2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A student uses E⃗ = (1/(4πε₀))∫(dq/r²)r̂ to find the electric field at a point P on the perpendicular bisector of a uniformly charged, straight rod of length L, a short distance from the rod. Which step of the method is correct?
Answer and reasoning
AIntegrate the magnitudes k dq/r² directly, because every element’s field has that form. A student who adds the contributions as numbers picks this. Integrating magnitudes treats every contribution as if it pointed the same way; at P, the contributions from different parts of the rod point in different directions, and their components along the rod cancel.
BTake r to be the distance from P to the rod’s center, the same for every element. A student who treats the rod as a point charge at its center picks this. In the integral, r is the distance from P to each element dq, and it changes from element to element.
CWrite each element’s charge as dq = Q dℓ, the total charge times its length dℓ. A student who uses the total charge in dq picks this. The charge on a length dℓ is λ dℓ, where λ = Q/L is the charge per unit length; Q dℓ has units of C·m, not C.
DResolve each element’s field dE⃗ into components, then integrate each component separately.Correct The contributions dE⃗ from different elements point in different directions, so they must be added as vectors. Resolving each into components (here, symmetry shows that the components along the rod cancel) turns the vector integral into integrals of numbers.
At which of the following points does symmetry show that the electric field is zero? In each case, the charge is spread uniformly over the objects named.
Answer and reasoning
AThe center of a thin semicircular arc with charge +Q along its whole length A student who thinks the field at the center of any uniform arc is zero, as for a ring, picks this. The semicircle has no charge on its missing half, so its contributions’ components along its line of symmetry do not cancel.
BThe center of a thin ring with charge +Q around its whole circumferenceCorrect Every element of the ring has a partner directly opposite it across the center, the same distance away with the same charge. Their fields at the center are equal and opposite, so the ring’s field there is zero.
CA point off a straight rod with charge +Q, on the rod’s perpendicular bisector A student who thinks the field is zero wherever a point is equally far from the charges on either side picks this. On the bisector the components along the rod cancel, but the perpendicular components add, pointing away from the rod.
DThe midpoint between two long parallel lines with charge densities +λ and −λ A student who thinks zero net charge gives zero field picks this. At the midpoint, the +λ line’s field points away from it and the −λ line’s field points toward the −λ line: the same direction, so the fields add.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.4.A.1 Continuous charge distribution Fix
Continuous charge distribution
Charge spread along a line or over a surface or volume rather than located at a few points. To find its field, it is divided into small elements, each with charge dq, and each element is treated as a point charge.
Linear charge density, λ
Charge per unit length of a thin rod, wire or arc: λ = Q/L when the charge is uniform, and λ(x) = dq/dx when it varies along the length. An element of length dℓ carries charge dq = λ dℓ. SI unit: C/m.
Principle of superposition
The electric field at a point is the vector sum of the fields produced there by every charge present. For a continuous distribution the sum becomes an integral of the elements' fields, E⃗ = ∫dE⃗.
Field of a continuous distribution
E⃗ = (1/(4πε₀))∫(dq/r²)r̂, where r is the distance from the element dq to the point where the field is found and r̂ is the unit vector pointing from dq toward that point. In practice each component of dE⃗ is integrated separately.
Field of an infinitely long line of charge
At a distance r from a very long, straight line with uniform density λ, E = 2kλ/r = λ/(2πε₀r), directed perpendicular to the line (away from it when λ is positive). It falls off as 1/r.
Field on the axis of a charged ring
For a thin ring of radius R with uniform charge Q, at a distance x from its center along its axis: E = kQx/(x² + R²)3/2, directed along the axis. It is zero at the center and approaches kQ/x² far from the ring.
Field of a finite, uniformly charged rod
For a rod of length L and charge Q: at a point on the rod's own line, a distance d beyond one end, E = kQ/(d(d + L)); on its perpendicular bisector, a distance y from the rod, E = kQ/(y√(y² + L²/4)). Both approach kQ/r² far from the rod.
Students often think The field of a continuous charge distribution is found by adding (integrating) the magnitudes k dq/r² of its elements’ fields. In fact Only where every element’s field points in the same direction, as at points on a straight rod’s own line. Otherwise the contributions dE⃗ point in different directions and must be added as vectors, by integrating each component separately.
Students often think A charged rod, ring or arc produces the field of a point charge, with the same total charge, located at its center. In fact No. Its elements are at different distances from the field point, and their fields point in different directions. The field approaches that of a point charge only at distances much larger than the object’s size.
