7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
Fixed charged objects produce an electric potential V at point P. A small particle with charge +q is placed at P and then replaced by a particle with charge −2q. Which statement about the electric potential at P is correct?
Answer and reasoning
AV reverses sign and doubles, since it is the particle's energy at P A student who takes the potential to be the potential energy of the particle placed at P picks this. That describes UE = qV, which does double and reverse sign; V is the energy per unit charge and does not depend on the particle.
BV is unchanged, since the fixed charges set it, not the particleCorrect The potential at P is the potential energy per unit charge, set by the fixed source charges and the position of P. Replacing +q by −2q changes UE = qV (it doubles in size and changes sign), but V itself is unchanged.
CV reverses sign and halves, because V = UE/q and q is now −2q A student who reads V = UE/q as saying V is inversely proportional to the particle's charge picks this. When q becomes −2q, UE becomes −2 times as large, so the ratio V stays the same.
DV changes sign but keeps its size, since the particle is negative A student who thinks the sign of the potential is set by the charge placed at P picks this. The sign of V comes from the fixed source charges; the particle's sign affects only the sign of UE.
The figure shows a thin ring of radius R with charge +Q spread uniformly around it, and a point P on the ring's axis. What is the electric potential at P? (k = 1/(4πε₀).)
Answer and reasoning
A√3kQ/(4R) A student who adds the contributions k dq/r as vectors, keeping only their components along the axis, picks this: each term is multiplied by cos θ = √3/2, giving (√3/2)(kQ/(2R)). Potential is a scalar; there are no components to take.
BkQ/(√3R) A student who treats the ring as a point charge at its center picks this, using the distance √3R from O to P. The charge is on the ring, and each element is 2R from P.
CkQ/(2R)Correct Every element dq of the ring is the same distance from P: r = √(R² + (√3R)²) = 2R. So V = k∫dq/r = (k/(2R))∫dq = kQ/(2R).
D(1/4)kQ/R² A student who uses 1/r² for the potential picks this: kQ/(2R)² = (1/4)kQ/R². The integrand for potential is dq/r; kQ/r² has units of V/m, a field.
Working Each element dq is at distance r = √(R² + x²) from P with x = √3R, so r = 2R for every element. V = (1/(4πε₀))∫dq/r = kQ/(2R). Distractors: ring as a point charge at O, r = √3R → kQ/(√3R); 1/r² → (1/4)kQ/R²; axial components (×cos θ, cos θ = √3R/2R) → √3kQ/(4R).
For two points A and B in an electric field, VB − VA = −50 V. Which statement is correct?
Answer and reasoning
AA charged object moved from A to B changes UE by −50 J, whatever its charge A student who thinks the change in potential energy is set by the points alone picks this. −50 V is −50 J per coulomb: ΔUE = qΔV depends on the charge moved.
BA −1.0 C charge moved from A to B lowers UE by 50 J, as B is at lower potential A student who thinks every charge loses potential energy moving toward lower potential picks this. For a negative charge ΔUE = (−1.0 C)(−50 V) = +50 J: UE increases.
CUE falls by 50 J per coulomb on the straight path, and by more on a longer, curved path A student who thinks the potential difference depends on the path picks this, expecting a longer path to change UE more. ΔV between two points, and so ΔUE per coulomb, is the same for every path.
DIf a +1.0 C charge were moved from A to B, by any path, UE would change by −50 JCorrect The potential difference is the change in UE per unit charge: ΔUE = qΔV = (+1.0 C)(−50 V) = −50 J, for any path from A to B.
Working ΔV = ΔUE/q, so ΔUE = qΔV. For q = +1.0 C: ΔUE = (1.0 C)(−50 V) = −50 J, the same for every path. For q = −1.0 C: ΔUE = +50 J (UE increases).
A battery maintains a potential difference between its terminals. Which statement correctly describes how it does this?
Answer and reasoning
AChemical reactions create new positive and negative charge at the two terminals A student who thinks a battery makes charge picks this. Charge is conserved: the reactions move existing charges apart, and the battery's net charge stays zero.
BIt releases stored charge, and it goes dead once the stored charge runs out A student who pictures a battery as a tank of charge picks this. A battery runs down when its chemical reactants are used up; it does not run out of charge.
CChemical reactions move positive and negative charges apart, to opposite terminalsCorrect Chemical processes inside the battery separate positive and negative charges, making one terminal positive and the other negative. That separation of charge produces the potential difference between the terminals.
DIt pushes out a fixed current, and that current sets the potential difference A student who thinks a battery supplies a fixed current picks this. An ideal battery maintains a potential difference; the current depends on what is connected to it.
In a region of space the electric potential depends only on x and is given by V(x) = V₀(x/a)³, where V₀ and a are positive constants. What is the x-component of the electric field at x = a?
Answer and reasoning
A−3V₀/aCorrect Ex = −dV/dx = −3V₀x²/a³. At x = a this is −3V₀/a: V increases with x, so the field points in the −x direction.
B3V₀/a A student who drops the minus sign in Ex = −dV/dx picks this: the slope is +3V₀/a, and the field points toward lower potential, in the −x direction.
CV₀/a A student who divides the potential by the position picks this: V(a)/a = V₀/a. The field is the negative of the rate of change of V, not V divided by x.
D−(1/4)V₀a A student who integrates the potential instead of differentiating it picks this: −∫V dx from 0 to a = −V₀a/4. That has units of V·m; the field is −dV/dx.
Working Ex = −dV/dx = −3V₀x²/a³; at x = a, Ex = −3V₀/a. Distractors: +dV/dx → +3V₀/a; V/x → V₀/a; −∫0a V dx → −V₀a/4.
