles. Electric energy conservation involves understanding how potential energy converts to kinetic energy as a charged particle moves through an electric field. In this scenario, the positive charge moves from higher to lower potential (300V to 100V), losing 4.0×10⁻⁴ J of potential energy which is converted entirely into kinetic energy. Choice B is correct because it accurately states that potential energy decreases by 4.0×10⁻⁴ J and this exact amount becomes kinetic energy, maintaining total mechanical energy conservation. Choice D is incorrect because it falsely claims energy disappears, violating conservation - the electric field does positive work on the charge, converting potential to kinetic energy. To help students: Use energy bar charts to visualize the transformation between potential and kinetic energy. Emphasize that for conservative forces like electric fields, ΔK = -ΔU always holds when no other forces do work.
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?
Potential difference causes energy loss, so $\Delta K< -\Delta U$ even without friction.
Potential difference sets the final speed directly, independent of charge and initial energy.
A negative $\Delta V$ implies $\Delta U>0$ and both $U$ and $K$ increase together.
A negative $\Delta V$ for a positive charge implies $\Delta U<0$ and $\Delta K=-\Delta U$.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy and the relationship between potential difference and energy changes. Electric energy conservation involves understanding how potential energy changes relate to kinetic energy changes when a charged particle moves through an electric field. In this scenario, the proton experiences a negative potential difference (-500V), causing its potential energy to decrease by 8.0×10⁻¹⁷ J, which becomes kinetic energy according to conservation of mechanical energy. Choice A is correct because it properly relates the signs: for a positive charge, negative ΔV means negative ΔU (potential energy decreases), and by conservation ΔK = -ΔU (kinetic energy increases). Choice B is incorrect because it claims both energies increase, which would violate conservation - total mechanical energy must remain constant in this conservative field. To help students: Use sign conventions carefully - positive charges naturally move toward lower potential, losing potential energy. Practice applying the work-energy theorem W = qΔV = ΔK to reinforce the energy transformation concept.
A $9.0,\text{V}$ ideal battery is connected to a $3.0,\Omega$ resistor using ideal wires, forming a closed circuit. Chemical energy in the battery becomes electrical energy that drives charge through a potential difference, and the moving charges transfer energy to the resistor’s lattice as thermal energy. The current is $I=\tfrac{V}{R}=\tfrac{9.0}{3.0}=3.0,\text{A}$. The electrical power delivered by the battery is $P_{\text{batt}}=IV= (3.0)(9.0)=27,\text{W}$, and the thermal power in the resistor is $P_R=I^2R=(3.0^2)(3.0)=27,\text{W}$. Over $t=10,\text{s}$, the battery’s chemical energy decrease is $E=Pt=270,\text{J}$ and the resistor gains $270,\text{J}$ of thermal energy, consistent with energy conservation for the closed system. Explain how energy is conserved in the given circuit.
Energy is conserved because $P_{\text{batt}}=P_R=27,\text{W}$, converting chemical energy to thermal energy.
Energy is not conserved because the resistor destroys $270,\text{J}$ of electrical energy over $10,\text{s}$.
Energy is conserved only if current is zero, since moving charges cannot transfer energy.
Energy is conserved because the battery’s voltage increases as it delivers energy to the resistor.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy in steady-state DC circuits. Electric energy conservation involves understanding how chemical energy from the battery transforms into electrical energy and then into thermal energy in the resistor. In this scenario, the battery delivers 27 W of electrical power, which exactly equals the 27 W of thermal power dissipated in the resistor, demonstrating perfect energy conservation. Choice A is correct because it identifies that power delivered equals power dissipated (both 27 W), showing energy conservation as chemical energy becomes thermal energy. Choice B is incorrect because it claims energy is destroyed rather than transformed - the 270 J of electrical energy becomes 270 J of thermal energy, not vanishing. To help students: Use power balance equations to show P_source = P_dissipated in steady-state circuits. Emphasize that resistors don't destroy energy but convert electrical energy into thermal energy through collisions between charge carriers and the lattice.
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?
It increases because $V$ decreases and $U=\tfrac12 C/V^2$ grows as discharge proceeds.
It stays equal to $U_0$ because the resistor only changes current, not energy storage.
It follows $U_C(t)=U_0 e^{-t/RC}$ because energy decays with the same exponent as voltage.
It follows $U_C(t)=U_0 e^{-2t/RC}$, decreasing as energy transfers to thermal energy in $R$.
Explanation
This question tests AP Physics C: Electricity and Magnetism concepts, specifically the conservation of electric energy and exponential decay in capacitor discharge. Electric energy conservation involves understanding how capacitor energy decreases exponentially while being converted to thermal energy in the resistor. In this scenario, the energy stored in the capacitor follows U_C(t) = U₀e^(-2t/RC), decreasing from 1.44×10⁻² J initially to 5.30×10⁻³ J after 0.030 s, with the difference becoming thermal energy. Choice A is correct because it gives the proper exponential decay formula with the factor of 2 in the exponent, which arises from squaring the exponentially decaying voltage in the energy formula. Choice B is incorrect because it omits the factor of 2 - since U = ½CV² and V decays as e^(-t/RC), the energy decays as e^(-2t/RC). To help students: Derive the energy decay formula step by step from the voltage decay to show where the factor of 2 originates. Plot both voltage and energy decay curves to visualize their different decay rates.