2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
An electron moves from point X to point Y in an electric field. The electric potential at Y is lower than at X. Which statement about the change ΔUE in the electric potential energy of the electron–field system is correct?
Answer and reasoning
AΔUE is negative, since the electron moves to a lower potential. A student who thinks a move to a lower potential always lowers the potential energy picks this. That holds for a positive charge. For a negative charge, ΔUE = qΔV is positive when ΔV is negative.
BΔUE is zero, since the energy of the system must be conserved. A student who thinks conservation of energy keeps each form of energy fixed picks this. The total energy is what is conserved; UE can change as long as another form, such as the electron's kinetic energy, changes to compensate.
CΔUE is positive, since the charge and ΔV are both negative.Correct ΔUE = qΔV. The electron's charge is negative, and ΔV = VY − VX is negative because Y is at the lower potential. The product of two negative numbers is positive, so the electric potential energy of the system increases.
DΔUE depends on the path the electron takes from X to Y. A student who thinks ΔUE depends on the route picks this. ΔUE = qΔV depends only on the potentials at X and Y, so every path between them gives the same ΔUE.
An electron is released from rest at a location where the electric potential is 50 V. Moving under the electric force alone, it reaches a location where the electric potential is 250 V. How much kinetic energy has it gained? Use e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
A3.2 × 10⁻¹⁷ JCorrect The electron moves through ΔV = 250 V − 50 V = +200 V, so ΔUE = qΔV = (−1.60 × 10⁻¹⁹ C)(200 V) = −3.2 × 10⁻¹⁷ J. The energy of the electron–field system is conserved, so the kinetic energy gained is ΔK = −ΔUE = 3.2 × 10⁻¹⁷ J.
B4.0 × 10⁻¹⁷ J A student who uses the potential at the final location, 250 V, instead of the potential difference picks this: (1.60 × 10⁻¹⁹ C)(250 V). The energy change depends on the difference, 250 V − 50 V = 200 V.
C1.6 × 10⁻¹⁷ J A student who carries the factor 1/2 of UC = (1/2)QΔV over to a single charge picks this: (1/2)(1.60 × 10⁻¹⁹ C)(200 V). A charge moving through a fixed potential difference changes the potential energy by qΔV, with no factor of 1/2.
D6.4 × 10⁻¹⁷ J A student who counts both the work done by the electric force and the loss of electric potential energy picks this: 2 × 3.2 × 10⁻¹⁷ J. These are two descriptions of the same energy transfer; with the electron–field system, ΔK = −ΔUE = 3.2 × 10⁻¹⁷ J.
Working ΔV = 250 V − 50 V = 200 V. ΔUE = qΔV = (−1.60 × 10⁻¹⁹ C)(200 V) = −3.2 × 10⁻¹⁷ J. Only the electric force does work, so ΔK = −ΔUE = 3.2 × 10⁻¹⁷ J.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
10.7.A.1 Object–field system Fix
Object–field system
The system made up of a charged object and the electric field it moves in (or the charges that produce that field). Electric potential energy belongs to this system, not to the object alone.
Change in electric potential energy, ΔUE = qΔV
When an object of charge q moves between two locations whose electric potentials differ by ΔV = Vfinal − Vinitial, the electric potential energy of the object–field system changes by ΔUE = qΔV. The signs of both q and ΔV matter. Unit: joule (J).
Potential and potential energy
Electric potential V is a property of a location, measured in volts (1 V = 1 J/C); electric potential energy UE belongs to an object–field system and is measured in joules. The same potential difference gives a different ΔUE for a different charge.
Independence of path
ΔUE depends only on the object's charge and on the potentials at the starting and ending locations, not on the path between them. A move along an equipotential line gives ΔUE = 0.
Students often think Moving to a higher electric potential always increases the electric potential energy, and moving to a lower potential always decreases it, whatever the sign of the charge. In fact Only for a positive charge. ΔUE = qΔV: for a negative charge, such as an electron, moving to a higher potential (ΔV > 0) decreases the electric potential energy of the object–field system, and moving to a lower potential increases it.
