Conservation of Electric Energy - AP Physics 2

1. 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
   Answer: smaller, because energy must transfer out of the system during redistribution. 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.
2. A proton and an electron are placed at rest at the exact midpoint between two large, parallel, oppositely charged conducting plates. The plates are separated by a distance $d$. Which particle has the greater kinetic energy upon striking a plate?
   Answer: They both have the same kinetic energy. 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.
3. An alpha particle, which has a charge of $+2e$, is accelerated from rest through an electric potential difference of $500 \text{ V}$. What is the final kinetic energy of the alpha particle?
   Answer: $1000 \text{ eV}$ 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 $1 \text{ V}$.
4. A particle with a positive charge moves from a region of low electric potential to a region of high electric potential. If the electric force is the only force acting on the particle, which statement correctly describes the changes in its energy?
   Answer: Its kinetic energy decreases, and the system's electric potential energy increases. The electric potential energy of a positive charge $q$ is given by $U_E = qV$. Since $q$ is positive, moving to a region of higher potential ($V$ increases) means its potential energy $U_E$ must increase. Because the electric force is conservative and is the only force acting, the total mechanical energy ($K + U_E$) is conserved. Therefore, if the potential energy increases, the kinetic energy must decrease by an equal amount.
5. 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?
   Answer: The final total stored energy is less than the initial, with the difference becoming thermal energy. 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.
6. An electron is released from rest in a region of space where there is a uniform electric field. It moves through a potential difference of $+100 \text{ V}$. What is the change in the kinetic energy of the electron?
   Answer: It increases by $1.6 \times 10^{-17} \text{ J}$. According to the conservation of energy, the change in kinetic energy is equal to the negative of the change in electric potential energy, $\Delta K = -\Delta U_E$. The change in potential energy is given by $\Delta U_E = q \Delta V$. For an electron, $q = -e = -1.6 \times 10^{-19} \text{ C}$. Thus, $\Delta U_E = (-1.6 \times 10^{-19} \text{ C})(+100 \text{ V}) = -1.6 \times 10^{-17} \text{ J}$. The change in kinetic energy is then $\Delta K = -(-1.6 \times 10^{-17} \text{ J}) = +1.6 \times 10^{-17} \text{ J}$. The positive sign indicates an increase in kinetic energy.
7. Two identical positive charges are released from rest at the same location in a uniform electric field. Charge 1 travels a distance $d_1$. Charge 2 travels a distance $d_2 = 2d_1$ along the same line. How does the kinetic energy gained by charge 2, $K_2$, compare to the kinetic energy gained by charge 1, $K_1$?
   Answer: $K_2 = 2K_1$ The kinetic energy gained is equal to the work done by the electric field, $K = W_E = F_E d$. In a uniform electric field, the force $F_E = qE$ is constant. Since the charges are identical, the force on each is the same. The work done is directly proportional to the distance traveled, $d$. Since charge 2 travels twice the distance of charge 1 ($d_2 = 2d_1$), the work done on it is twice as large, and it gains twice the kinetic energy: $K_2 = 2K_1$.
8. A charged particle moves with constant velocity along a path that is always perpendicular to the electric field lines in a certain region. What is the work done by the electric field on the particle?
   Answer: Zero. A path perpendicular to the electric field lines is an equipotential path, meaning the electric potential $V$ is constant along the path. The change in potential is $\Delta V = 0$. The work done by the electric field is $W_E = -\Delta U_E = -q \Delta V$. Since $\Delta V = 0$, the work done by the electric field is zero. This is consistent with the particle moving at constant velocity, as the electric force does no work to change its kinetic energy.
9. An electron and a proton are both accelerated from rest through the same potential difference, $|\Delta V|$. Which of the following correctly compares the final kinetic energy $K$ and final speed $v$ of the two particles?
   Answer: $K_e = K_p$ and $v_e > v_p$ The change in kinetic energy is given by $\Delta K = -q \Delta V$. Since the electron and proton have charges of the same magnitude (e) and are accelerated through the same potential difference, the magnitude of their change in kinetic energy will be the same: $|\Delta K| = e|\Delta V|$. Thus, their final kinetic energies are equal, $K_e = K_p$. Since kinetic energy is $K = \frac{1}{2}mv^2$, the particle with the smaller mass will have the greater final speed. The electron's mass is much smaller than the proton's mass, so $v_e > v_p$
10. An electron is moving in a region of space where the electric potential creates a 'potential well,' which is a local minimum of potential energy. If the electron's total energy is less than the potential energy of the surrounding regions ('hills'), what is the most likely behavior of the electron?
