AP Physics 2 Quiz: Conservation Of Electric Energy

What this quiz covers

This quiz focuses on Conservation Of Electric Energy, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It stays the same because total charge is conserved when the capacitors are connected.
2. It decreases because some energy must leave the capacitor system during charge redistribution. (correct answer)
3. It increases because the equivalent capacitance doubles when connected in parallel.
4. It becomes equal to the battery's power output because the battery set the initial voltage.

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.

Question 2

1. It increases by a factor of 3 because capacitance increases at constant charge.
2. It decreases by a factor of 3 because $U=\dfrac{Q^2}{2C}$ with constant $Q$. (correct answer)
3. It stays the same because energy cannot change without a battery present.
4. It becomes zero because the electric field must vanish when plates move closer.

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.

Question 3

1. $+10 \text{ J}$ (correct answer)
2. $-10 \text{ J}$
3. Zero, because the speed is constant.
4. The change cannot be determined without knowing the value of the charge.

Explanation: 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, $\Delta U_E = -(-10 \text{ J}) = +10 \text{ J}$.

Question 4

1. It decreases because increasing plate separation weakens the electric field.
2. It increases because $U=\dfrac{Q^2}{2C}$ and $C$ decreases while $Q$ stays constant. (correct answer)
3. It stays the same because the battery was disconnected so no energy can change.
4. 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.

Question 5

1. $\sqrt{\frac{kQq}{mR}}$
2. $\sqrt{\frac{2kQq}{mR}}$ (correct answer)
3. $\sqrt{\frac{kQq}{2mR}}$
4. $\frac{kQq}{mR}$

Explanation: 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.

Question 6

1. $\frac{mv^2}{2q}$ (correct answer)
2. $\frac{mv}{2q}$
3. $\frac{2q}{mv^2}$
4. $\frac{2q}{mv}$

Explanation: To bring the particle to rest, the change in its kinetic energy must be $\Delta K = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}mv^2$. By the work-energy theorem for conservative forces, $\Delta K = -\Delta U_E$. The change in potential energy is $\Delta U_E = q \Delta V$. Therefore, $-\frac{1}{2}mv^2 = -q \Delta V$. Solving for the magnitude of the potential difference $|\Delta V|$ gives $|\Delta V| = \frac{mv^2}{2q}$

Question 7

1. The proton, because it has a greater mass.
2. The electron, because it has a smaller mass and thus greater acceleration.
3. They both have the same kinetic energy. (correct answer)
4. It depends on the separation distance $d$ between the plates.

Explanation: 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.

Question 8

1. $\Delta U_1 > \Delta U_2$ because Path 2 is longer.
2. $\Delta U_1 < \Delta U_2$ because the force is averaged over a longer distance.
3. $\Delta U_1 = \Delta U_2$ because the electrostatic force is a conservative force. (correct answer)
4. The relationship cannot be determined without knowing the exact shape of Path 2.

Explanation: The electrostatic force is a conservative force. This means the work done by the force, and therefore the change in potential energy, depends only on the initial and final positions of the particle, not on the path taken between them. Since both paths start at A and end at B, the change in electric potential energy is the same for both paths: $\Delta U_1 = \Delta U_2$.

Question 9

1. $C_1$ stores three times the energy of $C_2$ because it has less charge.
2. They store equal energy because they are in parallel.
3. Neither stores energy because the battery provides constant power, not energy.
4. $C_2$ stores three times the energy of $C_1$ because both have the same voltage. (correct answer)

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.

Question 10

1. $eV$ (correct answer)
2. $-eV$
3. Zero, as the energy is stored as potential energy.
4. It depends on the distance between the plates.

Explanation: 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 $eV$.

Question 11

1. It increases by $1.6 \times 10^{-17} \text{ J}$. (correct answer)
2. It decreases by $1.6 \times 10^{-17} \text{ J}$.
3. It increases by $100 \text{ J}$.
4. It decreases by $100 \text{ J}$.