8.4.A.2 Symmetry cancellation Fix
Symmetry cancellation
If every element of a distribution has a mirror-image partner, with the same charge, across a line through the field point, the two elements' field components perpendicular to that line cancel. Only the components along the line then need to be integrated.
Field at the center of a semicircular arc
For a thin semicircular arc of radius R with uniform density λ (total charge Q = πRλ), E = 2kλ/R = 2kQ/(πR²) at its center, directed along the arc's line of symmetry, away from the arc when the charge is positive.
Students often think The field is zero at the center of any uniformly charged ring or arc, and at points near the center of a ring, because the contributions from all around cancel. In fact No. At the exact center of a complete ring, every element has an identical partner directly opposite, so the fields cancel. A semicircle has no charge on its missing half, so its field at the center is not zero; and at points on a ring’s axis near its center, the axial components add to a small but nonzero field.
Students often think At a point on the perpendicular bisector of a rod, the components along the rod cancel whatever the charge on each half, so the field is perpendicular to the rod. In fact Only if mirror-image elements carry the same charge. If the rod’s two halves carry opposite charges, the components perpendicular to the rod cancel instead, and the field on the bisector points along the rod.
14 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 14
A thin, insulating rod is bent into a quarter of a circle of radius R and carries charge +Q spread uniformly along its length. What is the magnitude of the electric field at the center of the circle? (k = 1/(4πε₀).)
Answer and reasoning
A1.00 kQ/R² A student who integrates the magnitudes of the contributions, as if they all pointed the same way, picks this: ∫kλR dθ/R² over the quarter circle gives kλ(π/2)/R = kQ/R². The contributions point in directions spread over 90°, so they must be added by components.
B0.64 kQ/R² A student who assumes that one component cancels, as it does at the center of a semicircle, picks this, keeping only kλ/R = 2kQ/(πR²). A quarter circle has no charge on the other side of either axis to cancel a component; here Ex = Ey.
C0.90 kQ/R²Correct Each element, dq = λR dθ with λ = 2Q/(πR), is a distance R from the center and gives dE = kλ dθ/R. Integrating over the quarter circle gives Ex = Ey = kλ/R; the arc is symmetric only about its 45° line, so neither component cancels. E = √(Ex² + Ey²) = √2 kλ/R = 2√2kQ/(πR²) ≈ 0.90 kQ/R².
D1.27 kQ/R² A student who finds both components correctly but adds them as numbers picks this: Ex + Ey = 2kλ/R = 4kQ/(πR²). The components are perpendicular, so the magnitude is √(Ex² + Ey²) = √2 kλ/R ≈ 0.90 kQ/R².
Working Center at the origin, arc from θ = 0 to θ = π/2. λ = Q/(πR/2) = 2Q/(πR); an element dq = λR dθ at distance R gives dE = kλ dθ/R, directed from the element toward the center. Ex = (kλ/R)∫0π/2 cosθ dθ = kλ/R and Ey = (kλ/R)∫0π/2 sinθ dθ = kλ/R. The arc is symmetric only about the 45° line, so neither component cancels. E = √2 kλ/R = 2√2kQ/(πR²) ≈ 0.90 kQ/R². Distractors: magnitudes integrated, kλ(π/2)/R = kQ/R²; one component only, as for a semicircle, kλ/R = 2kQ/(πR²) ≈ 0.64 kQ/R²; components added as numbers, 2kλ/R = 4kQ/(πR²) ≈ 1.27 kQ/R².
A thin rod of length L carries charge +Q spread uniformly along its length. Point P is on the line of the rod, a distance d beyond one end. What is the magnitude of the electric field at P? (k = 1/(4πε₀).)
Answer and reasoning
AkQ/(d+½L)² A student who treats the rod as a point charge at its center picks this. The elements nearer P have stronger fields and the farther ones weaker, and these do not average to the field of a point charge at the center; the two agree only when d ≫ L.
B2kQ/(dL) A student who uses the field of an infinitely long line, 2kλ/r, with λ = Q/L and r = d picks this. That result holds beside a very long line, at a distance measured perpendicular to it; P is on the rod’s own line, beyond its end.
CkQL/(d(L+d)) A student who writes dq = Q dx instead of dq = (Q/L) dx picks this. The charge on a length dx is λ dx, with λ = Q/L; the extra factor of L leaves the result in N·m/C instead of N/C.
DkQ/(d(L+d))Correct Every element lies on the line through P, so all the contributions point the same way and add as numbers. With dq = (Q/L) dx and x measured from P, E = ∫dd+L k(Q/L) dx/x² = (kQ/L)[1/d − 1/(d + L)] = kQ/(d(L + d)). For d ≫ L this becomes kQ/d², as it should.
