Electric Potential Energy - AP Physics C Electricity and Magnetism

1. Two point charges, $+2Q$ and $-Q$, are separated by a distance $d$. What is the change in electric potential energy of the system if the separation distance is decreased to $d/2$?
   Answer: The potential energy decreases by $k \frac{2Q^2}{d}$ The initial potential energy is $U_i = k \frac{(+2Q)(-Q)}{d} = -k \frac{2Q^2}{d}$. The final potential energy is $U_f = k \frac{(+2Q)(-Q)}{d/2} = -k \frac{4Q^2}{d}$. The change in potential energy is $\Delta U = U_f - U_i = \left(-k \frac{4Q^2}{d}\right) - \left(-k \frac{2Q^2}{d}\right) = -k \frac{2Q^2}{d}$. A decrease of $k \frac{2Q^2}{d}$.
2. System A consists of two protons separated by a distance $d$. System B consists of a proton and an electron separated by a distance $d$. System C consists of two electrons separated by a distance $d$. Let $U_A$, $U_B$, and $U_C$ be the electric potential energies. Which option correctly ranks the potential energies?
   Answer: $U_A = U_C > U_B$ Let the elementary charge be $e$. The potential energies are: $U_A = k \frac{(+e)(+e)}{d} = +k \frac{e^2}{d}$. $U_B = k \frac{(+e)(-e)}{d} = -k \frac{e^2}{d}$. $U_C = k \frac{(-e)(-e)}{d} = +k \frac{e^2}{d}$. Since $k, e, d$ are positive, $U_A$ and $U_C$ are equal and positive, while $U_B$ is negative. Therefore, the correct ranking from most positive to most negative is $U_A = U_C > U_B$.
3. The electric potential in a region of space is described by the function $V(x, y) = 4xy^2$, where $V$ is in volts and $x$ and $y$ are in meters. An external agent moves a proton (charge $e = 1.6 \times 10^{-19} \text{ C}$) from the origin ($0,0$) to the point ($2 \text{ m}, 1 \text{ m}$) at a constant velocity. What is the change in the electric potential energy of the proton?
   Answer: $1.28 \times 10^{-18} \text{ J}$ The change in electric potential energy is given by $\Delta U_E = q \Delta V = q(V_f - V_i)$. The initial potential at the origin ($0,0$) is $V_i = 4(0)(0)^2 = 0 \text{ V}$. The final potential at ($2 \text{ m}, 1 \text{ m}$) is $V_f = 4(2)(1)^2 = 8 \text{ V}$. The change in potential is $\Delta V = 8 \text{ V} - 0 \text{ V} = 8 \text{ V}$. The change in potential energy is $\Delta U_E = (1.6 \times 10^{-19} \text{ C})(8 \text{ V}) = 12.8 \times 10^{-19} \text{ J} = 1.28 \times 10^{-18} \text{ J}$.
4. Two point charges are held a fixed distance apart. If the magnitude of each charge is doubled, how does the electric potential energy of the system change?
   Answer: It is quadrupled. The electric potential energy of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = k \frac{q_1 q_2}{r}$. If the magnitude of each charge is doubled, the new charges are $2q_1$ and $2q_2$. The new potential energy $U'$ is $U' = k \frac{(2q_1)(2q_2)}{r} = 4 \left(k \frac{q_1 q_2}{r}\right) = 4U$. Thus, the potential energy is quadrupled.
5. An electron is accelerated from rest through a potential difference of $V$. It then enters a region of uniform electric field where it is brought to rest. The change in the electron's electric potential energy during the initial acceleration phase is $\Delta U_1$, and during the deceleration phase is $\Delta U_2$. Which of the following is correct?
   Answer: $\Delta U_1 < 0$, $\Delta U_2 > 0$, and $|\Delta U_1| = |\Delta U_2|$ During acceleration from rest, the electron's kinetic energy increases. By conservation of energy, its potential energy must decrease. So, $\Delta U_1 < 0$. The change is $\Delta U_1 = (-e)V$. During deceleration to rest, its kinetic energy decreases back to zero. Thus, its potential energy must increase, so $\Delta U_2 > 0$. By the work-energy theorem, the total change in kinetic energy is zero, so the net work done by the electric field is zero. This means the total change in potential energy is zero: $\Delta U_1 + \Delta U_2 = 0$, which implies $\Delta U_2 = -\Delta U_1$. Therefore, their magnitudes are equal.