    Answer: It will be trapped in the well, oscillating back and forth. This situation is analogous to a ball rolling in a valley. The electron's total energy is conserved. Since its total energy is less than the potential energy required to 'climb the hills,' it cannot escape the well. It will move back and forth, with its kinetic energy converting into potential energy as it moves away from the minimum, and potential energy converting back into kinetic energy as it moves toward the minimum. This constitutes oscillatory motion.
11. If the experiment is repeated with a new particle of the same mass and charge, but with an initial speed of $2v_i$, what is the new distance of closest approach?
    Answer: $r_{min}/4$ By conservation of energy, the initial kinetic energy is converted into electric potential energy at the closest approach. $\frac{1}{2}mv_i^2 = k\frac{Qq}{r_{min}}$. When the initial speed is doubled, the initial kinetic energy becomes $\frac{1}{2}m(2v_i)^2 = 4 \times \frac{1}{2}mv_i^2$. For this quadrupled energy to equal the potential energy at closest approach, we need $4 \times k\frac{Qq}{r_{min}} = k\frac{Qq}{r_{new}}$, which gives $r_{new} = r_{min}/4$.
12. A charge of $+2.0 \text{ C}$ is moved from a point with an electric potential of $50 \text{ V}$ to a point with an electric potential of $20 \text{ V}$. What is the work done by the electric field on the charge?
    Answer: $+60 \text{ J}$ The work done by the electric field is equal to the negative of the change in electric potential energy: $W_E = -\Delta U_E$. The change in potential energy is $\Delta U_E = q \Delta V = q(V_f - V_i)$. Here, $q = +2.0 \text{ C}$, $V_i = 50 \text{ V}$, and $V_f = 20 \text{ V}$. So, $\Delta U_E = (2.0 \text{ C})(20 \text{ V} - 50 \text{ V}) = -60 \text{ J}$. Therefore, the work done by the field is $W_E = -(-60 \text{ J}) = +60 \text{ J}$.
13. 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?
    Answer: The battery's energy decrease equals $CV^2$, twice the capacitor's stored energy, with the rest becoming thermal. 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.
14. A particle of charge $+q$ and mass $m$ is released from rest at a point A where the electric potential is $V_A$. It then accelerates to a point B where the potential is $V_B$. Which expression represents the speed of the particle at point B?
    Answer: $\sqrt{\frac{2q(V_A - V_B)}{m}}$ From the conservation of energy, the gain in kinetic energy equals the loss in potential energy. $K_f - K_i = -(U_{E,f} - U_{E,i})$. Since the particle starts from rest, $K_i = 0$. So, $\frac{1}{2}mv_f^2 = -(qV_B - qV_A) = q(V_A - V_B)$. Solving for the final speed $v_f$ gives $v_f = \sqrt{\frac{2q(V_A - V_B)}{m}}$. Note that for a positive charge to accelerate from rest, it must move to a region of lower potential, so $V_A > V_B$, making the term under the square root positive.
15. 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?
    Answer: It decreases by a factor of 3 because $U=\dfrac{Q^2}{2C}$ with constant $Q$. 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.
16. A small particle with charge $-q$ and mass $m$ is in a circular orbit of radius $R$ around a large, fixed particle with charge $+Q$. What is the minimum speed the small particle must be given to escape to an infinite distance away?
    Answer: $\sqrt{\frac{2kQq}{mR}}$ 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.
17. 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?
    Answer: $C_2$ stores three times the energy of $C_1$ because both have the same voltage. 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.
18. An electron is released from rest very close to the negative plate of a parallel-plate capacitor with a potential difference of $V$ between the plates. What is the kinetic energy of the electron just as it reaches the positive plate?
    Answer: $eV$ The electron moves from the negative plate (lower potential) to the positive plate (higher potential). The potential difference it moves through is $+\Delta V$. The charge of the electron is $q=-e$. The change in kinetic energy is $\Delta K = -\Delta U_E = -q \Delta V = -(-e)(V) = eV$. Since it starts from rest, its final kinetic energy is $eV$.
19. An external agent performs $+10 \text{ J}$ of work to move a charge at a constant speed from point X to point Y in an electric field. What is the change in the electric potential energy of the charge-field system?
    Answer: $+10 \text{ J}$ 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, $\Delta U_E = -(-10 \text{ J}) = +10 \text{ J}$.
20. 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 $C_2$?
    Answer: $U_1>U_2$ because both carry the same charge and $U=\dfrac{Q^2}{2C}$. 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.

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