Working Measure x from P along the rod’s line: the rod runs from x = d to x = d + L. dq = (Q/L) dx; every element’s field at P points along the line, away from the rod, so the magnitudes add. E = ∫dd+L k(Q/L) dx/x² = (kQ/L)[1/d − 1/(d + L)] = kQ/(d(L + d)). Check (sympy): for d ≫ L, E → kQ/d². Distractors: point charge at the rod’s center, kQ/(d + ½L)²; infinite-line result 2kλ/d with λ = Q/L, 2kQ/(dL); dq = Q dx, ∫dd+L kQ dx/x² = kQL/(d(L + d)), in N·m/C rather than N/C.
A very long, thin, straight wire has uniform linear charge density λ. A student finds the magnitude of the electric field at a perpendicular distance a from the wire by integrating E⃗ = (1/(4πε₀))∫(dq/r²)r̂ along the whole wire. What result should the student obtain?
Answer and reasoning
Aλ/(4ε₀a) A student who integrates the magnitudes kλ dx/(x² + a²) without resolving them picks this: the integral is πkλ/a = λ/(4ε₀a). The contributions of different elements point in different directions; only their components perpendicular to the wire add.
Bλ/(2πε₀a)Correct Elements at equal distances on either side of the point nearest P cancel each other’s components along the wire, so only the perpendicular components add: E = ∫ kλa dx/(x² + a²)3/2, from −∞ to ∞, = 2kλ/a = λ/(2πε₀a). The field falls off as 1/a, because the wire extends in both directions.
Cλ/(4πε₀a²) A student who thinks a line of charge produces a field that falls off as 1/r², like a point charge, picks this: kλ/a². The units show the error, N/(C·m) instead of N/C. Adding the contributions along an infinite line gives a field proportional to 1/a.
D0 A student who integrates the component of each dE⃗ along the wire picks this. That component does cancel between elements on either side of the point nearest P, but the perpendicular component does not, and it gives 2kλ/a.
Working Wire along the x-axis; P a distance a from the origin, on a line perpendicular to the wire. dq = λ dx at r = √(x² + a²). Components along the wire from elements at x and −x cancel; perpendicular components add, each dE · a/r. E = ∫−∞∞ kλa dx/(x² + a²)3/2 = kλa[x/(a²√(x² + a²))]−∞∞ = 2kλ/a = λ/(2πε₀a). Distractors (sympy-checked): magnitudes integrated, ∫kλ dx/(x² + a²) = πkλ/a = λ/(4ε₀a); point-charge form with λ in place of q, kλ/a² = λ/(4πε₀a²); component along the wire integrated, 0.
A thin ring of radius R carries charge +Q spread uniformly around it. Point P is on the axis of the ring, a distance R from its center. What is the magnitude of the electric field at P? (k = 1/(4πε₀).)
Answer and reasoning
A0.35 kQ/R²Correct Each element is √2 R from P. Elements on opposite sides of the ring cancel each other’s components perpendicular to the axis, so only the axial components add, each k dq/(2R²) times cosθ = 1/√2. So E = kQ/(2√2 R²) ≈ 0.35 kQ/R².
B0.50 kQ/R² A student who takes the field to be kQ/r² because every element is the same distance, √2 R, from P picks this: kQ/(2R²). The contributions point in different directions around a cone; only their axial components, a fraction 1/√2 of each, add.
C1.00 kQ/R² A student who treats the ring as a point charge at its center, a distance R from P, picks this. None of the ring’s charge is at the center: every element is √2 R from P, and the components perpendicular to the axis cancel.
D2.22 kQ/R² A student who writes dq = Q dφ, where φ is the angle around the ring, picks this: integrating from 0 to 2π gives 2π times the correct field. The charge on an element is dq = (Q/2π) dφ, the fraction of the ring it occupies times Q.
Working Every element is r = √(R² + R²) = √2 R from P. Elements on opposite sides of the ring cancel each other’s components perpendicular to the axis; the axial components add, each k dq/r² × cosθ with cosθ = x/r = 1/√2. E = kQx/(x² + R²)3/2 at x = R: kQR/(2R²)3/2 = kQ/(2√2 R²) ≈ 0.35 kQ/R². Distractors: all the charge treated as a point charge at distance √2 R, kQ/(2R²) = 0.50 kQ/R²; point charge at the ring’s center, kQ/R² = 1.00 kQ/R²; dq = Q dφ (φ the angle around the ring), which gives 2π times the correct field, ≈ 2.22 kQ/R².
