Electric Potential Energy
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AP Physics 2 › Electric Potential Energy
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 _____?
remains the same because electric potential is independent of charge.
increases because the charge moves to a higher electric potential.
decreases because the charge moves to a higher electric potential.
decreases because a negative charge always loses potential energy when it moves.
Explanation
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.
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?
It decreases because the charge moves to a lower electric potential.
It remains the same because only the electric field strength matters.
It increases because a positive charge always gains potential energy when it moves.
It increases because the charge moves to a lower electric potential.
Explanation
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.
A $-1.0,\mu\text{C}$ charge moves from point $L$ at $V_L=+12,\text{V}$ to point $M$ at $V_M=+12,\text{V}$. The charge is negative, and the initial and final potentials are equal. As the charge moves from $L$ to $M$, the electric potential energy _____?
increases because electric potential energy equals electric potential.
increases because a negative charge gains energy when it moves.
decreases because the charge is negative.
remains the same because the electric potential is the same at both points.
Explanation
Electric potential energy. When a charge moves between two points at the same potential (V_L = V_M = +12 V), the change in potential energy is ΔU = q(V_f - V_i) = (-1.0 μC)(+12 V - 12 V) = 0 μJ. The potential energy remains constant because U = qV depends only on the potential at a location, not on the path taken or the sign of the charge when potentials are equal. This is true regardless of whether the charge is positive or negative. Choice D incorrectly equates electric potential energy with electric potential, failing to recognize that U = qV includes the charge as a factor. Track both charge sign and potential when reasoning about electric potential energy.
A charge $q=-2.0,\mu\text{C}$ is moved from point P at $V_P=10,\text{V}$ to point Q at $V_Q=70,\text{V}$. As the charge moves, the electric potential energy ____.
decreases because electric potential increased, independent of charge sign
decreases because $\Delta U=q\Delta V<0$
increases because the charge moves to higher potential
remains the same because the charge magnitude is constant
Explanation
Electric potential energy. The potential energy of a charge in an electric field is U = qV, and the change in potential energy when moving between points is ΔU = qΔV. For this negative charge (-2.0 μC) moving from V_P = 10 V to V_Q = 70 V, the change in potential is ΔV = 70 V - 10 V = +60 V. Therefore, ΔU = (-2.0 μC)(+60 V) = -120 μJ, which is negative, indicating a decrease in potential energy. Choice A incorrectly assumes that moving to higher potential always increases energy, failing to account for the negative charge which experiences the opposite effect. Track both charge sign and potential when reasoning about electric potential energy.
A $-3.0,\mu\text{C}$ charge is moved slowly from $V=+5,\text{V}$ to $V=+25,\text{V}$ in a region of electric field. As the charge moves, which statement correctly describes the change in electric potential energy?
It increases because the charge is negative and the potential increases.
It decreases because the charge is negative and the potential is higher.
It decreases because the charge is negative and the potential increases.
It stays the same because only the electric field strength matters.
Explanation
This problem 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 = +5 V to V = +25 V, the change in potential energy is ΔU = q(Vf - Vi) = (-3.0 μC)(+25 V - 5 V) = (-3.0 μC)(+20 V) = -60 μJ. Since ΔU is negative, the potential energy decreases. Choice B incorrectly claims energy increases, likely confusing the fact that negative charges naturally move toward higher potential with the energy change. Remember: track both charge sign and potential when reasoning about electric potential energy.
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 ____.
decreases because negative charges must lose potential energy when moving
increases because $\Delta U=q\Delta V>0$
remains the same because the potential decreases by $60,\text{V}$
decreases because $\Delta V<0$ always implies $\Delta U<0$
Explanation
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.
A charge $q=-1.0,\text{nC}$ moves from point A at $V_A=200,\text{V}$ to point B at $V_B=50,\text{V}$. As the charge moves, the electric potential energy ____.
increases because negative charges gain energy when potential decreases
remains the same because $\Delta V\neq 0$ only changes potential
decreases because the charge moved to lower potential
increases because $\Delta U=q\Delta V>0$
Explanation
Electric potential energy. The electric potential energy of a charge-field system is U = qV, and changes according to ΔU = qΔV when the charge moves. For this negative charge (-1.0 nC) moving from 200 V to 50 V, the potential change is ΔV = 50 V - 200 V = -150 V. With a negative charge and negative ΔV, we calculate ΔU = (-1.0 nC)(-150 V) = +150 nJ, which is positive, indicating an increase in energy. Choice D contains the correct physical insight but uses incorrect terminology - negative charges do gain energy when moving to lower potential, but this is because ΔU = qΔV becomes positive when both q and ΔV are negative. Track both charge sign and potential when reasoning about electric potential energy.
A charge $q=+6.0,\text{nC}$ is moved from point M at $V_M=0,\text{V}$ to point N at $V_N=120,\text{V}$. As the charge moves, the electric potential energy ____.
increases because $\Delta U=q\Delta V>0$
remains the same because $V_M$ is zero
decreases because the charge is positive
decreases because only the distance moved determines $\Delta U$
Explanation
Electric potential energy. The potential energy of a charge in an electric field is U = qV, where q is the charge and V is the potential. When a charge moves between points, the energy change is ΔU = qΔV = q(V_N - V_M). For this positive charge (+6.0 nC) moving from 0 V to 120 V, we have ΔV = 120 V - 0 V = +120 V, so ΔU = (+6.0 nC)(+120 V) = +720 nJ, which is positive. Choice C incorrectly assumes that because the initial potential is zero, the energy remains unchanged, missing that what matters is the change in potential, not whether one value is zero. Track both charge sign and potential when reasoning about electric potential energy.
A $-0.80,\mu\text{C}$ charge is moved from point E at $V_E=-50,\text{V}$ to point F at $V_F=-20,\text{V}$ in a static electric field. As the charge moves from E to F, the electric potential energy _____.
decreases because the charge moves to a higher electric potential
increases because the charge moves to a higher electric potential
increases because the charge moves to a lower electric potential
remains the same because both potentials are negative
Explanation
This question tests understanding of electric potential energy. Electric potential energy is U = qV, where q is the charge and V is the potential at that point. For a negative charge (-0.80 μC) moving from V_E = -50 V to V_F = -20 V, we calculate ΔU = q(V_F - V_E) = (-0.80 μC)(-20 V - (-50 V)) = (-0.80 μC)(+30 V) = -24 μJ. Since ΔU is negative, the potential energy decreases as the negative charge moves to higher (less negative) potential. Choice C incorrectly assumes that both potentials being negative means no change, ignoring the actual calculation. Track both charge sign and potential when reasoning about electric potential energy: ΔU = qΔV.
A $+2.0,\mu\text{C}$ charge moves from point A at $V_A=+80,\text{V}$ to point B at $V_B=+20,\text{V}$ in an external electric field. As the charge moves from A to B, the electric potential energy of the charge in the field will:
decrease because the charge is positive and the potential decreases
increase because the charge is positive and the potential increases
increase because the charge is positive and the potential decreases
remain the same because only the electric field strength matters
Explanation
This problem 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), the initial energy is U_A = (+2.0 μC)(+80 V) = +160 μJ, and the final energy is U_B = (+2.0 μC)(+20 V) = +40 μJ. Since U_B < U_A, the potential energy decreases by 120 μJ. Choice A incorrectly assumes that positive charges gain energy when moving to lower potential, confusing this with the behavior of negative charges. When solving electric potential energy problems, remember that U = qV directly gives the energy, and positive charges naturally move toward lower potential, losing potential energy in the process.