Students often think The change in electric potential energy depends on the path taken or on the distance moved: a longer path means a larger change. In fact No. ΔUE = qΔV depends only on the charge and on the potentials at the starting and ending locations. A charge moved along an equipotential, however far, has ΔUE = 0, and any two paths between the same two points give the same ΔUE.
10.7.A.2 Energy conservation for a charged object Fix
Energy conservation for a charged object
If the electric force is the only force that does work on a charged object, the total energy of the object–field system is constant: Ki + UE,i = Kf + UE,f, so ΔK = −ΔUE = −qΔV.
Kinetic energy gained through a potential difference
A particle of charge q that starts from rest and moves through a potential difference ΔV under the electric force alone gains kinetic energy −qΔV, whatever its mass. Its final speed then depends on its mass, through K = (1/2)mv².
Students often think Because energy is conserved, each form of energy of the system, its kinetic energy and its potential energy, stays the same. In fact No. Conservation of energy means that the TOTAL energy of a system with no external work done on it is constant. Within the object–field system, the kinetic energy and the electric potential energy can each change, as long as the changes add to zero: ΔK + ΔUE = 0.
Students often think The energy a charge gains in moving through a potential difference ΔV is (1/2)qΔV, as in the energy stored in a capacitor. In fact No. A charge q moving through a potential difference ΔV has ΔUE = qΔV, and if only the electric force does work, ΔK = −qΔV. The factor 1/2 in UC = (1/2)QΔV arises because a capacitor's potential difference grows from zero as it charges; a single charge moving through a fixed ΔV has no such factor.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The diagram shows equipotential lines in a region of electric field, labeled with their potentials. A particle with a negative charge is moved from point P, in turn, to points X, Y and Z along the straight paths shown. Which ranking of the changes in the electric potential energy of the particle–field system, ΔUX, ΔUY and ΔUZ, is correct?
Answer and reasoning
AΔUY > ΔUX > ΔUZ A student who thinks a move to a higher potential always raises the potential energy picks this, the ranking for a positive charge. The charge is negative, so every sign reverses: the move to Y, at higher potential, lowers UE, and the move to Z raises it.
BΔUZ > ΔUX > ΔUYCorrect ΔUE = qΔV with q negative, and P is on the 20 V line. To X: ΔV = 0, since X is on the same equipotential, so ΔUX = 0. To Y: ΔV = +20 V, so ΔUY is negative. To Z: ΔV = −10 V, so ΔUZ is positive. Hence ΔUZ > ΔUX > ΔUY.
CΔUY > ΔUZ > ΔUX A student who uses only the size of each potential difference picks this: 20 V to Y, 10 V to Z and 0 to X. The signs matter: for a negative charge, ΔUY = q(+20 V) is negative, the smallest of the three.
DΔUX > ΔUY > ΔUZ A student who thinks ΔUE grows with the distance moved picks this, because the path to X is the longest. X is on the same equipotential as P, so ΔUX = 0 however far the particle moves.
Working q < 0. P is at 20 V. ΔV to X = 20 V − 20 V = 0, so ΔUX = 0. ΔV to Y = 40 V − 20 V = +20 V, so ΔUY = q(20 V) < 0. ΔV to Z = 10 V − 20 V = −10 V, so ΔUZ = q(−10 V) > 0. Ranking: ΔUZ > ΔUX > ΔUY.
A particle with charge −q (where q > 0) moves from a location where the electric potential is V₁ to a location where the electric potential is V₂. Which expression gives the change in the electric potential energy of the particle–field system?
Answer and reasoning
A+q·(V₂ − V₁) A student who drops the sign of the particle's charge picks this, the change for a positive charge. The charge is −q, so ΔUE = −q(V₂ − V₁): for a negative charge, a move to higher potential lowers UE.
B+q·|V₂ − V₁| A student who uses only the size of the potential difference picks this, so that UE always increases. Both ΔV = V₂ − V₁ and the charge have signs, and ΔUE can be negative.