A very long, straight line of charge has uniform linear charge density λ > 0. Point a is a perpendicular distance ra from the line and point b is a perpendicular distance rb from it, with rb > ra. What is the potential difference Vb − Va?
Answer and reasoning
A−(λ/(2πε₀)) ln(rb/ra)Correct By Gauss's law the field is radial with E = λ/(2πε₀r). Vb − Va = −∫E dr from ra to rb = −(λ/(2πε₀)) ln(rb/ra), which is negative, as it should be for moving away from a positive line along the field.
B(λ/(2πε₀)) ln(rb/ra) A student who drops the minus sign in ΔV = −∫E⃗·dr⃗ picks this. Moving from a to b is moving along the field, away from the positive line, so the potential falls: Vb − Va must be negative.
C(λ/(4πε₀))(1/rb − 1/ra) A student who uses a 1/r² field, as for a point charge, picks this: −∫λ/(4πε₀r²) dr from ra to rb. The field of a long line falls as 1/r, and integrating 1/r gives a logarithm.
D−(λ/(2πε₀))(rb − ra)/ra A student who multiplies the field at a by the distance moved picks this, as if the field were uniform. The field weakens with distance, so ΔV must be found by integrating.
Working Gauss's law, cylinder of radius r and length ℓ: E(2πrℓ) = λℓ/ε₀, E = λ/(2πε₀r) radially outward. Vb − Va = −∫rarb λ/(2πε₀r) dr = −(λ/(2πε₀)) ln(rb/ra) < 0. Distractors: no minus sign; 1/r² field → (λ/(4πε₀))(1/rb − 1/ra); field at a times distance → −(λ/(2πε₀))(rb − ra)/ra.
Points P and Q lie on the same equipotential line in an electric field; the line is at a potential of 30 V. Which statement must be true?
Answer and reasoning
AThe electric field has the same magnitude at point P as at point Q A student who thinks equal potential means equal field picks this. The field depends on how quickly V changes across the line, which can differ from P to Q (the neighbouring lines can be closer at one point than at the other). It can be true, for example in a uniform field, but it need not be.
BMoving a charge from P to Q by any path changes UE by zeroCorrect P and Q are at the same potential, so ΔV = 0 and ΔUE = qΔV = 0 for any charge. The potential difference does not depend on the path, so this holds for any path, not only along the line.
CA charge placed at P or at Q has zero electric potential energy A student who reads 'equipotential' as 'zero potential energy' picks this. A charge q at P or Q gives the system UE = q(30 V), which is not zero; what is zero is the change in UE between P and Q.
DThe electric field is zero at point P and is also zero at point Q A student who thinks the field vanishes on an equipotential line picks this. The field has no component along the line, but V changes across it, so the field there is perpendicular to the line and not zero in general.
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.2.A.1 Electric potential, V Fix
Electric potential, V
The electric potential energy per unit charge at a point: V = UE/q, where UE is the electric potential energy the system would have with a test charge q at that point. It is a scalar set by the source charges and the position, not by the test charge. Unit: volt (V), 1 V = 1 J/C.
Zero of electric potential
Only differences in potential have physical meaning, so a zero must be chosen. For an isolated point charge or a bounded charge distribution the convention is V = 0 infinitely far away; for an infinitely long line or cylinder of charge that choice fails, and V is zero at a chosen finite distance or only differences are used.
Students often think The electric potential at a point is the electric potential energy of the charge placed there, so it changes when that charge changes: doubling the charge doubles the potential, and reversing its sign reverses the poten… In fact No. V is the potential energy per unit charge. A charge q at a point where the potential is V gives the system UE = qV, which depends on q; V is set by the other charges and the position.
Students often think Because V = UE/q, the potential at a point is inversely proportional to the charge placed there: doubling that charge halves the potential. In fact No. V = UE/q defines V. When the test charge changes, UE changes in proportion, so the ratio V does not change.
9.2.A.2 Potential of a continuous charge distribution Fix
Potential of a continuous charge distribution
V = (1/(4πε0)) ∫ dq/r: divide the distribution into charge elements dq, add the scalar contributions dq/r, where r is the distance from each element to the point, and integrate over the distribution.
Charge element, dq
A small piece of charge in an integral. On a line with linear charge density λ (unit: C/m), dq = λ dℓ; on a circular arc of radius R, dℓ = R dθ, so dq = λR dθ.
Electric potential of a point charge
V = q/(4πε0 r) = kq/r, with q inserted with its sign and r the distance from the charge to the point. V is positive near a positive charge, negative near a negative one, and falls off as 1/r (the field falls off as 1/r²).
Scalar superposition of potential
The potential due to several point charges is the sum of the individual potentials added as signed numbers: V = (1/(4πε0)) Σi qi/ri, with each ri measured from charge i to the point. No directions or components are involved.
Students often think When finding the potential due to several charges, the signs of the charges can be ignored: every charge contributes a positive amount. In fact No. Potential is a signed scalar: a negative charge contributes a negative potential, kq/r with q < 0, which partly cancels positive contributions.
Students often think Potentials from different charges, or from different charge elements, are added as vectors: only their components along a chosen direction add, and components in opposite directions cancel. In fact No. Potential is a scalar. The contributions kqi/ri, or k dq/r for charge elements, are added as signed numbers whatever the directions from the charges to the point.
9.2.A.3 Electric potential difference, ΔV Fix
Electric potential difference, ΔV
The change in electric potential energy per unit charge when a test charge is moved between two points: ΔV = Vb − Va = ΔUE/q. It depends only on the two points, not on the path or on the test charge. Unit: volt (V).