The figure shows a thin, straight rod with charge spread uniformly along its length, and a point P on the rod’s perpendicular bisector. What is the magnitude of the electric field at P? Use k = 9.0 × 10⁹ N·m²/C².
Answer and reasoning
A3.2 × 10³ N/C A student who treats the rod as a point charge at its midpoint picks this: k(8.0 × 10⁻⁹ C)/(0.15 m)². Most of the rod’s charge is farther from P than the midpoint is, and its field points partly along the rod, so the rod’s field at P is smaller.
B1.9 × 10³ N/CCorrect Pairs of elements at equal distances on either side of the midpoint cancel each other’s components along the rod, so only the perpendicular components add. Integrating gives E = kQ/(y√(y² + L²/4)); here √(y² + L²/4) = √((0.15 m)² + (0.20 m)²) = 0.25 m, so E = (9.0 × 10⁹)(8.0 × 10⁻⁹ C)/((0.15 m)(0.25 m)) = 1.9 × 10³ N/C.
C2.4 × 10³ N/C A student who uses the field of an infinitely long line, 2kλ/r with λ = Q/L = 2.0 × 10⁻⁸ C/m, picks this. The rod is only 0.40 m long, not much longer than P’s distance from it, so it produces less field than an infinite line would.
D2.2 × 10³ N/C A student who integrates the magnitudes k dq/r² without resolving them picks this: (2kQ/(Ly))arctan(L/(2y)) = 2.2 × 10³ N/C. The components along the rod cancel in pairs and only the perpendicular components add, so the field is smaller.
Working Rod from x = −L/2 to L/2 with L = 0.40 m; P at y = 0.15 m on the perpendicular bisector. Components along the rod cancel in pairs; perpendicular components add: E = ∫ kλy dx/(x² + y²)3/2 from −L/2 to L/2 = kQ/(y√(y² + L²/4)) = (9.0 × 10⁹)(8.0 × 10⁻⁹ C)/((0.15 m)(0.25 m)) = 1.9 × 10³ N/C. Distractors: point charge at the midpoint, kQ/y² = 3.2 × 10³ N/C; infinite line with λ = Q/L = 2.0 × 10⁻⁸ C/m, 2kλ/y = 2.4 × 10³ N/C; magnitudes integrated, (2kQ/(Ly))arctan(L/(2y)) = 2.2 × 10³ N/C.
A thin, insulating rod is bent into a semicircle of radius 0.15 m and carries charge +4.0 nC spread uniformly along its length. What is the magnitude of the electric field at the center of the semicircle? Use k = 9.0 × 10⁹ N·m²/C².
Answer and reasoning
A1.0 × 10³ N/CCorrect Elements on either side of the semicircle’s line of symmetry cancel each other’s components across it, so only the components along it add: E = (kλ/R)∫0π sinθ dθ = 2kλ/R, with λ = Q/(πR). So E = 2kQ/(πR²) = 2(9.0 × 10⁹)(4.0 × 10⁻⁹ C)/(π(0.15 m)²) = 1.0 × 10³ N/C.
B1.6 × 10³ N/C A student who takes the field to be kQ/R² because all the charge is a distance R from the center picks this. The elements’ fields point in different directions and the components across the line of symmetry cancel, so the field is smaller by a factor of 2/π.
C3.2 × 10³ N/C A student who writes dq = Q dθ picks this: (kQ/R²)∫0π sinθ dθ = 2kQ/R². An element’s charge is λR dθ = (Q/π) dθ, so this result is too large by a factor of π.
D6.8 × 10³ N/C A student who writes dq = λ dθ, leaving out the radius in the arc length R dθ, picks this: 2kλ/R² = 2kQ/(πR³). An element’s length is R dθ; without the R the result is in N/(C·m), not N/C.
Working λ = Q/(πR). Element dq = λR dθ at distance R: dE = kλ dθ/R. The components across the semicircle’s line of symmetry cancel in pairs; E = (kλ/R)∫0π sinθ dθ = 2kλ/R = 2kQ/(πR²) = 2(9.0 × 10⁹)(4.0 × 10⁻⁹ C)/(π(0.15 m)²) = 1.0 × 10³ N/C. Distractors: kQ/R² = 1.6 × 10³ N/C (all charge at distance R, directions ignored); dq = Q dθ, 2kQ/R² = 3.2 × 10³ N/C; dq = λ dθ, 2kλ/R² = 2kQ/(πR³) = 6.8 × 10³ (units N/(C·m)).