6. Based on the scenario, calculate the change in electric potential energy for a $+1.5\,\mu\text{C}$ charge moved through a battery from 0 V to 12 V.
   Answer: $+1.8\times10^{-5}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge moved through a battery from 0 V to 12 V experiences a potential difference of ΔV = 12 - 0 = 12 V. The correct answer, A, is derived by applying ΔU = qΔV = (1.5×10⁻⁶ C)(12 V) = 1.8×10⁻⁵ J. A common distractor, B, arises from incorrect unit conversion, treating microCoulombs as Coulombs. To assist students, emphasize careful unit conversion (1 μC = 10⁻⁶ C) and practice dimensional analysis to ensure energy units (Joules) result from charge (Coulombs) times potential (Volts).
7. Three identical point charges, each with charge $+q$, are fixed at the vertices of an equilateral triangle with side length $s$. What is the total electric potential energy of this system of charges?
   Answer: $3k \frac{q^2}{s}$ The total electric potential energy of a system of charges is the scalar sum of the potential energies of all unique pairs of charges. In this system, there are three unique pairs of charges. Since all charges are $+q$ and all separation distances are $s$, the potential energy of each pair is $U_{pair} = k \frac{q \cdot q}{s} = k \frac{q^2}{s}$. The total potential energy is the sum for the three pairs: $U_{total} = 3 \times U_{pair} = 3k \frac{q^2}{s}$.
8. Based on the scenario, calculate the change in electric potential energy for a $+4.0\,\mu\text{C}$ charge moving from 120 V to 60 V in a circuit.
   Answer: $-2.4\times10^{-4}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge moving from 120 V to 60 V experiences a potential difference of ΔV = 60 - 120 = -60 V. The correct answer, A, is derived by applying ΔU = qΔV = (4.0×10⁻⁶ C)(-60 V) = -2.4×10⁻⁴ J. A common distractor, B, arises from incorrectly assigning the sign, thinking that moving to lower potential should increase energy. To assist students, emphasize that positive charges naturally move toward lower potentials, losing potential energy in the process, similar to objects falling in gravitational fields.
9. An isolated system consists of two particles with charges $+q$ and $+4q$ and masses $m$ and $4m$, respectively. They are released from rest when separated by a distance $d$. What is the ratio of the kinetic energy of charge $+q$ to the kinetic energy of charge $+4q$ long after their release?
   Answer: 4:1 The system is isolated, so momentum is conserved. Since they start from rest, the total momentum is always zero. Therefore, their momenta must be equal in magnitude and opposite in direction: $|p_1| = |p_2|$, so $mv_1 = (4m)v_2$, which gives $v_1 = 4v_2$. The kinetic energy is $K = \frac{1}{2}mv^2$. The ratio of their kinetic energies is $\frac{K_1}{K_2} = \frac{\frac{1}{2}m v_1^2}{\frac{1}{2}(4m)v_2^2} = \frac{1}{4} \left(\frac{v_1}{v_2}\right)^2$. Substituting $v_1 = 4v_2$, we get $\frac{K_1}{K_2} = \frac{1}{4} (4)^2 = \frac{16}{4} = 4$. The ratio is 4:1.
10. In the context of the electric field, calculate the change in electric potential energy for a $-1.0\,\mu\text{C}$ charge moving 0.40 m along a uniform $300\,\text{V/m}$ field.
    Answer: $+1.2\times10^{-4}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the negative charge moving 0.40 m along a 300 V/m field experiences a potential difference of ΔV = -Ed = -300 × 0.40 = -120 V. The correct answer, B, is derived by applying ΔU = qΔV = (-1.0×10⁻⁶ C)(-120 V) = 1.2×10⁻⁴ J. A common distractor, A, arises from forgetting that negative charges gain energy when moving along field lines (toward lower potential). To assist students, emphasize that negative charges behave oppositely to positive charges in terms of energy changes, gaining energy when moving in the field direction.