Students often think The change in electric potential energy between two points is set by the two points alone, so it is the same whatever charge is moved between them. In fact No. The potential difference ΔV between two points is the same for any test charge, but ΔUE = qΔV scales with the charge moved and changes sign with it.
Students often think Every charged object, positive or negative, loses electric potential energy when it moves toward lower potential, just as every object loses gravitational potential energy when it moves down. In fact No. ΔUE = qΔV. A positive charge moving to lower potential loses UE; a negative charge moving to lower potential gains UE.
9.2.A.4 Potential difference from a battery Fix
Potential difference from a battery
Chemical processes inside a battery move positive and negative charges apart, to opposite terminals, producing a potential difference between the terminals. The battery neither creates charge nor holds a supply of charge that is used up.
Students often think A battery creates electric charge, producing new positive and negative charge at its terminals. In fact No. Charge is conserved. Chemical processes in the battery separate charges that are already present, moving positive charge to one terminal and negative charge to the other.
Students often think A battery is a store of charge that it gives out, and it goes dead when the stored charge has been used up. In fact No. A battery maintains a potential difference by separating charge through chemical reactions. It runs down when its chemical reactants are used up, not because it runs out of charge.
9.2.B.1 Field component from potential, Ex = −dV/dx Fix
Field component from potential, Ex = −dV/dx
The component of the electric field in any direction equals the negative of the rate of change of the potential with position in that direction. On a V–x graph, Ex is the negative of the slope at the point; where V is constant, Ex = 0. Unit: V/m, which equals N/C.
Students often think The electric potential and the electric field are zero at the same places: where one is zero, so is the other. In fact No. The field depends on how V changes with position, not on its value. Midway between equal and opposite charges V = 0 but the field is not zero; midway between equal like charges the field is zero but V is not.
Students often think The electric field at a point equals the potential there divided by the position or distance: E = V/x (as in E = V/d for parallel plates). In fact No. Ex = −dV/dx is the negative of the slope of V at the point. V/x is a ratio of coordinates, not a rate of change, and has nothing to do with the field unless V happens to be proportional to x from the origin.
9.2.B.2 Potential difference from the field, ΔV = −∫E⃗·dr⃗ Fix
Potential difference from the field, ΔV = −∫E⃗·dr⃗
Vb − Va = −∫ab E⃗·dr⃗. Only the displacement component along the field contributes; for a uniform field and a displacement Δx along it, ΔV = −ExΔx. The result is the same for every path between a and b.
Students often think The potential difference is the integral of the field along the path without a minus sign: ΔV = +∫E⃗·dr⃗, so V increases along the field. In fact No. ΔV = −∫ab E⃗·dr⃗. Moving along the field lowers the potential, so the minus sign is needed.
Students often think When finding ΔV from an Ex–x graph, the areas above and below the axis are both added as positive amounts. In fact No. ΔV = −∫Ex dx uses signed areas: where Ex is negative the area is negative and contributes a positive change in V.
9.2.B.3 Electric field vector map Fix
Electric field vector map
A diagram of arrows at sample points, each showing the direction of the electric field there and, by its length, the field's magnitude. It can be used to predict the direction of the force on a charge: along the arrow for a positive charge, opposite for a negative one.
Equipotential line (isoline)
A line joining points at the same electric potential. Moving a test charge between any two points on it changes UE by zero, whatever path is taken.
Isolines and field lines
Isolines are perpendicular to electric field vectors at every point. A field map can be drawn from an isoline map, and the reverse. For isolines drawn at equal steps of V, closer spacing means a stronger field.
Direction of the field and potential
The electric field vector points in the direction in which the potential decreases most rapidly: from higher to lower potential.
No field component along an isoline
Because V does not change along an isoline, the component of E⃗ along it is zero; the electric force on a charge moving along an isoline is perpendicular to its displacement and does no work.
Students often think The electric field points toward higher potential, so Ex has the same sign as the slope dV/dx. In fact No. The field points toward lower potential: Ex = −dV/dx, so where V increases with x, Ex is negative.
Students often think Equipotential lines and electric field lines are the same kind of line, so the field points along equipotential lines and V is constant along field lines. In fact No. Equipotential lines are perpendicular to field lines. Moving along a field line changes V; moving along an equipotential line does not.
20 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 20
A particle with charge +Q is fixed in place. Point X is a distance r from it and point Y is a distance 2r from it. A test particle with charge +q is placed at X; a test particle with charge +2q is placed at Y. What is the ratio VX/VY of the electric potentials at X and Y?
Answer and reasoning
A1.00 A student who takes the potential at a point to be the energy of the particle placed there picks this: UX = kQq/r and UY = kQ(2q)/(2r) are equal. V is set by +Q and the distance.
B4.00 A student who uses V = kQ/r² picks this: (1/r²)/(1/(2r)²) = 4. The potential of a point charge falls as 1/r.
C0.50 A student who thinks potential grows in proportion to distance from a positive charge, like mgh with height, picks this: r/(2r). For a positive charge V = kQ/r falls with distance.
D2.00Correct V = kQ/r depends only on the source charge and the distance: VX = kQ/r and VY = kQ/(2r), so VX/VY = 2.00. The test charges change UE = qV, not V.
A thin rod is bent into a semicircular arc of radius R centered on point C. The rod carries charge Q spread uniformly along it. What is the electric potential at C? (k = 1/(4πε₀).)
Answer and reasoning
AkQ/RCorrect Every element is the same distance R from C. With dq = λR dθ and λ = Q/(πR): V = k∫(λR dθ)/R from 0 to π = kλπ = kQ/R. Equivalently, V = (k/R)∫dq = kQ/R.