A thin, insulating rod lies along the x-axis. The graph shows the rod’s linear charge density λ as a function of position x; the graph begins and ends at the ends of the rod. What is the magnitude of the electric field at the origin, x = 0? Use k = 9.0 × 10⁹ N·m²/C².
Answer and reasoning
A4.5 × 10² N/C A student who finds the total charge correctly, ∫αx dx = 2.0 × 10⁻⁹ C, and treats it as a point charge at the rod’s midpoint, 0.20 m from the origin, picks this. Each element’s field depends on 1/x², so charge near the origin counts for more than charge far from it, and the rod is not equivalent to a point charge at its midpoint.
B6.0 × 10² N/C A student who treats the density as uniform at its average value, 1.0 × 10⁻⁸ C/m, and uses the result for a uniform rod picks this. Here the density is smallest where the elements are closest to the origin, where they count for most, so the field is less than for a uniform rod.
C1.8 × 10³ N/C A student who thinks the field depends only on the total charge and the distance to the nearest point of the rod picks this: k(2.0 × 10⁻⁹ C)/(0.10 m)². Only the charge at x = 0.10 m is that close; the rest is farther away and contributes less.
D4.9 × 10² N/CCorrect The graph gives λ = αx with α = 5.0 × 10⁻⁸ C/m², from x = 0.10 m to x = 0.30 m. Every element lies on the x-axis on the same side of the origin, so the contributions point the same way and add: E = ∫ kαx dx/x² = kα ln(0.30/0.10) = (9.0 × 10⁹)(5.0 × 10⁻⁸)(1.10) = 4.9 × 10² N/C.
Working The graph is a straight line that, extended, passes through the origin: λ = αx with α = (15 × 10⁻⁹ C/m)/(0.30 m) = 5.0 × 10⁻⁸ C/m², for x from 0.10 m to 0.30 m. All the elements are on the x-axis on the same side of the origin, so their fields there all point along −x and the magnitudes add: E = ∫0.100.30 kαx dx/x² = kα ln(0.30/0.10) = (9.0 × 10⁹)(5.0 × 10⁻⁸)(ln 3) = 4.9 × 10² N/C. Distractors: total charge Q = ∫αx dx = 2.0 × 10⁻⁹ C as a point charge at the midpoint (0.20 m), 4.5 × 10² N/C; uniform rod at the average density 1.0 × 10⁻⁸ C/m, kλL/(d(d + L)) = 6.0 × 10² N/C; total charge at the nearest end (0.10 m), 1.8 × 10³ N/C.
A thin ring of radius R carries charge +Q spread uniformly around it. Point P is on the ring’s axis, a distance x from its center, where x is much smaller than R. P is moved along the axis to a distance 2x from the center, still much smaller than R. How does the magnitude of the electric field at P change?
Answer and reasoning
AIt decreases to one-quarter of the value it had at distance x. A student who treats the ring as a point charge at its center picks this, using E ∝ 1/x². No charge is at the center; every element is about R from P, and near the center the axial components grow with x.
BIt stays about the same, as every element is still about R away. A student who takes the field to be kQ/(x² + R²) ≈ kQ/R², because every element is about the same distance away, picks this. The elements’ fields point in different directions; only their small axial components add, and those grow in proportion to x.
CIt increases to about twice the value it had at distance x.Correct Near the center, x² + R² ≈ R², so the on-axis field E = kQx/(x² + R²)3/2 becomes E ≈ kQx/R³. The perpendicular components always cancel, and the axial components grow in proportion to x, so doubling x doubles the field.
DIt stays zero, since all the contributions cancel near the center. A student who thinks the ring’s field is zero near its center, as it is at the center itself, picks this. Only at the center are the contributions exactly opposite; slightly off center along the axis they all have axial components in the same direction.
Working On the axis, E = kQx/(x² + R²)3/2. For x ≪ R, x² + R² ≈ R², so E ≈ kQx/R³: near the center the field is proportional to x (zero at the center itself, where the contributions cancel). Doubling x doubles E; sympy: lim(x→0) E(2x)/E(x) = 2. Distractors: point charge at the center, E ∝ 1/x² → one-quarter; charge all at about distance R, kQ/(x² + R²) ≈ kQ/R² → about the same; contributions assumed to cancel near the center → stays zero.
A thin rod of length L, with charge +Q spread uniformly along it, lies on the x-axis from x = 0 to x = L. The magnitude of the electric field at point P, at x = −L, is E₁. The rod is replaced by a rod of length 2L, with the same charge +Q spread uniformly, lying from x = 0 to x = 2L. The magnitude of the field at P is now E₂. What is the ratio E₂/E₁?
