Conservation of Electric Energy
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AP Physics 2 › Conservation of Electric Energy
A $1.5,\text{V}$ ideal battery charges a capacitor $C=2.0,\mu\text{F}$, then is removed. Define the system as the capacitor only. The capacitor is then connected to an identical uncharged capacitor in parallel. Which statement correctly describes what happens to the electric potential energy stored in the capacitors?
It stays the same because total charge is conserved when the capacitors are connected.
It becomes equal to the battery’s power output because the battery set the initial voltage.
It decreases because some energy must leave the capacitor system during charge redistribution.
It increases because the equivalent capacitance doubles when connected in parallel.
Explanation
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.
A capacitor is charged to voltage $V_0$ and then disconnected from any battery. Define the system as capacitor + external agent. The agent slowly decreases the plate separation, increasing the capacitance by a factor of 3 while keeping charge on the plates constant. Which statement correctly describes the capacitor’s stored energy?
It increases by a factor of 3 because capacitance increases at constant charge.
It stays the same because energy cannot change without a battery present.
It becomes zero because the electric field must vanish when plates move closer.
It decreases by a factor of 3 because $U=\dfrac{Q^2}{2C}$ with constant $Q$.
Explanation
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.
A capacitor $C=2.0,\mu\text{F}$ is charged to $V=8.0,\text{V}$ by a battery and then disconnected. The plates are pulled farther apart so the capacitance becomes $1.0,\mu\text{F}$ while charge cannot leave. The system is the capacitor plus field. Which statement correctly describes the stored energy?
It stays the same because the battery was disconnected so no energy can change.
It decreases because increasing plate separation weakens the electric field.
It increases because $U=\dfrac{Q^2}{2C}$ and $C$ decreases while $Q$ stays constant.
It becomes zero because the voltage must remain fixed at $8.0,\text{V}$.
Explanation
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.
A $1.5\ \text{V}$ ideal battery is connected to a parallel combination of $C_1=2.0\ \mu\text{F}$ and $C_2=6.0\ \mu\text{F}$ until steady state. System boundary: both capacitors only. Which statement correctly describes the energy distribution between the capacitors?
They store equal energy because they are in parallel.
$C_1$ stores three times the energy of $C_2$ because it has less charge.
Neither stores energy because the battery provides constant power, not energy.
$C_2$ stores three times the energy of $C_1$ because both have the same voltage.
Explanation
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.
A capacitor $C=4.0\ \mu\text{F}$ is charged to $V_0=10\ \text{V}$, then disconnected from the battery. It is then connected in parallel to an identical uncharged capacitor. System boundary: both capacitors + connecting wires. Which conclusion follows from energy conservation for the final stored energy?
The final total stored energy is less than the initial, with the difference becoming thermal energy.
The final total stored energy is zero because the final voltage is zero.
The final total stored energy is greater than the initial because charge spreads out.
The final total stored energy equals the initial because no battery is present.
Explanation
This problem tests conservation of electric energy. Initially, the charged capacitor stores energy U₁ = ½CV₀² = ½(4.0×10⁻⁶)(10)² = 200 μJ. When connected to an identical uncharged capacitor, charge redistributes until both reach the same voltage V = V₀/2 = 5 V (since Q = CV₀ splits equally between 2C total). The final energy is U_f = 2×(½CV²) = CV²/2 = ½(4.0×10⁻⁶)(25) = 100 μJ, which is half the initial energy. The missing 100 μJ becomes thermal energy in the connecting wires during the rapid charge redistribution. Choice C incorrectly assumes energy is conserved just because no battery is present, forgetting that charge motion through resistance dissipates energy. The strategy is to calculate initial and final stored energies, then attribute any decrease to thermal dissipation.
An initially uncharged capacitor $C$ is connected to an ideal battery of emf $V$ and reaches steady state. System boundary: battery + capacitor + wires. Which statement correctly relates the battery’s energy decrease to the capacitor’s stored energy?
The battery’s energy decrease is zero because current stops at steady state.
The battery’s energy decrease equals the capacitor’s charge divided by time.
The battery’s energy decrease equals the capacitor’s stored energy $\tfrac12 CV^2$.
The battery’s energy decrease equals $CV^2$, twice the capacitor’s stored energy, with the rest becoming thermal.
