Electric Potential Energy - AP Physics 2

1. A $-2.5\,\mu\text{C}$ charge moves from G at $V_G=-100\,\text{V}$ to H at $V_H=0\,\text{V}$. As the charge moves from G to H, the electric potential energy will:
   Answer: decrease because the charge is negative and the potential increases This problem examines electric potential energy when a negative charge moves to zero potential. Using U = qV, the initial energy is $U_G$ = (-2.5 μC)(-100 V) = +250 μJ, and the final energy is $U_H$ = (-2.5 μC)(0 V) = 0 μJ. Since the energy changes from +250 μJ to 0 μJ, the potential energy decreases by 250 μJ. Choice D incorrectly assumes that V = 0 means U = 0 for all charges, forgetting that U = qV depends on both charge and potential. When dealing with zero potential, remember it's just a reference point; the actual energy U = qV can be positive, negative, or zero depending on the charge.
2. A $+3.0\,\mu\text{C}$ charge moves from point $S$ at $V_S=+8\,\text{V}$ to point $T$ at $V_T=-2\,\text{V}$. The charge is positive, and the final potential is lower than the initial potential. As the charge moves from $S$ to $T$, the electric potential energy _____?
   Answer: decreases because the charge moves to a lower electric potential. Electric potential energy. A positive charge (+3.0 μC) moving from $V_S$ = +8 V to $V_T$ = -2 V experiences a decrease in potential of 10 V. The change in potential energy is ΔU = q($V_f$ - $V_i$) = (+3.0 μC)(-2 V - 8 V) = (+3.0 μC)(-10 V) = -30 μJ. The negative result shows that potential energy decreases as the positive charge moves to lower potential, consistent with the charge moving in the direction the field would naturally push it. Choice A incorrectly states that energy increases, missing that positive charges lose potential energy when moving to lower potentials. Track both charge sign and potential when reasoning about electric potential energy.
3. A charge $q=-6.0\,\mu\text{C}$ is moved from point 1 at $V_1=+15\,\text{V}$ to point 2 at $V_2=-5\,\text{V}$. As the charge moves from 1 to 2, the electric potential energy _____.
   Answer: increases because the charge moves to a lower electric potential This question tests understanding of electric potential energy. Electric potential energy is U = qV, where q is the charge and V is the potential. For a negative charge (-6.0 μC) moving from $V_1$ = +15 V to $V_2$ = -5 V, we find ΔU = q($V_2$ - $V_1$) = (-6.0 μC)(-5 V - 15 V) = (-6.0 μC)(-20 V) = +120 μJ. Since ΔU is positive, the potential energy increases as the negative charge moves to lower potential. Choice C incorrectly focuses on the sign difference between potentials rather than calculating the actual energy change. To determine potential energy changes, always use ΔU = qΔV and check the resulting sign.
4. A $-6.0\,\mu\text{C}$ charge is moved from point $Q$ at $V_Q=+50\,\text{V}$ to point $R$ at $V_R=+20\,\text{V}$. The charge is negative, and the final potential is lower than the initial potential. As the charge moves from $Q$ to $R$, the electric potential energy _____?
   Answer: increases because the charge moves to a lower electric potential. Electric potential energy. For a negative charge (-6.0 μC) moving from $V_Q$ = +50 V to $V_R$ = +20 V, the potential decreases by 30 V. The change in potential energy is ΔU = q($V_f$ - $V_i$) = (-6.0 μC)(+20 V - 50 V) = (-6.0 μC)(-30 V) = +180 μJ. The positive change indicates an increase in potential energy because negative charges have higher potential energy at lower positive potentials (less negative U = qV values). Choice A incorrectly predicts a decrease, failing to account for the negative sign of the charge. Track both charge sign and potential when reasoning about electric potential energy.
5. A charge $q=-8.0\,\text{nC}$ moves from point J at $V_J=30\,\text{V}$ to point K at $V_K=90\,\text{V}$. As the charge moves, the electric potential energy ____.
   Answer: decreases because $\Delta U=q\Delta V<0$ Electric potential energy. The electric potential energy of a charge q at potential V is U = qV, and energy changes follow ΔU = qΔV. For this negative charge (-8.0 nC) moving from 30 V to 90 V, the potential increases by ΔV = 90 V - 30 V = +60 V. With a negative charge and positive ΔV, we calculate ΔU = (-8.0 nC)(+60 V) = -480 nJ, which is negative, indicating a decrease in potential energy. Choice A incorrectly assumes that positive ΔV always means positive ΔU, ignoring that the sign of the charge reverses this relationship for negative charges. Track both charge sign and potential when reasoning about electric potential energy.
