Electric Potential Energy
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AP Physics 2 › Electric Potential Energy
A $-0.80,\mu\text{C}$ charge moves from point $U$ at $V_U=-3,\text{V}$ to point $V$ at $V_V=-18,\text{V}$. The charge is negative, and the final potential is lower than the initial potential. As the charge moves from $U$ to $V$, the electric potential energy _____?
remains the same because both potentials are negative.
decreases because electric potential equals electric potential energy.
increases because the charge moves to a lower electric potential.
decreases because the charge moves to a lower electric potential.
Explanation
Electric potential energy. A negative charge (-0.80 μC) moving from V_U = -3 V to V_V = -18 V sees the potential become more negative (decrease). The change in potential energy is ΔU = q(V_f - V_i) = (-0.80 μC)(-18 V - (-3 V)) = (-0.80 μC)(-15 V) = +12 μJ. The positive change indicates an increase in potential energy because negative charges have higher potential energy at lower (more negative) potentials. Choice C incorrectly assumes that both potentials being negative means no change, ignoring the actual difference in values. 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 _____.
increases because the charge moves to a higher electric potential
decreases 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 charge $q=-2.0,\mu\text{C}$ is moved from point X where $V_X=-5,\text{V}$ to point Y where $V_Y=-35,\text{V}$ in an electrostatic field. As the charge moves from X to Y, the electric potential energy _____.
increases because the charge moves to a higher electric potential
increases because the charge moves to a lower electric potential
decreases 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 electric potential. For a negative charge (-2.0 μC) moving from V_X = -5 V to V_Y = -35 V, the change is ΔU = q(V_Y - V_X) = (-2.0 μC)(-35 V - (-5 V)) = (-2.0 μC)(-30 V) = +60 μJ. Since ΔU is positive, the potential energy increases as the negative charge moves to lower (more negative) potential. Choice C incorrectly assumes that having both negative potentials means no energy change, missing that the magnitude difference matters. When analyzing potential energy changes, always compute ΔU = qΔV carefully, paying attention to all signs.
A $+4.0,\mu\text{C}$ charge is released from rest and moves from point M at $V_M=+60,\text{V}$ to point N at $V_N=+20,\text{V}$. As the charge moves from M to N, the electric potential energy _____.
decreases because the charge moves to a lower electric potential
remains the same because the charge is released from rest
increases because the charge moves to a lower electric potential
decreases because the charge moves to a higher electric potential
Explanation
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 potential. For a positive charge (+4.0 μC) moving from V_M = +60 V to V_N = +20 V, we calculate ΔU = q(V_N - V_M) = (+4.0 μC)(+20 V - 60 V) = (+4.0 μC)(-40 V) = -160 μJ. Since ΔU is negative, the potential energy decreases as the positive charge moves to lower potential. Choice C incorrectly assumes that being released from rest affects the potential energy change, when initial motion state doesn't change the energy calculation. Track potential energy changes using ΔU = qΔV, remembering positive charges lose potential energy moving to lower potentials.
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 _____.
decreases because the charge moves to a lower electric potential
increases because the charge moves to a lower electric potential
decreases because the charge moves to a higher electric potential
remains the same because only the electric field strength matters
Explanation
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.
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 _____.
remains constant because potential is a property of space, not energy
increases because the charge moves to a higher electric potential
decreases because the electric field strength must be smaller at location 2
decreases because the charge moves to a higher electric potential
Explanation
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.
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 _____.
increases because the charge is positive
decreases because the electric field does work along the path
remains the same because the electric field strength must be zero
remains the same because the electric potential is the same at R and S
Explanation
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.
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 _____.
remains the same because the potentials have opposite signs
decreases because the charge moves to a lower electric potential
increases because the charge moves to a higher electric potential
increases because the charge moves to a lower electric potential
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. 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.
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 _____.
decreases because the charge moves to a higher electric potential
remains the same because only the charge sign matters
increases because the charge moves to a higher electric potential
decreases because the electric field must be weaker at D
Explanation
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.
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 _____.
remains the same because the potential at A is zero
increases because the charge moves to a higher electric potential
decreases because the charge moves to a higher electric potential
decreases because the charge moves to a lower electric potential
Explanation
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.