BkQ/R² A student who uses 1/r² for the potential, as for the field, picks this: k∫dq/R² = kQ/R². The integrand for potential is dq/r; kQ/R² has units of V/m.
C2kQ/(πR) A student who adds the contributions as vectors, keeping only components along the arc's axis of symmetry, picks this: k∫λ cos θ dθ from −π/2 to π/2 = 2kλ = 2kQ/(πR). Potential is a scalar; no cosine factor belongs in the integral.
D2kQ/(πR²) A student who takes the potential to be the electric field at C picks this: 2kQ/(πR²) is the magnitude of the field at the center of the arc. That has units of V/m; the potential integral k∫dq/r gives kQ/R.
Working dq = λR dθ, λ = Q/(πR), r = R for every element. V = k∫0π λR dθ/R = kλπ = kQ/R. Distractors: 1/r² integrand → kQ/R²; components 2kλ = 2kQ/(πR); field at C = 2kQ/(πR²).
Arc 1 is a semicircular arc of radius R with charge Q spread uniformly along it; the electric potential at its center is V₁. Arc 2 is a quarter-circle arc of radius R/2 with charge Q/2 spread uniformly along it; the electric potential at its center is V₂. What is V₂/V₁?
Answer and reasoning
A0.50 A student who writes dq = (total charge) dθ picks this: arc 1 gives πkQ/R and arc 2 gives k(Q/2)(π/2)/(R/2) = πkQ/(2R). With the correct element dq = λr dθ, each arc gives k × (its charge)/(its radius).
B1.00Correct Every element of an arc is one radius from its center, so V = k∫dq/r = k(total charge)/(radius). V₁ = kQ/R; V₂ = k(Q/2)/(R/2) = kQ/R. The ratio is 1.00: the arc's angle does not matter.
C2.83 A student who uses the field at each center as the potential picks this: the field at the center of arc 1 is 2kQ/(πR²) and at the center of arc 2 is 4√2kQ/(πR²), a ratio of 2√2 ≈ 2.83. The potential integral k∫dq/r gives k(charge)/(radius) for each arc.
D1.41 A student who adds components along each arc's symmetry axis picks this: arc 1 gives 2kQ/(πR) and arc 2 gives k[2Q/(πR)]√2, so the ratio is √2 ≈ 1.41. Potential is a scalar: no cosine factors.
Working For an arc of charge q and radius a, every dq is at distance a from the center: V = k∫dq/a = kq/a. V₁ = kQ/R; V₂ = k(Q/2)/(R/2) = kQ/R; V₂/V₁ = 1.00. Distractors: fields at the centers used as potentials → [√2k(2Q/(πR))/(R/2)]/[2kQ/(πR²)] = 2√2 ≈ 2.83; dq = q dθ → [k(Q/2)(π/2)/(R/2)]/[πkQ/R] = 0.50; vector components → [√2 k·2Q/(πR)]/[2kQ/(πR)] = 1.41.
The figure shows a thin rod with charge +6.0 nC spread uniformly along its length and a point P on the line of the rod, beyond its right end. What is the electric potential at P? (k = 9.0 × 10⁹ N·m²/C²; take V = 0 infinitely far away.)
Answer and reasoning
A2.7 × 10² V A student who treats the rod as a point charge at its midpoint picks this: kQ/(0.20 m) = 270 V. The nearer half of the rod contributes more than the farther half, so the true potential is larger.
B5.4 × 10² V A student who places all the charge at the end nearest P picks this: kQ/(0.10 m) = 540 V. Most of the rod is farther than 0.10 m from P, so the true potential is smaller.
C1.8 × 10³ V A student who integrates dq/r² picks this: kλ(1/0.10 m − 1/0.30 m) = 1800. That integral gives a field (in V/m), not a potential; the potential integrand is dq/r.
D3.0 × 10² VCorrect With x measured from P along the rod, λ = Q/L = 3.0 × 10⁻⁸ C/m and V = kλ∫dx/x from 0.10 m to 0.30 m = kλ ln 3 = (9.0 × 10⁹)(3.0 × 10⁻⁸)(1.10) ≈ 3.0 × 10² V.
Working λ = 6.0 × 10⁻⁹ C/0.20 m = 3.0 × 10⁻⁸ C/m. Element dq = λ dx at distance x from P, x from 0.10 m to 0.30 m. V = kλ∫dx/x = kλ ln(0.30/0.10) = (9.0 × 10⁹)(3.0 × 10⁻⁸)(ln 3) = 270 × 1.099 ≈ 297 V ≈ 3.0 × 10² V.
A thin rod of length L carries a uniform linear charge density λ and lies along the x-axis from x = −L/2 to x = +L/2. Point P is on the rod's perpendicular bisector, a distance y from its midpoint. Which integral gives the electric potential at P? (k = 1/(4πε₀).)
Answer and reasoning
AV = ∫ kλ dx/(x² + y²), from −L/2 to L/2 A student who uses 1/r² for the potential picks this. That integrand has units of V/m, a field; the potential integrand is dq/r.
BV = ∫ kλy dx/√(x² + y²)³, from −L/2 to L/2 A student who sets up the field integral when asked for the potential picks this: kλy dx/√(x² + y²)³ is the bisector component of the field from each element. Potential is the scalar sum of k dq/r, and this integrand has units of V/m, not V.
CV = ∫ kλ dx/√(x² + y²), from −L/2 to L/2Correct The element dq = λ dx at position x is a distance √(x² + y²) from P. Adding the scalar contributions k dq/r over the whole rod gives V = ∫ kλ dx/√(x² + y²) from −L/2 to L/2.