6. A particle with charge $+2.0\,\mu\text{C}$ moves from point A at $V_A=+30\,\text{V}$ to point B at $V_B=-10\,\text{V}$ in an external electric field. Ignore nonconservative forces. As the charge moves from A to B, the electric potential energy of the charge-field system _____.
   Answer: decreases because the charge moves to a lower electric potential This question tests understanding of electric potential energy. Electric potential energy is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC) moving from $V_A$ = +30 V to $V_B$ = -10 V, we calculate the change: ΔU = q($V_B$ - $V_A$) = (+2.0 μC)(-10 V - 30 V) = (+2.0 μC)(-40 V) = -80 μJ. Since ΔU is negative, the potential energy decreases. Choice C incorrectly assumes that only electric field strength matters, ignoring that potential energy depends on both charge and potential difference. When solving potential energy problems, always calculate ΔU = qΔV and check the sign to determine if energy increases or decreases.
7. A $+0.50\,\mu\text{C}$ charge moves from point $N$ at $V_N=-5\,\text{V}$ to point $P$ at $V_P=+15\,\text{V}$. The charge is positive, and the final potential is higher than the initial potential. As the charge moves from $N$ to $P$, the electric potential energy _____?
   Answer: increases because the charge moves to a higher electric potential. Electric potential energy. A positive charge (+0.50 μC) moving from $V_N$ = -5 V to $V_P$ = +15 V experiences an increase in potential of 20 V. The change in potential energy is ΔU = q($V_f$ - $V_i$) = (+0.50 μC)(+15 V - (-5 V)) = (+0.50 μC)(+20 V) = +10 μJ. The positive result confirms that potential energy increases when a positive charge moves to higher potential, requiring work to be done against the electric field. Choice C incorrectly assumes crossing zero potential has special significance, when only the potential difference matters. Track both charge sign and potential when reasoning about electric potential energy.
8. A $+4.0\,\mu\text{C}$ charge is moved slowly from point $J$ at $V_J=0\,\text{V}$ to point $K$ at $V_K=+40\,\text{V}$. The charge is positive, and the final potential is greater than the initial potential. As the charge moves from $J$ to $K$, the electric potential energy _____?
   Answer: increases because the charge moves to a higher electric potential. Electric potential energy. When a positive charge (+4.0 μC) moves from $V_J$ = 0 V to $V_K$ = +40 V, the change in potential energy is ΔU = q($V_f$ - $V_i$) = (+4.0 μC)(+40 V - 0 V) = +160 μJ. The positive change indicates an increase in potential energy, as positive charges have higher potential energy at higher potentials. This is analogous to lifting a mass against gravity - work must be done to move a positive charge to higher potential. Choice B incorrectly assumes that moving slowly means no energy change, confusing kinetic energy considerations with potential energy. Track both charge sign and potential when reasoning about electric potential energy.
9. A $-3.0\,\mu\text{C}$ charge is moved slowly from point P at $V_P=10\,\text{V}$ to point Q at $V_Q=70\,\text{V}$ in an electric field. As the charge moves from P to Q, its electric potential energy will:
   Answer: decrease because the charge is negative and the potential increases This problem involves electric potential energy of a negative charge. Electric potential energy is calculated as U = qV, where the sign of the charge matters crucially. For the -3.0 μC charge, the initial energy is $U_P$ = (-3.0 μC)(+10 V) = -30 μJ, and the final energy is $U_Q$ = (-3.0 μC)(+70 V) = -210 μJ. Since -210 μJ < -30 μJ, the potential energy decreases (becomes more negative). Choice C incorrectly states that energy increases, failing to recognize that for negative charges, moving to higher potential means moving to more negative energy values. To avoid confusion with negative charges, always calculate U = qV numerically, keeping track of all signs throughout the calculation.
10. A charge $q=-1.0\,\mu\text{C}$ is moved from point A at $V_A=0\,\text{V}$ to point B at $V_B=+40\,\text{V}$ in a static electric field. As the charge moves from A to B, the electric potential energy _____.
    Answer: decreases because the charge moves to a higher electric potential This question tests understanding of electric potential energy. Electric potential energy equals U = qV, where q is the charge and V is the electric potential. For a negative charge (-1.0 μC) moving from $V_A$ = 0 V to $V_B$ = +40 V, the change is ΔU = q($V_B$ - $V_A$) = (-1.0 μC)(+40 V - 0 V) = (-1.0 μC)(+40 V) = -40 μJ. Since ΔU is negative, the potential energy decreases as the negative charge moves to higher potential. Choice C incorrectly assumes zero potential means no energy change, failing to recognize that potential differences drive energy changes. Always calculate ΔU = qΔV to determine energy changes, regardless of whether one potential is zero.