Answer and reasoning
A0.67Correct At a point a distance d beyond the near end of a uniform rod of length ℓ, integration gives E = kQ/(d(d + ℓ)). Here d = L both times: E₁ = kQ/(2L²) and E₂ = kQ/(3L²), so E₂/E₁ = 2/3 ≈ 0.67. Spreading the same charge over a longer rod puts less of it near P.
B0.56 A student who treats each rod as a point charge at its midpoint picks this: the midpoint moves from 1.5L to 2L from P, giving (1.5/2)² ≈ 0.56. For a point on the rod’s own line, the charge near P counts for more than a point charge at the midpoint would suggest.
C0.50 A student who uses the infinite-line field 2kλ/d, with λ = Q/ℓ, picks this: doubling the length halves λ, which would halve the field. That formula applies beside a very long line, not at a point on the rod’s own line beyond its end.
D1.00 A student who thinks the field depends only on the total charge and the distance to the rod’s nearest point picks this, since neither changes. Spreading the charge over a longer rod moves part of it farther from P, so the field decreases.
Working For a uniform rod of length ℓ and charge Q, at a point on its line a distance d beyond its near end, E = ∫dd+ℓ k(Q/ℓ) dx/x² = kQ/(d(d + ℓ)). Here d = L both times: E₁ = kQ/(L · 2L) = kQ/(2L²) and E₂ = kQ/(L · 3L) = kQ/(3L²). E₂/E₁ = 2/3 ≈ 0.67. Distractors: point charge at each rod’s midpoint, (1.5L/2L)² = 9/16 ≈ 0.56; infinite-line result 2kλ/d with λ = Q/ℓ, which halves: 0.50; total charge and nearest distance both unchanged: 1.00.
The figure shows an end view of two very long, parallel lines of charge, perpendicular to the page, each with the same uniform positive linear charge density, and three points P₁, P₂ and P₃ in the plane of the page. E₁, E₂ and E₃ are the magnitudes of the electric field at the three points. Which ranking is correct?
Answer and reasoning
AE₁ > E₃ > E₂ A student who adds the magnitudes of the two fields at each point picks this: 4kλ/a at P₁, about 2.8kλ/a at P₃ and 2.7kλ/a at P₂. The fields are vectors: at P₁ they are opposite and cancel, and at P₃ they partly cancel.
BE₂ > E₁ = E₃ A student who thinks the field is zero everywhere on the perpendicular bisector between two like charges picks this. Only at P₁ are the two fields exactly opposite; at P₃ their components along the bisector both point away from the lines and add to 2kλ/a.
CE₁ = E₂ > E₃ A student who ranks the points by their distance from the nearest line picks this: P₁ and P₂ are each a from a line, P₃ is √2 a from both. What matters is how the fields of both lines combine as vectors; at P₁ they cancel completely.
DE₂ > E₃ > E₁Correct Each line’s field is 2kλ/r, pointing away from it. At P₁ the two fields are equal and opposite: E₁ = 0. At P₂ both point the same way: E₂ = 2kλ/a + 2kλ/(3a) ≈ 2.7kλ/a. At P₃, √2 a from each line, the components along the line joining the charges cancel and the components along the bisector add: E₃ = 2kλ/a. So E₂ > E₃ > E₁.
Working Each line gives 2kλ/r, directed away from it. P₁ (a from each line, fields opposite): E₁ = 0. P₂ (a and 3a from the lines, both fields pointing to the right): E₂ = 2kλ/a + 2kλ/(3a) = (8/3)kλ/a ≈ 2.7kλ/a. P₃ (√2 a from each line): components along the line joining the charges cancel; components along the bisector add: E₃ = 2 × (2kλ/(√2 a)) × (1/√2) = 2kλ/a. So E₂ > E₃ > E₁. Distractors: magnitudes added (4, 2.7, 2.8 kλ/a) → E₁ > E₃ > E₂; field zero all along the bisector → E₂ > E₁ = E₃; ranked by distance to the nearest line (a, a, √2 a) → E₁ = E₂ > E₃.
The figure shows a thin rod whose two halves carry the charges labeled, each spread uniformly, and a point P on the rod’s perpendicular bisector. Which claim about the direction of the electric field at P, with its justification, is correct?
Answer and reasoning
AZero: the rod’s net charge is zero, so the fields produced by its two halves cancel each other at point P. A student who thinks zero net charge means zero field picks this. The +Q and −Q are in different places, so their fields at P are not opposite: both have components along the rod toward the −Q half, and those add.