C−(V₂ − V₁)/q A student who rearranges ΔV = ΔUE/q wrongly, dividing the potential difference by the charge, picks this: (V₂ − V₁)/(−q) = −(V₂ − V₁)/q. ΔUE = qΔV, so multiply by the charge; volts divided by coulombs is not an energy.
D−q·(V₂ − V₁)Correct ΔUE = (charge)(ΔV), with the particle's charge −q and ΔV = Vfinal − Vinitial = V₂ − V₁. So ΔUE = −q(V₂ − V₁): if V₂ > V₁, the negative particle's electric potential energy decreases.
A small charged sphere moves from one location to another in an electric field, and the electric force is the only force that does work on it. The bar chart shows the sphere's kinetic energy K and the electric potential energy UE of the sphere–field system at the first location, and UE at the second location. What is the sphere's kinetic energy at the second location?
Answer and reasoning
A2.0 μJ A student who takes the final kinetic energy to be just the potential energy lost picks this: 4.0 μJ − 2.0 μJ. That is the kinetic energy GAINED; the sphere already had 1.0 μJ, so Kf = 3.0 μJ.
B5.0 μJ A student who counts the energy transfer twice, once as work done by the field and once as lost potential energy, picks this: 1.0 μJ + 2(2.0 μJ). With the sphere–field system, the 2.0 μJ drop in UE is the only transfer to kinetic energy.
C3.0 μJCorrect Only the electric force does work, so the total energy of the sphere–field system is constant: Ki + UE,i = Kf + UE,f. From the chart, 1.0 μJ + 4.0 μJ = Kf + 2.0 μJ, so Kf = 3.0 μJ. The 2.0 μJ drop in UE appears as a 2.0 μJ gain in K.
D1.0 μJ A student who thinks conservation of energy keeps the kinetic energy fixed picks this. It is the total, K + UE = 5.0 μJ, that stays constant; since UE falls by 2.0 μJ, K rises by 2.0 μJ.
Working Ki + UE,i = Kf + UE,f: 1.0 μJ + 4.0 μJ = Kf + 2.0 μJ, so Kf = 3.0 μJ.
A proton and an electron are each released from rest beside one plate of a pair of oppositely charged parallel plates in vacuum: the proton beside the positive plate and the electron beside the negative plate. Each moves under the electric force alone to the opposite plate. Which particle arrives with the greater speed, and why?
Answer and reasoning
ANeither: charges of equal magnitude give them equal final speeds. A student who equates equal energy with equal speed picks this. The kinetic energies are equal, but K = (1/2)mv², so the far less massive electron is much faster.
BThe electron: both gain equal kinetic energy, and its mass is smaller.Correct Each particle crosses the same potential difference and has a charge of magnitude e, so each system loses the same electric potential energy, e|ΔV|, and each particle gains the same kinetic energy. Since K = (1/2)mv², the electron, with about 1/1800 of the proton's mass, reaches a much greater speed (about 43 times as great).
CThe proton: the electron moves to higher potential, raising its UE. A student who thinks a move to a higher potential always raises the potential energy picks this. For the negative electron, the move to the higher-potential plate lowers UE, by the same amount as the proton's UE falls, so both gain the same kinetic energy.
DThe electron: its larger acceleration gives it more kinetic energy. A student who links kinetic energy to acceleration picks this. The electron is faster, but not because it gains more energy: both gain e|ΔV|. Its larger acceleration and greater final speed come from its smaller mass.
Working |ΔUE| = e|ΔV| for each, so Ke = Kp. v = √(2K/m): ve/vp = √(mp/me) = √(1.67 × 10⁻²⁷/9.11 × 10⁻³¹) ≈ 43.