11. A $-3.0\,\mu\text{C}$ charge is moved from point $C$ at $V_C=+5\,\text{V}$ to point $D$ at $V_D=+25\,\text{V}$. The charge is negative, and the final potential is higher than the initial potential. As the charge moves from $C$ to $D$, the electric potential energy of the charge _____?
    Answer: decreases because the charge moves to a higher electric potential. Electric potential energy. The electric potential energy relationship U = qV shows that for a negative charge (-3.0 μC), the potential energy is negative when at positive potential. Moving from $V_C$ = +5 V to $V_D$ = +25 V, the change in potential energy is ΔU = q($V_f$ - $V_i$) = (-3.0 μC)(+25 V - 5 V) = -60 μJ, indicating a decrease. A negative charge at higher positive potential has more negative (lower) potential energy, similar to how a negatively charged particle is repelled from positive regions. Choice D incorrectly assumes negative charges always lose energy when moving, without considering the potential difference. Track both charge sign and potential when reasoning about electric potential energy.
12. A charge $q=-4.0\,\mu\text{C}$ moves from point X at $V_X=-20\,\text{V}$ to point Y at $V_Y=-80\,\text{V}$. As the charge moves, the electric potential energy ____.
    Answer: increases because $\Delta U=q\Delta V>0$ Electric potential energy. The electric potential energy of a charge q at potential V is U = qV, and changes in energy follow ΔU = qΔV. For this negative charge (-4.0 μC) moving from $V_X$ = -20 V to $V_Y$ = -80 V, the potential change is ΔV = -80 V - (-20 V) = -60 V. With a negative charge and negative ΔV, we get ΔU = (-4.0 μC)(-60 V) = +240 μJ, which is positive, indicating an increase in potential energy. Choice B incorrectly assumes that a decrease in potential (ΔV < 0) always means a decrease in energy, failing to recognize that for negative charges, the relationship is reversed. Track both charge sign and potential when reasoning about electric potential energy.
13. A $-1.0\,\mu\text{C}$ charge moves from A at $V_A=+30\,\text{V}$ to B at $V_B=+90\,\text{V}$. As the charge moves from A to B, the electric potential energy will:
    Answer: decrease because the charge is negative and the potential increases This problem examines electric potential energy for a negative charge moving to higher potential. Using U = qV, the initial energy is $U_A$ = (-1.0 μC)(+30 V) = -30 μJ, and the final energy is $U_B$ = (-1.0 μC)(+90 V) = -90 μJ. Since -90 μJ < -30 μJ, the potential energy decreases (becomes more negative). Choice A incorrectly states that energy increases, confusing the increase in potential with an increase in energy for a negative charge. For negative charges, remember that moving to higher potential means moving to more negative energy values, so always calculate U = qV explicitly rather than relying on intuition.
14. A $+0.50\,\mu\text{C}$ charge moves from point R at $V_R=-12\,\text{V}$ to point S at $V_S=-12\,\text{V}$ along a path in an electrostatic field. As the charge moves from R to S, the electric potential energy _____.
    Answer: remains the same because the electric potential is the same at R and S This question tests understanding of electric potential energy. Electric potential energy is U = qV, and for a charge moving between two points, the change is ΔU = q(V_final - V_initial). Here, a positive charge (+0.50 μC) moves from $V_R$ = -12 V to $V_S$ = -12 V, so ΔU = (+0.50 μC)(-12 V - (-12 V)) = (+0.50 μC)(0 V) = 0 μJ. Since the potentials are equal, there is no change in potential energy. Choice B incorrectly assumes the electric field does work along any path, but in electrostatics, work depends only on endpoints, not path. Remember that electric potential energy change depends only on the potential difference between endpoints: ΔU = qΔV.
15. A $+3.0\,\mu\text{C}$ charge moves from point E at $V_E=120\,\text{V}$ to point F at $V_F=120\,\text{V}$ along a different path in the same electric field. As the charge moves from E to F, the electric potential energy will:
    Answer: remain the same because the initial and final potentials are equal This question tests understanding of electric potential energy as a state function. Since electric potential energy depends only on position (through U = qV), not on the path taken, and both initial and final potentials are 120 V, we have $U_E$ = (+3.0 μC)(120 V) = +360 μJ and $U_F$ = (+3.0 μC)(120 V) = +360 μJ. Therefore, ΔU = 0 and the energy remains the same. Choice D incorrectly suggests that path length matters for potential energy, confusing this with work done against friction or other non-conservative forces. Remember that electric potential energy is path-independent; only the initial and final positions (potentials) determine the energy change.
16. A particle with charge $+3.0\,\mu\text{C}$ moves from point C where $V_C=+8\,\text{V}$ to point D where $V_D=+18\,\text{V}$ in an electrostatic field. As the charge moves from C to D, the electric potential energy _____.