Explanation
This problem tests conservation of electric energy. When charging a capacitor from zero to voltage V, the battery moves charge Q = CV, doing work W = QV = CV². However, the capacitor only stores energy U = ½CV², which is half the battery's work. The other half (½CV²) becomes thermal energy in the wire resistance during charging. This 50-50 split is fundamental to RC charging: half the battery's energy always becomes heat regardless of resistance value. Choice A incorrectly assumes all battery energy goes to the capacitor, missing the inevitable thermal dissipation. The key principle is that charging a capacitor through any finite resistance converts exactly half the battery's energy to heat.
A capacitor $C=8.0,\mu\text{F}$ is charged to $V_0=12,\text{V}$ and then disconnected from the battery. It is then connected in parallel (like plates together) to an uncharged capacitor $2C$. Define the system as both capacitors + wires (ignore heating). Compared to the initial energy, the final total stored energy is
smaller, because energy must transfer out of the system during redistribution.
zero, because the charged capacitor fully discharges into the uncharged one.
larger, because adding a capacitor increases equivalent capacitance.
the same, because total charge is conserved in the system.
Explanation
This problem tests conservation of electric energy. Initially, the 8.0μF capacitor stores U₁ = ½(8.0μF)(12V)² = 576μJ. When connected to the uncharged 16μF capacitor, charge redistributes until both reach the same voltage. Total charge Q = (8.0μF)(12V) = 96μC is conserved, and the final voltage is Vf = Q/Ctotal = 96μC/24μF = 4V. The final energy is Uf = ½(24μF)(4V)² = 192μJ, which is less than the initial 576μJ. Choice B incorrectly assumes charge conservation implies energy conservation, missing that energy is dissipated during charge redistribution. Choice D wrongly thinks all energy disappears, not recognizing that some energy remains stored at the new equilibrium. The strategy is to calculate initial and final energies separately, expecting energy loss whenever charge redistributes between capacitors at different voltages.
A $9.0,\text{V}$ ideal battery is connected to two capacitors in parallel, $C_1=2.0,\mu\text{F}$ and $C_2=6.0,\mu\text{F}$, until fully charged, then disconnected. Define the system as the two capacitors only. Which statement correctly compares the energies stored in the two capacitors?
$U_1=U_2$ because both have the same voltage in parallel.
$U_1<U_2$ because the larger capacitor draws more power while charging.
$U_1<U_2$ because both have the same voltage and $U=\tfrac12CV^2$.
$U_1>U_2$ because the smaller capacitor must store more energy.
Explanation
This problem tests conservation of electric energy. When capacitors are connected in parallel to a battery, they reach the same voltage V = 9.0V, but store different amounts of energy based on their capacitance. The energy stored in each capacitor is U = ½CV², so U₁ = ½(2.0μF)(9.0V)² = 81μJ and U₂ = ½(6.0μF)(9.0V)² = 243μJ, making U₁ < U₂. Choice B incorrectly assumes equal voltage means equal energy, missing that energy depends on both voltage and capacitance. Choice C reverses the relationship, wrongly thinking smaller capacitors store more energy. The key strategy is to remember that for capacitors at the same voltage, energy is proportional to capacitance: larger C means more stored energy.
A $3.0,\text{V}$ ideal battery is connected across two capacitors in series, $C_1=2.0,\mu\text{F}$ and $C_2=6.0,\mu\text{F}$, until fully charged. Define the system as the two capacitors only (battery excluded). Which statement correctly compares the energies stored in $C_1$ and $C_2$?
$U_1<U_2$ because the larger capacitor must store more energy in series.
$U_1=U_2$ because capacitors in series share the same voltage.
$U_1>U_2$ because both carry the same charge and $U=\dfrac{Q^2}{2C}$.
$U_1>U_2$ because the smaller capacitor receives more power during charging.
Explanation
This problem tests conservation of electric energy. In a series connection, capacitors carry the same charge Q but have different voltages inversely proportional to their capacitance. The total charge is Q = CₑqV = (1.5μF)(3.0V) = 4.5μC, where Cₑq = C₁C₂/(C₁+C₂) = 1.5μF. Using U = Q²/2C, we get U₁ = (4.5μC)²/(2×2.0μF) = 5.06μJ and U₂ = (4.5μC)²/(2×6.0μF) = 1.69μJ, so U₁ > U₂. Choice B incorrectly claims series capacitors share the same voltage (they share the same charge). Choice C reverses the relationship, missing that at constant charge, smaller capacitance means larger energy. The strategy is to remember that in series, equal charge means energy is inversely proportional to capacitance.