A proton has 8.0 × 10⁻¹⁷ J of kinetic energy and moves directly toward a region where the electric potential is 400 V higher than at its present location. Only the electric force does work on it. Can the proton reach that region, and why? Use e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
AYes: reaching it needs 6.4 × 10⁻¹⁷ J, which is less than its kinetic energy.Correct Moving to the region raises the electric potential energy of the proton–field system by ΔUE = qΔV = (1.60 × 10⁻¹⁹ C)(400 V) = 6.4 × 10⁻¹⁷ J. The proton has 8.0 × 10⁻¹⁷ J of kinetic energy, which is enough, so it arrives with 8.0 × 10⁻¹⁷ J − 6.4 × 10⁻¹⁷ J = 1.6 × 10⁻¹⁷ J of kinetic energy left.
BYes: its kinetic energy is conserved, so it keeps moving at the same speed. A student who thinks conservation of energy keeps the kinetic energy fixed picks this. The total energy is conserved: as the proton moves to higher potential, the system's electric potential energy rises and the proton slows down. It does get there, but with less kinetic energy.
CNo: reaching it needs 1.3 × 10⁻¹⁶ J, which is greater than its kinetic energy. A student who counts the energy change twice, as work done against the field and as potential energy gained, picks this: 2 × 6.4 × 10⁻¹⁷ J. The kinetic energy needed is just ΔUE = 6.4 × 10⁻¹⁷ J.
DNo: reaching it needs 400 J, which is far more than its kinetic energy. A student who treats the 400 V potential difference as 400 J of energy picks this. A volt is a joule per coulomb; for the proton's charge, 400 V corresponds to (1.60 × 10⁻¹⁹ C)(400 V) = 6.4 × 10⁻¹⁷ J.
Working ΔUE = qΔV = (1.60 × 10⁻¹⁹ C)(400 V) = 6.4 × 10⁻¹⁷ J < 8.0 × 10⁻¹⁷ J, so the proton reaches the region with K = 8.0 × 10⁻¹⁷ J − 6.4 × 10⁻¹⁷ J = 1.6 × 10⁻¹⁷ J.
A particle with mass m and charge −q (where q > 0) passes point A with speed v₀. Moving under the electric force alone, it reaches point B with speed v₀/2. Which expression gives VB − VA, the electric potential at B minus the electric potential at A?
Answer and reasoning
A−0.375mv₀²/qCorrect The particle's kinetic energy falls from (1/2)mv₀² to (1/2)m(v₀/2)² = (1/8)mv₀², a loss of (3/8)mv₀², so the system's electric potential energy rises by (3/8)mv₀². With ΔUE = (−q)(VB − VA), VB − VA = −(3/8)mv₀²/q: a negative particle slows as it moves toward lower potential.
B−0.750mv₀²/q A student who takes the energy change to be (1/2)qΔV, as for a capacitor, picks this: (1/2)q|ΔV| = (3/8)mv₀² gives |ΔV| = (3/4)mv₀²/q. For a single charge moving through a potential difference, ΔUE = qΔV, with no factor of 1/2.
C−0.188mv₀²/q A student who counts both the work done by the field and the change in UE, so that the kinetic energy changes by 2q|ΔV|, picks this: 2q|ΔV| = (3/8)mv₀² gives |ΔV| = (3/16)mv₀²/q ≈ 0.188mv₀²/q. For the particle–field system, ΔK = −ΔUE, counted once.
D−0.250mv₀²/q A student who thinks halving the speed halves the kinetic energy takes the loss to be (1/2)(1/2)mv₀² = (1/4)mv₀² and picks this. K is proportional to v², so at half the speed the particle keeps only a quarter of its kinetic energy and loses (3/8)mv₀².
Working Energy of the particle–field system is conserved, so ΔUE = −ΔK = (1/2)mv₀² − (1/2)m(v₀/2)² = (3/8)mv₀². ΔUE = (−q)(VB − VA), so VB − VA = −(3/8)mv₀²/q = −0.375mv₀²/q.
A proton passes point A with speed v₀. Moving under the electric force alone, it reaches point B, where the electric potential is lower than at A by ΔV. The quantity eΔV is equal to the proton's kinetic energy at A. What is the proton's speed at B?