BParallel to the rod, toward the −Q half: mirror-image pairs’ components cancel across the rod and add along it.Correct Pair each element of the +Q half with its mirror image in the −Q half. The positive element’s field at P points away from it, up and toward the −Q half; the negative element’s field points toward it, down and toward the −Q half. The two are equal in magnitude, so their perpendicular components cancel and their components along the rod add.
CPerpendicular to the rod, away from it: at a point on the perpendicular bisector, the components along the rod cancel. A student who applies the result for a uniformly charged rod picks this. The components along the rod cancel only when mirror-image elements carry the same charge; here they carry opposite charges, so the perpendicular components cancel instead.
DParallel to the rod, toward the +Q half: the field at P points toward the half of the rod that holds positive charge. A student who thinks electric fields point toward positive charge picks this. The field points away from positive charge and toward negative charge, so each pair of elements gives a field at P toward the −Q half.
Working Pair an element of the +Q half at distance x left of the midpoint with the mirror-image element of the −Q half at x to the right. The + element’s field at P points away from it (up and to the right); the − element’s field points toward it (down and to the right), with equal magnitude. The perpendicular components cancel; the components along the rod, both toward the −Q half, add. Every pair does the same, so E⃗ at P is parallel to the rod, toward the −Q half.
The figure shows three thin, insulating arcs of the same radius, each with charge spread uniformly over the parts labeled. E₁, E₂ and E₃ are the magnitudes of the electric field at the centers of Arcs 1, 2 and 3. Which ranking is correct?
Answer and reasoning
AE₂ > E₁ = E₃ A student who thinks zero net charge gives zero field picks this for Arc 3. The +Q/2 and −Q/2 are on different quarters, so their fields at the center are not opposite: both have components pointing from the positive quarter toward the negative one.
BE₃ > E₁ = E₂ A student who thinks the field at the center of any uniformly charged arc is zero, as for a ring, picks this. Only the ring has an identical element opposite every element; the semicircle’s contributions all have components along its line of symmetry, and those add.
CE₂ = E₃ > E₁Correct In the ring, elements on opposite sides cancel: E₁ = 0. In Arc 2 the components across the line of symmetry cancel and those along it add, giving 2kQ/(πR²). In Arc 3 each quarter has the same size of density, Q/(πR), but mirror-image elements have opposite signs, so the components along the line of symmetry cancel and those across it add, again 2kQ/(πR²).
DE₁ = E₂ = E₃ A student who adds the magnitudes k dq/R² of the elements’ fields picks this: every element of each arc is R from its center and each arc carries charge of total size Q, so each sum is kQ/R². The contributions point in different directions: in the ring they cancel completely, and in Arcs 2 and 3 only the components along one direction add, giving 2kQ/(πR²).
Working Arc 1 (ring): every element has an identical partner opposite: E₁ = 0. Arc 2 (semicircle, +Q, λ = Q/(πR)): components across its line of symmetry cancel; E₂ = 2kλ/R = 2kQ/(πR²). Arc 3: each quarter has |λ| = (Q/2)/(πR/2) = Q/(πR); mirror-image elements have opposite charges, so the components along the line of symmetry cancel and the components across it add: E₃ = (kλ/R)[∫π/2π (−cosθ)dθ + ∫0π/2 cosθ dθ] = 2kλ/R = 2kQ/(πR²), pointing from the + quarter toward the − quarter (sympy-checked). So E₂ = E₃ > E₁. Distractor: magnitudes integrated → kQ/R² for every arc → E₁ = E₂ = E₃.
A thin, insulating rod is bent into a semicircle of radius R. One quarter of the semicircle carries charge +Q and the other quarter carries charge +3Q, each spread uniformly along its own length. What is the magnitude of the electric field at the center of the semicircle? (k = 1/(4πε₀).)
Answer and reasoning
A2.55 kQ/R² A student who assumes the components along the diameter cancel, as they do for a uniformly charged semicircle, picks this, keeping only |Ey| = 8kQ/(πR²) ≈ 2.55 kQ/R². That cancellation needs mirror-image elements with equal charge; here one quarter carries three times the charge of the other, leaving |Ex| = 4kQ/(πR²).
B2.85 kQ/R²Correct Each quarter's field at the center is found by integrating the components of k dq/R², with λ = 6Q/(πR) on one quarter and 2Q/(πR) on the other. The components along the diameter do not cancel, because the quarters carry different charges: |Ex| = 4kQ/(πR²) and |Ey| = 8kQ/(πR²), so |E| = 4√5kQ/(πR²) ≈ 2.85 kQ/R².