DV = ∫ (kλ/y) dx, from −L/2 to L/2 A student who takes the perpendicular distance y as r for every element picks this. Only the element at the midpoint is a distance y from P; an element at x is √(x² + y²) away.
Working dq = λ dx at (x, 0); distance to P(0, y) is r = √(x² + y²). V = k∫dq/r = ∫−L/2L/2 kλ dx/√(x² + y²) (= 2kλ ln[(L/2 + √(L²/4 + y²))/y]). Distractors: 1/r² integrand; the bisector-component (field) integrand kλy dx/r³ (field taken as potential); r = y for every element, giving ∫(kλ/y)dx.
A particle with charge −4.0 nC is fixed in place. Point A is 0.20 m from the particle and point B is 0.60 m from it. What is the potential difference VB − VA? (k = 9.0 × 10⁹ N·m²/C²; take V = 0 infinitely far away.)
Answer and reasoning
A+1.2 × 10² VCorrect VA = kq/rA = (9.0 × 10⁹)(−4.0 × 10⁻⁹)/0.20 = −180 V and VB = (9.0 × 10⁹)(−4.0 × 10⁻⁹)/0.60 = −60 V. VB − VA = −60 V − (−180 V) = +120 V: moving away from a negative charge raises the potential.
B−6.0 × 10¹ V A student who takes the potential difference to be the potential at the final point picks this: VB = k(−4.0 nC)/(0.60 m) = −60 V. The difference also needs VA = −180 V: VB − VA = −60 V − (−180 V) = +120 V.
C−1.2 × 10² V A student who subtracts in the reverse order, VA − VB, picks this. VB − VA is final minus initial: +120 V.
D+8.0 × 10² V A student who uses V = kq/r² picks this: k(−4.0 nC)(1/0.60² − 1/0.20²) m⁻² = +800. The potential of a point charge falls as 1/r, not 1/r².
Working VA = (9.0 × 10⁹)(−4.0 × 10⁻⁹ C)/(0.20 m) = −180 V; VB = (9.0 × 10⁹)(−4.0 × 10⁻⁹ C)/(0.60 m) = −60 V. VB − VA = +120 V.
The figure shows two particles with charges +4.0 nC and −4.0 nC and a point P. What is the electric potential at P? (k = 9.0 × 10⁹ N·m²/C²; take V = 0 infinitely far away.)
Answer and reasoning
A1.9 × 10² V A student who ignores the sign of the −4.0 nC particle picks this: 120 V + 72 V = 192 V. The negative particle contributes −72 V, which partly cancels the +120 V.
B9.6 × 10¹ V A student who adds the two contributions as vectors picks this: +120 V along the vertical side and −72 V along the line from the −4.0 nC particle to P combine, component by component, to a magnitude of 96 V. Potential is a scalar: add +120 V and −72 V as numbers.
C2.6 × 10² V A student who uses V = kq/r² picks this: k(4.0 nC)(1/0.30² − 1/0.50²) m⁻² = 256. The potential of a point charge falls as 1/r.
D4.8 × 10¹ VCorrect The −4.0 nC particle is √(0.30² + 0.40²) = 0.50 m from P. V = k(+4.0 × 10⁻⁹)/0.30 + k(−4.0 × 10⁻⁹)/0.50 = 120 V − 72 V = 48 V = 4.8 × 10¹ V.
Working r₁ = 0.30 m; r₂ = √(0.30² + 0.40²) = 0.50 m. V = (9.0 × 10⁹)[(4.0 × 10⁻⁹)/0.30 + (−4.0 × 10⁻⁹)/0.50] = 120 V − 72 V = 48 V.
Two particles, each with charge +3q, are fixed on the x-axis at x = −a and x = +a. Point P is on the y-axis at y = √3a. What is the electric potential at P? (k = 1/(4πε₀); take V = 0 infinitely far away.)
Answer and reasoning
A(3√3/2)kq/a A student who adds the two contributions as vectors picks this: each 3kq/(2a) is resolved along the lines from the particles to P, the x-components cancel and the y-components, (√3/2)(3kq/(2a)) each, add. Potential is a scalar: the two contributions simply add.
B3kq/aCorrect Each particle is √(a² + 3a²) = 2a from P. V = 3kq/(2a) + 3kq/(2a) = 3kq/a.
C2√3kq/a A student who replaces the pair by a single charge 6q at their midpoint, the origin, picks this: k(6q)/(√3a) = 2√3kq/a. Each particle is 2a from P, not √3a.
D3kq/(2a²) A student who uses 1/r² picks this: 2(3kq)/(2a)² = 3kq/(2a²). That has units of V/m; the potential of each particle is kq/r.
Working r = √(a² + (√3a)²) = 2a for each particle. V = Σkqi/ri = 2(3kq)/(2a) = 3kq/a. Distractors: vector addition → 2(3kq/(2a))(√3/2) = (3√3/2)kq/a; 6q at the origin → k(6q)/(√3a) = 2√3kq/a; 1/r² → 2(3kq)/(2a)² = 3kq/(2a²).
Point M₁ is midway between two particles with equal charges +Q. Point M₂ is midway between a particle with charge +Q and a particle with charge −Q. The particles in each pair are a distance d apart. Which statement about the electric potential V and the electric field E⃗ at M₁ and M₂ is correct? (Take V = 0 infinitely far away.)
Answer and reasoning
AAt M₁, V ≠ 0 and E⃗ ≠ 0; at M₂, V = 0 and E⃗ = 0 A student who computes the potentials correctly but thinks V and E⃗ vanish together picks this: V = 0 at M₂ is taken to mean E⃗ = 0 there, and V ≠ 0 at M₁ to mean E⃗ ≠ 0. At M₂ both fields point toward −Q and add; at M₁ the equal and opposite fields cancel.