11. The work required to assemble a system of four identical positive charges $+q$ at the corners of a square of side $s$ is $W_A$. The work required to assemble a system of four identical positive charges $+q$ at the corners of a square of side $2s$ is $W_B$. Which statement correctly compares the work done?
    Answer: $W_A = 2 W_B$ The work to assemble a configuration is equal to its total electric potential energy. The potential energy is the sum of the potential energies of all pairs. For a square of side length $L$, there are 4 pairs at distance $L$ and 2 pairs at distance $L\sqrt{2}$. The total energy is $U = 4k\frac{q^2}{L} + 2k\frac{q^2}{L\sqrt{2}} = \frac{kq^2}{L}(4 + \sqrt{2})$. The work is proportional to $1/L$. Thus, $W_A \propto 1/s$ and $W_B \propto 1/(2s)$. This means $W_B = W_A / 2$, or $W_A = 2W_B$.
12. The electric potential energy of a system consisting of a point charge $q$ and a fixed point charge $Q$ is $U(r) = kQq/r$. Which of the following describes the work done by the electric field as the charge $q$ is moved from an initial distance $R$ to a final distance $3R$?
    Answer: $+k \frac{2Qq}{3R}$ The work done by the conservative electric field is the negative of the change in potential energy: $W_E = -\Delta U = -(U_f - U_i)$. The initial potential energy is $U_i = kQq/R$. The final potential energy is $U_f = kQq/(3R)$. The change in potential energy is $\Delta U = U_f - U_i = \frac{kQq}{3R} - \frac{kQq}{R} = \frac{kQq - 3kQq}{3R} = -\frac{2kQq}{3R}$. The work done by the field is $W_E = -(-\frac{2kQq}{3R}) = +\frac{2kQq}{3R}$.
13. In the context of the electric field, calculate the change in electric potential energy for a $+1.0\,\mu\text{C}$ charge moved 0.50 m across a uniform $200\,\text{V/m}$ field.
    Answer: $-1.0\times10^{-4}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the charge moved 0.50 m across a 200 V/m field, experiencing a potential difference of ΔV = -Ed = -200 × 0.50 = -100 V (negative because moving along field direction). The correct answer, B, is derived by applying ΔU = qΔV = (1.0×10⁻⁶ C)(-100 V) = -1.0×10⁻⁴ J. A common distractor, A, arises from sign confusion about field direction. To assist students, emphasize that "across" typically means along the field direction (high to low potential), and practice visualizing field lines and equipotential surfaces.
14. In the context of the electric field, calculate the change in electric potential energy for a $+2.5\,\mu\text{C}$ charge moving from 10 V to 6 V.
    Answer: $-1.0\times10^{-5}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge moving from 10 V to 6 V experiences a potential difference of ΔV = 6 - 10 = -4 V. The correct answer, A, is derived by applying ΔU = qΔV = (2.5×10⁻⁶ C)(-4 V) = -1.0×10⁻⁵ J. A common distractor, B, arises from sign confusion when potential decreases. To assist students, emphasize that positive charges lose energy when moving to lower potentials, and practice calculating potential differences as final minus initial values.
15. A system in a stable equilibrium configuration is disturbed by a small displacement. Which statement must be true about the system's electric potential energy and the work done by the conservative electric forces within the system?
    Answer: The potential energy increases, and the electric forces do negative work. A stable equilibrium corresponds to a local minimum in the potential energy function. When the system is slightly displaced from this minimum, its potential energy must increase. The work done by a conservative force, such as the electric force, is equal to the negative of the change in potential energy: $W_E = -\Delta U_E$. Since the potential energy increases ($\Delta U_E > 0$), the work done by the electric forces must be negative. This negative work corresponds to a restoring force that tends to bring the system back to equilibrium.
16. An external agent moves a $-3.0 \text{ C}$ point charge at constant velocity from a point with an electric potential of $+20 \text{ V}$ to a point with an electric potential of $-10 \text{ V}$. What is the work done on the charge by the external agent?