C3.82 kQ/R² A student who adds the two components, 4kQ/(πR²) + 8kQ/(πR²) = 12kQ/(πR²) ≈ 3.82 kQ/R², picks this. Perpendicular components combine as √(Ex² + Ey²), giving 4√5kQ/(πR²) ≈ 2.85 kQ/R².
D4.00 kQ/R² A student who integrates the magnitudes k dq/R² of all the elements' fields picks this: every element is R from the center, so the sum is k(4Q)/R². The elements' fields point in different directions and partly cancel; their components must be integrated separately.
Working Put the center at the origin with the diameter along the x-axis; the +3Q quarter runs from θ = 0 to π/2 and the +Q quarter from π/2 to π. Each element dq = λR dθ is R from the center, and its field there points along −(cos θ, sin θ). On each quarter λ = (its charge)/(πR/2). +3Q quarter, λ = 6Q/(πR): E⃗ = −(kλ/R)(∫cos θ dθ, ∫sin θ dθ) from 0 to π/2 = −(6kQ/(πR²))(1, 1). +Q quarter, λ = 2Q/(πR): E⃗ = −(kλ/R)(∫cos θ dθ, ∫sin θ dθ) from π/2 to π = −(2kQ/(πR²))(−1, 1). Sum: Ex = −4kQ/(πR²), Ey = −8kQ/(πR²); |E| = (kQ/(πR²))√(16 + 64) = 4√5kQ/(πR²) ≈ 2.85 kQ/R². The components along the diameter do not cancel, because the two quarters carry different charges. Distractors (sympy-checked): components along the diameter assumed to cancel, as for a uniform semicircle, |Ey| = 8kQ/(πR²) ≈ 2.55 kQ/R²; magnitude taken as |Ex| + |Ey| = 12kQ/(πR²) ≈ 3.82 kQ/R²; magnitudes k dq/R² integrated, k(4Q)/R² = 4.00 kQ/R².
A very long, straight wire has uniform linear charge density +λ. A thin, insulating rod of length L carries charge +Q spread uniformly along its length. The rod lies on a line that meets the wire at a right angle, with the rod's near end a distance a from the wire. What is the magnitude of the electric force exerted on the rod by the wire?
Answer and reasoning
A(λQ/(4ε₀L))ln(1+L/a) A student who finds the wire's field by integrating the magnitudes k dq/r² along the wire, getting λ/(4ε₀r), picks this. The components along the wire cancel and only the perpendicular components add, giving λ/(2πε₀r) and a force (λQ/(2πε₀L))ln(1 + L/a).
B(λQ/(2πε₀L))ln(1+L/a)Correct The wire's field at distance r is λ/(2πε₀r), along the rod. Each element dq = (Q/L) dr experiences dF = λ dq/(2πε₀r), all in the same direction, so F = (λQ/(2πε₀L))∫dr/r from a to a + L = (λQ/(2πε₀L))ln(1 + L/a).
C(λQ/(4πε₀L))(1/a−1/(a+L)) A student who takes the wire's field to fall off as 1/r², like a point charge's, λ/(4πε₀r²), picks this: integrating along the rod gives (λQ/(4πε₀L))(1/a − 1/(a + L)), which is in N/m, not N. The field of a long wire falls off as 1/r.
D(λQ/(8πε₀L))(8L/(2a+L)) A student who treats the rod as a point charge Q at its center, a + L/2 from the wire, picks this: λQ/(2πε₀(a + L/2)), written here as (λQ/(8πε₀L))(8L/(2a+L)) in the same form as the other options. The wire's field is stronger near the wire, so each element's force must use the field at its own position, and the forces must be integrated.
Working Integrating along the whole wire gives its field at perpendicular distance r: E = λ/(2πε₀r), directed away from the wire, which is along the rod. An element of the rod of length dr at distance r carries dq = (Q/L) dr and experiences dF = E dq; all these forces point the same way, so F = ∫aa+L (λ/(2πε₀r))(Q/L) dr = (λQ/(2πε₀L))ln((a + L)/a) = (λQ/(2πε₀L))ln(1 + L/a). Distractors (sympy-checked): the wire's field from integrated magnitudes, λ/(4ε₀r) ⇒ (λQ/(4ε₀L))ln(1 + L/a); the wire's field taken ∝ 1/r², λ/(4πε₀r²) ⇒ (λQ/(4πε₀L))(1/a − 1/(a + L)), in N/m rather than N; the rod's charge placed at its center, a + L/2 from the wire ⇒ λQ/(2πε₀(a + L/2)) = (λQ/(8πε₀L))(8L/(2a + L)). Units: λQ/(ε₀L) is (C/m)(C)/((C²/(N·m²))(m)) = N.
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