BAt M₁, V = 0 and E⃗ = 0; at M₂, V ≠ 0 and E⃗ ≠ 0 A student who adds potentials as vectors, like fields, picks this: at M₁ the two equal contributions are taken to point in opposite directions and cancel, and at M₂ to point the same way and add. Potentials are scalars: at M₁ they add, and at M₂, +kQ/(d/2) and −kQ/(d/2) cancel.
CAt M₁, V ≠ 0 and E⃗ = 0; at M₂, V = 0 and E⃗ ≠ 0Correct At M₁ the equal and opposite field vectors cancel, but the scalar potentials add: V = 2kQ/(d/2) = 4kQ/d. At M₂ the potentials +kQ/(d/2) and −kQ/(d/2) cancel, but both fields point toward −Q and add.
DAt M₁, V ≠ 0 and E⃗ = 0; at M₂, V ≠ 0 and E⃗ ≠ 0 A student who ignores the sign of −Q when adding potentials picks this: at M₂ both contributions are taken as positive and add to 4kQ/d. With the sign kept, they cancel and V = 0 at M₂.
Working M₁ (charges +Q, +Q, each d/2 away): V = kQ/(d/2) + kQ/(d/2) = 4kQ/d ≠ 0; the two fields have equal magnitude and opposite directions, so E⃗ = 0. M₂ (+Q and −Q): V = kQ/(d/2) − kQ/(d/2) = 0; both fields point toward −Q, E = 2kQ/(d/2)² = 8kQ/d² ≠ 0.
A particle with charge +q is moved from point A to point B along a straight path of length d, and the electric potential energy of the system increases by 6.00 μJ. A particle with charge −2q is then moved from A to B along a curved path of length 2d. What is the change in electric potential energy of the system for the second trip?
Answer and reasoning
A−12.0 μJCorrect ΔV = ΔUE/q = (6.00 μJ)/q, the same for any path from A to B. For the second particle ΔUE = (−2q)ΔV = −2(6.00 μJ) = −12.0 μJ.
B+12.0 μJ A student who treats the negative particle as if it changed energy in the same way as the positive one picks this, doubling +6.00 μJ. ΔUE = qΔV with q = −2q reverses the sign.
C+6.00 μJ A student who thinks the change in UE is set by the points A and B alone picks this. It is ΔV that depends only on the points; ΔUE = qΔV changes with the charge moved.
D−24.0 μJ A student who thinks the change in UE grows with the length of the path picks this, doubling −12.0 μJ for the path of length 2d. The electric field is conservative: only the endpoints matter.
Working ΔV = VB − VA = (+6.00 μJ)/(+q). Second trip: ΔUE = (−2q)ΔV = −2 × 6.00 μJ = −12.0 μJ, independent of path.
The graph shows the electric potential V as a function of position x along the x-axis, in a region where the electric field is directed along the x-axis. What is the x-component of the electric field at x = 0.20 m?
Answer and reasoning
A+2.0 × 10³ V/m A student who thinks the field points toward higher potential picks this, taking Ex equal to the slope, +2.0 × 10³ V/m. Ex = −dV/dx: the field points toward lower potential, the −x direction.
B−2.0 × 10³ V/mCorrect At x = 0.20 m the graph is on its straight rising part, from 100 V at x = 0.10 m to 500 V at x = 0.30 m. Slope = 400 V/0.20 m = 2.0 × 10³ V/m, so Ex = −dV/dx = −2.0 × 10³ V/m.
C+1.5 × 10³ V/m A student who divides the potential at the point by its position picks this: 300 V/0.20 m. The field depends on how V changes with x, not on V/x.
D−1.0 × 10³ V/m A student who uses the chord from x = 0 picks this: −(300 V − 100 V)/(0.20 m). That averages over the level part of the graph, where Ex = 0; the field at x = 0.20 m comes from the slope there.
Working Straight segment from (0.10 m, 100 V) to (0.30 m, 500 V): dV/dx = 400 V/0.20 m = 2000 V/m. Ex = −dV/dx = −2.0 × 10³ V/m.
The graph shows the x-component Ex of the electric field as a function of position x along the x-axis; the field has no other components. What is the potential difference V(0.40 m) − V(0)?
Answer and reasoning
A−2.0 × 10¹ V A student who drops the minus sign in ΔV = −∫E⃗·dr⃗ picks this, taking the signed area, −20 V, as ΔV. Moving along the field lowers V, so the minus sign is needed: ΔV = +20 V.
B−1.0 × 10² V A student who adds the two areas as magnitudes picks this: −(40 V + 60 V). From 0.20 m to 0.40 m, Ex is negative, so that area is negative and raises the potential.
C+2.0 × 10¹ VCorrect ΔV = −∫Ex dx = −(signed area under the graph from 0 to 0.40 m) = −[(200 N/C)(0.20 m) + (−300 N/C)(0.20 m)] = −(40 V − 60 V) = +20 V.
D+1.2 × 10² V A student who multiplies the field at one point by the whole distance picks this: −(−300 N/C)(0.40 m). The field is not the same over the whole interval, so the integral must be taken piece by piece.
Working ΔV = −∫00.40 Ex dx = −[(200)(0.20) + (−300)(0.20)] V = −(40 − 60) V = +20 V.
The figure shows a region of uniform electric field and a path from point A through point C to point B. What is the potential difference VB − VA?
Answer and reasoning
A−1.0 × 10² V A student who multiplies the field by the straight-line distance from A to B, √(0.40² + 0.30²) m = 0.50 m, picks this. Only the displacement along the field changes V: 0.30 m.