    Answer: increases because the charge moves to a higher electric potential This question tests understanding of electric potential energy. Electric potential energy is calculated as U = qV, where q is the charge and V is the electric potential. For a positive charge (+3.0 μC) moving from $V_C$ = +8 V to $V_D$ = +18 V, the change is ΔU = q($V_D$ - $V_C$) = (+3.0 μC)(+18 V - 8 V) = (+3.0 μC)(+10 V) = +30 μJ. Since ΔU is positive, the potential energy increases as the positive charge moves to higher potential. Choice D incorrectly relates energy change to field strength rather than potential difference, missing that potential energy depends on V, not E. Always calculate potential energy changes using ΔU = qΔV, with positive charges gaining energy at higher potentials.
17. A particle with charge $-3.0\,\mu\text{C}$ is moved slowly from point P at $V_P=+5\,\text{V}$ to point Q at $V_Q=+25\,\text{V}$. As the charge moves from P to Q, the electric potential energy of the system _____.
    Answer: decreases because the charge moves to a higher electric potential This question tests understanding of electric potential energy. Electric potential energy is calculated as U = qV, where q is the charge and V is the electric potential. For a negative charge (-3.0 μC) moving from $V_P$ = +5 V to $V_Q$ = +25 V, the change in potential energy is ΔU = q($V_Q$ - $V_P$) = (-3.0 μC)(+25 V - 5 V) = (-3.0 μC)(+20 V) = -60 μJ. Since ΔU is negative, the potential energy decreases as the negative charge moves to higher potential. Choice C incorrectly assumes that being negative means the energy stays constant, failing to recognize that negative charges lose potential energy when moving to higher potentials. Remember that negative charges naturally move toward higher potentials, losing potential energy in the process.
18. A charge $q=+1.5\,\mu\text{C}$ moves from a location at $V_1=-30\,\text{V}$ to $V_2=-10\,\text{V}$. As the charge moves, the electric potential energy ____.
    Answer: increases because $\Delta U=q\Delta V>0$ Electric potential energy. The potential energy of a charge-field system depends on both the charge q and the electric potential V through U = qV. When a charge moves, the change in potential energy is ΔU = qΔV = q($V_2$ - $V_1$). For this positive charge (+1.5 μC) moving from -30 V to -10 V, we calculate ΔV = -10 V - (-30 V) = +20 V, giving ΔU = (+1.5 μC)(+20 V) = +30 μJ, which is positive. Choice B incorrectly assumes that because both potentials are negative, the energy remains constant, missing that what matters is the change in potential, not the absolute values. Track both charge sign and potential when reasoning about electric potential energy.
19. A $+2.0\,\mu\text{C}$ charge moves from point $A$ at $V_A=+30\,\text{V}$ to point $B$ at $V_B=+10\,\text{V}$ in a region where electric potential is defined. The charge's sign is positive, and the final potential is lower than the initial potential. As the charge moves from $A$ to $B$, how does the electric potential energy of the charge change?
    Answer: It decreases because the charge moves to a lower electric potential. Electric potential energy. The electric potential energy of a charge is given by U = qV, where q is the charge and V is the electric potential at that location. For a positive charge (+2.0 μC), moving from higher potential (+30 V) to lower potential (+10 V) means ΔU = q($V_f$ - $V_i$) = (+2.0 μC)(+10 V - 30 V) = -40 μJ, indicating a decrease in potential energy. When a positive charge moves to lower potential, it moves in the direction the electric field would naturally push it, like a ball rolling downhill, thus losing potential energy. Choice D incorrectly assumes positive charges always gain energy when moving, ignoring that the direction relative to potential matters. Track both charge sign and potential when reasoning about electric potential energy.
20. A $+1.5\,\mu\text{C}$ charge moves from location 1 at $V_1=-20\,\text{V}$ to location 2 at $V_2=+10\,\text{V}$, with only the electric force doing work. As the charge moves, the electric potential energy _____.
    Answer: increases because the charge moves to a higher electric potential This question tests understanding of electric potential energy. The electric potential energy of a charge is U = qV, where q is the charge and V is the potential at that location. For a positive charge (+1.5 μC) moving from $V_1$ = -20 V to $V_2$ = +10 V, we find ΔU = q($V_2$ - $V_1$) = (+1.5 μC)(+10 V - (-20 V)) = (+1.5 μC)(+30 V) = +45 μJ. Since ΔU is positive, the potential energy increases. Choice D incorrectly relates potential energy change to electric field strength, when potential energy depends only on charge and potential difference. To solve potential energy problems, use ΔU = qΔV and remember that positive charges gain energy moving to higher potentials.

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