    Answer: $+90 \text{ J}$ The work done by an external agent to move a charge at constant velocity is equal to the change in the electric potential energy of the charge: $W_{ext} = \Delta U_E = q \Delta V = q(V_{final} - V_{initial})$. Here, $q = -3.0 \text{ C}$, $V_{initial} = +20 \text{ V}$, and $V_{final} = -10 \text{ V}$. So, $W_{ext} = (-3.0 \text{ C})(-10 \text{ V} - 20 \text{ V}) = (-3.0 \text{ C})(-30 \text{ V}) = +90 \text{ J}$.
17. Based on the scenario, calculate the change in electric potential energy for a $+6.0\,\mu\text{C}$ charge moved through a $-8\,\text{V}$ potential difference.
    Answer: $-4.8\times10^{-5}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the positive charge experiences a potential difference of ΔV = -8 V (given directly). The correct answer, A, is derived by applying ΔU = qΔV = (6.0×10⁻⁶ C)(-8 V) = -4.8×10⁻⁵ J. A common distractor, B, arises from misunderstanding that a negative potential difference means the charge moves to a lower potential. To assist students, emphasize that potential difference signs indicate direction of change, and positive charges lose energy when experiencing negative potential differences.
18. Three point charges $+Q$, $-Q$, and $+2Q$ are placed at the vertices of an equilateral triangle of side length $s$. An external agent disassembles the system, moving each charge to an infinite distance from the others. What is the total work done by the external agent?
    Answer: $k \frac{Q^2}{s}$ The work done by an external agent to disassemble the system is $W_{ext} = U_{final} - U_{initial}$. The final potential energy, with all charges at infinite separation, is $U_{final} = 0$. The initial potential energy is the sum of the energies of the three pairs: $U_{initial} = k\frac{(+Q)(-Q)}{s} + k\frac{(+Q)(+2Q)}{s} + k\frac{(-Q)(+2Q)}{s} = k\frac{-Q^2+2Q^2-2Q^2}{s} = -k\frac{Q^2}{s}$. The work done is $W_{ext} = 0 - \left(-k\frac{Q^2}{s}\right) = +k\frac{Q^2}{s}$.
19. A proton is released from rest in a region of space with a non-uniform electric field. As the proton moves from point A to point B, its kinetic energy increases by $5.0 \times 10^{-16} \text{ J}$. What is the change in the electric potential energy of the proton-field system?
    Answer: $-5.0 \times 10^{-16} \text{ J}$ According to the principle of conservation of energy for the proton-field system, the total energy remains constant. Since only the conservative electric force does work, the sum of the changes in kinetic energy ($\Delta K$) and electric potential energy ($\Delta U_E$) must be zero. $\Delta K + \Delta U_E = 0$. Given that the kinetic energy increases by $5.0 \times 10^{-16} \text{ J}$, we have $\Delta K = +5.0 \times 10^{-16} \text{ J}$. Therefore, $\Delta U_E = -\Delta K = -5.0 \times 10^{-16} \text{ J}$.
20. Based on the scenario, calculate the change in electric potential energy for a $+2.0\,\mu\text{C}$ charge moving 0.30 m along a uniform $400\,\text{V/m}$ field between capacitor plates.
    Answer: $-2.4\times10^{-4}\,\text{J}$ This question tests AP Physics C skills in understanding electric potential energy within electric fields. Electric potential energy (U) is the energy a charge has due to its position in an electric field, calculated by U = qV. In this scenario, the charge moving 0.30 m along a 400 V/m field experiences a potential difference of ΔV = -Ed = -400 × 0.30 = -120 V (negative because moving along the field direction decreases potential). The correct answer, D, is derived by applying ΔU = qΔV = (2.0×10⁻⁶ C)(-120 V) = -2.4×10⁻⁴ J. A common distractor, A, arises from forgetting the negative sign when moving along the field direction. To assist students, emphasize that positive charges lose potential energy when moving along electric field lines, and practice identifying the sign of potential changes based on movement direction relative to the field.

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