B−6.0 × 10¹ VCorrect From A to C the displacement is perpendicular to the field, so V does not change. From C to B the displacement is 0.30 m along the field: ΔV = −E Δx = −(200 N/C)(0.30 m) = −60 V. So VB − VA = −6.0 × 10¹ V.
C−1.4 × 10² V A student who multiplies the field by the total length of the path, 0.40 m + 0.30 m, picks this. The potential difference does not depend on the path; only the displacement along the field counts.
D−8.0 × 10¹ V A student who treats the field lines as lines of constant potential picks this, taking the 0.40 m segment across the lines as the one that changes V: −(200 N/C)(0.40 m). V changes along the field, not across it.
Working ΔV = −∫E⃗·dr⃗. A→C: displacement ⊥ E⃗, contribution 0. C→B: −(200 N/C)(0.30 m) = −60 V. VB − VA = −60 V.
A very long, straight line carries a uniform linear charge density λ. The potential difference between two points at distances r and 2r from the line has magnitude ΔV₀. The magnitude of the potential difference between two points at distances 2r and 4r from the line is written as a multiple of ΔV₀. What is that multiple?
Answer and reasoning
A2.00 A student who thinks the potential difference grows in proportion to the distance between the points picks this: the second pair is 2r apart, the first r apart. The field is not uniform; it is weaker farther out.
B0.25 A student who takes the potential of the line to fall off as 1/r², like the field of a point charge, picks this: (1/(2r)² − 1/(4r)²)/(1/r² − 1/(2r)²) = 1/4. The potential of a long line is not of that form: its field falls as 1/r, so V depends on ln r.
C0.50 A student who uses a 1/r² field for the line picks this: then |ΔV| ∝ 1/rinner − 1/router, which is 1/(2r) for the first pair and 1/(4r) for the second. The field of a long line falls as 1/r, and ΔV depends on the ratio of distances.
D1.00Correct E = λ/(2πε₀r), so |ΔV| = (λ/(2πε₀)) ln(router/rinner). Both pairs have the same ratio of distances, 2, so both potential differences equal (λ/(2πε₀)) ln 2: the multiple is 1.00.
Working |ΔV| = ∫E dr = (λ/(2πε₀)) ln(r2/r1). First pair: ln 2; second pair: ln(4r/2r) = ln 2. Ratio 1.00. Distractors: ∝ separation → 2r/r = 2.00; 1/r² field → (1/(2r) − 1/(4r))/(1/r − 1/(2r)) = 0.50; V ∝ 1/r² → (3/16)/(3/4) = 0.25.
The figure shows four equally spaced equipotential lines and a proton at point P moving with velocity v⃗ parallel to the lines. Only the electric force acts on the proton. Which describes the proton's motion after it leaves P?
Answer and reasoning
AIt moves straight along the 30 V line at constant speed A student who thinks the field is zero on an equipotential line picks this. V changes across the line, from 40 V to 20 V, so there is a field at P, perpendicular to the lines, and a force on the proton.
BIt curves toward the 40 V line, speeding up as it goes A student who thinks the field points toward higher potential picks this, sending the force on the proton toward the 40 V line. The field, and the force on a positive charge, point toward lower potential.
CIt follows a curved path toward the 20 V line, gaining speedCorrect The field is perpendicular to the lines and points toward lower potential, from the 40 V side toward the 20 V side; the force on the proton is along the field. The force is perpendicular to the initial velocity, so the path curves toward the 20 V line and the proton gains speed as it moves toward lower potential.
DIt moves straight to the 20 V line along a field line A student who thinks a charge moves along the field line through its position picks this. The force changes the velocity gradually; the proton keeps its upward velocity component and follows a curved path.
Working Lines 40 V, 30 V, 20 V, 10 V from left to right, equally spaced: uniform E⃗ pointing right, toward lower V. The force on the proton, eE⃗, is to the right, perpendicular to v⃗ (up). A constant force perpendicular to the initial velocity gives a parabolic path bending right, toward the 20 V line; the proton moves to lower potential, so UE falls and its kinetic energy rises.
The figure shows four equipotential lines and three points A, B and C. How do the magnitudes of the electric field at A, B and C compare?
Answer and reasoning
AEA > EB = ECCorrect The lines are drawn at equal steps of 10 V, so the field magnitude is about 10 V divided by the gap between neighbouring lines. The gap is smallest near A, and the gaps near B and near C are equal: EA > EB = EC.
BEA = EB > EC A student who takes the field to be strongest where the potential is highest picks this: A and B are on the 0 V line and C is on the −10 V line. The field depends on the spacing of the lines, not on the value of V.
CEB = EC > EA A student who thinks the field is stronger where equipotential lines are farther apart picks this. The same 10 V change over a smaller distance means a stronger field, so the field is strongest at A.
DEC > EA = EB A student who thinks the field is zero wherever the potential is zero picks this: A and B are on the 0 V line. The potential changes across that line, so the field at A and B is not zero.
The figure is an electric field vector map with two points P and Q. Which claim about the electric potentials at P and Q is correct, with correct reasoning?
Answer and reasoning
AVQ > VP, because the field points toward the region of higher potential A student who thinks the field points toward higher potential picks this. The field points toward lower potential, so going from P to Q along the field, V falls.
BVQ > VP, because the field is stronger at Q than it is at P A student who thinks potential is highest where the field is strongest picks this. The size of the field gives how fast V changes, not how large V is; its direction shows that V falls from P to Q.
CVP = VQ, because P and Q both lie on the very same field line A student who treats field lines as lines of constant potential picks this. V changes along a field line; it is constant along lines perpendicular to the field.