Answer and reasoning
A1.00v₀ A student who sets the final kinetic energy equal to the potential energy lost, leaving out the kinetic energy the proton already had, gets KB = eΔV = KA and picks this. The energy lost is ADDED to the initial kinetic energy: KB = 2KA.
B1.73v₀ A student who counts the field's work and the loss of UE as two separate gains, each eΔV, gets KB = KA + 2eΔV = 3KA and picks this: √3·v₀ ≈ 1.73v₀. For the proton–field system the gain in kinetic energy is eΔV, counted once.
C1.41v₀Correct The system's electric potential energy falls by eΔV, which equals KA, so the proton's kinetic energy doubles: KB = 2KA. Since K = (1/2)mv², the speed changes by the square root of that factor: vB = √2·v₀ ≈ 1.41v₀.
D2.00v₀ A student who finds that the kinetic energy doubles but takes the speed to double with it picks this. Kinetic energy is proportional to v², so doubling K multiplies v by √2.
Working KB = KA − ΔUE = KA + eΔV = KA + KA = 2KA. K ∝ v², so vB = √2·v₀ ≈ 1.41v₀.
An electron passes point A with a speed of 5.0 × 10⁶ m/s. Moving under the electric force alone, it reaches point B, where the electric potential is 50 V lower than at A. What is the electron's speed at B? Use e = 1.60 × 10⁻¹⁹ C and me = 9.11 × 10⁻³¹ kg.
Answer and reasoning
A2.7 × 10⁶ m/sCorrect For the electron, ΔUE = qΔV = (−e)(−50 V) = +8.0 × 10⁻¹⁸ J: moving to lower potential raises the electron–field system's potential energy. Its kinetic energy falls by the same amount, from (1/2)me vA² = 1.14 × 10⁻¹⁷ J to 3.4 × 10⁻¹⁸ J, so vB = √(2KB/me) = 2.7 × 10⁶ m/s.
B6.5 × 10⁶ m/s A student who thinks moving to a lower potential always lowers the electric potential energy picks this, adding 8.0 × 10⁻¹⁸ J to the kinetic energy. The electron's charge is negative, so ΔUE = (−e)(−50 V) is positive; the electron slows down, to 2.7 × 10⁶ m/s.
C1.5 × 10⁶ m/s A student who takes kinetic energy to be proportional to speed picks this: KB is 0.30 of KA, so the speed would be 0.30 × 5.0 × 10⁶ m/s. Kinetic energy is proportional to v², so vB = vA√(KB/KA) = 2.7 × 10⁶ m/s.
D4.0 × 10⁶ m/s A student who takes the change in energy to be (1/2)qΔV, as for a capacitor, picks this: the kinetic energy would fall by only 4.0 × 10⁻¹⁸ J. The change in the system's electric potential energy is qΔV = 8.0 × 10⁻¹⁸ J, and KB = 3.4 × 10⁻¹⁸ J gives 2.7 × 10⁶ m/s.
Working ΔUE = qΔV = (−1.60 × 10⁻¹⁹ C)(−50 V) = +8.0 × 10⁻¹⁸ J. KA = (1/2)(9.11 × 10⁻³¹ kg)(5.0 × 10⁶ m/s)² = 1.14 × 10⁻¹⁷ J. Energy conservation: KB = KA − ΔUE = 1.14 × 10⁻¹⁷ J − 0.80 × 10⁻¹⁷ J = 3.4 × 10⁻¹⁸ J. vB = √(2KB/me) = 2.7 × 10⁶ m/s. Errors: potential energy taken to fall at lower potential (m01): KB = KA + 8.0 × 10⁻¹⁸ J, vB = 6.5 × 10⁶ m/s; K ∝ v (m13): vB = vA(KB/KA) = 1.5 × 10⁶ m/s; energy change (1/2)qΔV (m08): KB = KA − 4.0 × 10⁻¹⁸ J, vB = 4.0 × 10⁶ m/s.
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