Explanation: 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.

Question 12

1. The kinetic energy increases because the electric field does positive work on the proton.
2. The kinetic energy decreases because the proton's electric potential energy increases. (correct answer)
3. The kinetic energy remains constant because the electric field is conservative.
4. The change in kinetic energy cannot be determined without knowing the path taken.

Explanation: For a positive charge like a proton, electric potential energy is given by $U_E = qV$. As the proton moves to a region of increasing electric potential ($V$ increases), its electric potential energy $U_E$ also increases. By the law of conservation of energy, the total energy ($K + U_E$) must remain constant. Therefore, if $U_E$ increases, the kinetic energy $K$ must decrease.

Question 13

1. $K_e > K_p$ and $v_e > v_p$
2. $K_e < K_p$ and $v_e < v_p$
3. $K_e = K_p$ and $v_e > v_p$ (correct answer)
4. $K_e = K_p$ and $v_e < v_p$

Explanation: 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 $v_e > v_p$

Question 14

1. $+60 \text{ J}$ (correct answer)
2. $-60 \text{ J}$
3. $+140 \text{ J}$
4. $-140 \text{ J}$

Explanation: 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 $W_E = -(-60 \text{ J}) = +60 \text{ J}$.

Question 15

1. $r_{min}/4$ (correct answer)
2. $r_{min}/2$
3. $2r_{min}$
4. $4r_{min}$

Explanation: 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 $r_{new} = r_{min}/4$.

Question 16

1. $\sqrt{\frac{2q(V_B - V_A)}{m}}$
2. $\sqrt{\frac{2q(V_A - V_B)}{m}}$ (correct answer)
3. $\frac{2q(V_A - V_B)}{m}$
4. $\frac{q(V_A - V_B)}{2m}$

Explanation: 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.

Question 17

1. The total kinetic energy increases, and the total electric potential energy increases.
2. The total kinetic energy decreases, and the total electric potential energy increases.
3. The total kinetic energy increases, and the total electric potential energy decreases. (correct answer)
4. The total kinetic energy decreases, and the total electric potential energy decreases.

Explanation: The system is isolated, so its total energy is conserved. The total energy is the sum of the total kinetic energy and the total electric potential energy. Initially, the charges are at rest, so the kinetic energy is zero, and the potential energy is positive due to repulsion. As they fly apart, their speeds increase, so the total kinetic energy increases. Because total energy must be conserved, the increase in kinetic energy must be accompanied by a decrease in electric potential energy.

Question 18

1. $120 \text{ eV}$ (correct answer)
2. $160 \text{ eV}$
3. $200 \text{ eV}$
4. $400 \text{ eV}$

Explanation: The initial potential at $x_i = 4.0\text{ m}$ is $V_i = (10)(4.0)^2 = 160 \text{ V}$. The final potential at $x_f = 2.0\text{ m}$ is $V_f = (10)(2.0)^2 = 40 \text{ V}$. The change in potential is $\Delta V = V_f - V_i = 40 \text{ V} - 160 \text{ V} = -120 \text{ V}$. The change in kinetic energy is $\Delta K = -\Delta U_E = -q \Delta V$. For a proton, $q = +e$. So, $\Delta K = -(+e)(-120 \text{ V}) = +120 \text{ eV}$. Since it started from rest, its final kinetic energy is $120 \text{ eV}$.

Question 19

1. $K_2 = 4K_1$
2. $K_2 = 2K_1$ (correct answer)
3. $K_2 = K_1$
4. $K_2 = K_1/2$

Explanation: 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: $K_2 = 2K_1$.

Question 20

1. It will be trapped in the well, oscillating back and forth. (correct answer)
2. It will escape from the well with a constant velocity.
3. It will move to the point of lowest potential energy and come to rest.
4. Its total energy will gradually decrease until it stops.

Explanation: 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.