DVP > VQ, because the field points from P toward Q, the way V decreasesCorrect The arrows point from P toward Q, and an electric field vector points in the direction of decreasing potential. Moving from P to Q along the field lowers V, so VP > VQ.
A small charged object is moved slowly along an equipotential line from point X to point Y. A student claims that the electric force on the object does no work during this motion. Which reasoning correctly supports the claim?
Answer and reasoning
AThe field has no component along the line, so the force is perpendicular to each displacementCorrect Along an equipotential line V does not change, so the field has no component along the line. The electric force qE⃗ is then perpendicular to every small displacement, and F⃗·dr⃗ = 0 at each step.
BThe field is zero along an equipotential line, so no electric force acts on the object A student who thinks the field vanishes on an equipotential line picks this. The field there is generally not zero; it is perpendicular to the line.
CThe object's speed stays constant, so no force on it does any work, the electric force included A student who applies the work–energy theorem to one force picks this. Constant speed means zero net work; the electric force could do work balanced by another force. It does none here because it is perpendicular to the path.
DThe object's UE is zero at every point of the line, so UE has nothing to change A student who reads 'equipotential' as 'zero potential energy' picks this. UE = qV is generally not zero on the line; it is constant along it, which is why its change is zero.
A thin rod is bent into a semicircular arc of radius R centered on point C. The charge on the rod is not uniform: its linear charge density is λ = λ₀ sinθ, where θ is the angle at C measured from one end of the arc (0 ≤ θ ≤ π) and λ₀ is a positive constant. What is the electric potential at C? (k = 1/(4πε₀); take V = 0 infinitely far away.)
Answer and reasoning
A2kλ₀Correct Each element, of length R dθ, carries dq = λ₀R sinθ dθ and is a distance R from C. Potential is a scalar, so every contribution adds in full: V = k∫0π λ₀ sinθ dθ = 2kλ₀. This is kQ/R for the rod’s total charge, Q = 2λ₀R.
B2kλ₀/R A student who uses 1/r², as for the field, picks this: k∫dq/R² = 2kλ₀/R. Each element’s potential is k dq/r, so the factor R in dq cancels the distance R and leaves 2kλ₀.
C(π/2)kλ₀ A student who adds the elements’ potentials as vectors, keeping only their components along the arc’s axis of symmetry, picks this: k∫λ₀ sinθ · sinθ dθ = (π/2)kλ₀. Potential is a scalar, so each element’s k dq/R adds in full.
D(π/2)kλ₀/R A student who takes the potential at C to equal the electric field there picks this: the field at C has magnitude (π/2)kλ₀/R. The potential is the scalar sum V = k∫dq/r, in volts, not the field, in newtons per coulomb.
Working An element of arc of length R dθ carries dq = λR dθ = λ₀R sinθ dθ, and every element is a distance R from C. V = k∫dq/R = k∫0π λ₀ sinθ dθ = kλ₀[−cosθ]0π = 2kλ₀. (Equivalently, the total charge is Q = ∫λ₀R sinθ dθ = 2λ₀R, all a distance R from C.) Distractors (sympy-checked): 1/r² in the integrand, k∫dq/R² = 2kλ₀/R; only the components along the arc’s axis of symmetry kept, k∫0π λ₀ sinθ · sinθ dθ = (π/2)kλ₀; the field at C taken as the potential: the x-components cancel and E = k∫0π (λ₀/R) sin²θ dθ = (π/2)kλ₀/R.
A very long, solid, insulating cylinder of radius R has a uniform positive volume charge density ρ. Vs is the electric potential at the cylinder’s surface and V₀ is the electric potential on its axis. What is the potential difference Vs − V₀?
Answer and reasoning
A−ρR²/(4ε₀)Correct Gauss’s law with a coaxial cylinder of radius r inside the charge gives E(2πrℓ) = ρπr²ℓ/ε₀, so E = ρr/(2ε₀), pointing outward. Then Vs − V₀ = −∫0R ρr/(2ε₀) dr = −ρR²/(4ε₀). The potential falls from the axis to the surface, in the direction of the field.
B+ρR²/(4ε₀) A student who leaves out the minus sign in ΔV = −∫E⃗ · dr⃗ picks this. Moving outward, along the field, lowers the potential, so Vs − V₀ is negative.
C−ρR²/(2ε₀) A student who multiplies the field at the surface, ρR/(2ε₀), by the distance R picks this. The field grows from zero on the axis to ρR/(2ε₀) at the surface, so the integral −∫E dr is half of that product.
D+ρR²/(2ε₀) A student who takes the field at a point to be the potential there divided by its distance from the axis (E = V/r) picks this: Vs = (ρR/(2ε₀))R and V₀ = 0. The field is the negative rate of change of V with position; V is found by integrating E along a path, not by multiplying E by a distance.
Working Inside the cylinder, apply Gauss’s law to a coaxial cylinder of radius r and length ℓ: E(2πrℓ) = ρπr²ℓ/ε₀, so E = ρr/(2ε₀), directed radially outward. Along a radial path from the axis to the surface: Vs − V₀ = −∫0R E dr = −∫0R ρr/(2ε₀) dr = −ρR²/(4ε₀). Distractors (sympy-checked): minus sign left out, +ρR²/(4ε₀); field at the surface, ρR/(2ε₀), times R, with the minus sign, −ρR²/(2ε₀); potential taken as field times distance from the axis (E = V/r), Vs = (ρR/(2ε₀))R and V₀ = 0, giving +ρR²/(2ε₀).
Compiled from the AP Physics C: Electricity and Magnetism 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
