5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A small charged sphere is fixed inside a chamber from which all the air has been pumped out. No other charged objects are nearby. Which statement about the electric field produced by the sphere is correct?
Answer and reasoning
AIt appears at a point once a second charge is placed there to feel it A student who thinks a field needs a charge to be present picks this. The sphere's field is already at every point; a second charge placed at a point experiences a force because the field is there.
BIt is absent from the chamber, as a field needs air or other matter to carry it A student who thinks a field needs a material to pass through picks this. Electric fields exist in a vacuum; electric forces act across empty space, for example between the particles of an atom.
CIt exists only along the field lines that can be drawn outward from the sphere A student who reads field lines as the only places where the field exists picks this. Field lines are a model drawn at a sample of places; the field exists at every point around the sphere, including between them.
DIt exists at every point around the sphere, with or without another charge thereCorrect A charged object produces an electric field in the space around it, at every point, whether or not a charge is there to experience it; a charge placed at a point only reveals the field by experiencing a force. The field does not need air: it exists in a vacuum.
A small sphere with a negative charge is isolated, far from other charged objects. An electron is placed at point P, a short distance east of the sphere. Which describes the electric field produced by the sphere at P and the electric force on the electron?
Answer and reasoning
AField away from the sphere; force on the electron toward the sphere A student who thinks every field points away from its source picks this. The field of a negative charge points toward it. The force on the electron is repulsive, away from the negatively charged sphere.
BField toward the sphere; force on the electron away from itCorrect The field of an isolated negative charge points toward it, the direction a positive test charge would be pulled; at P that is west, toward the sphere. The electron is negative, so the force on it is opposite to the field: east, away from the sphere. Like charges repel.
CField toward the sphere; force on the electron toward the sphere A student who thinks every charge is pushed along the field picks this. The field direction is right, but the force on a negative charge is opposite to the field: the electron is repelled, away from the sphere.
DField away from the sphere; force on the electron away from the sphere A student who takes the field's direction from the force on the electron actually at P picks this. The force is away from the sphere, but the field is defined with a positive test charge, which would be pulled toward the sphere.
The diagram shows two small charged spheres, A and B, fixed in place, and point P. Treat the spheres as point charges, and use k = 9.0 × 10⁹ N·m²/C². What is the magnitude of the net electric field at P?
Answer and reasoning
A1.1 × 10³ N/C A student who adds the magnitudes of the two fields picks this: 4.0 × 10² + 6.75 × 10² N/C. The fields point in perpendicular directions, so they combine as vectors, by the Pythagorean theorem.
B3.5 × 10² N/C A student who uses the distance between A and B, 0.36 m, for both fields picks this: about 2.8 × 10² and 2.1 × 10² N/C, combined to 3.5 × 10² N/C. Each field depends on the distance from its own sphere to P: 0.30 m for A and 0.20 m for B.
C7.8 × 10² N/CCorrect A is 0.30 m from P: EA = kq/r² = (9.0 × 10⁹)(4.0 × 10⁻⁹)/(0.30)² = 4.0 × 10² N/C, pointing away from A (to the right). B is 0.20 m from P: EB = (9.0 × 10⁹)(3.0 × 10⁻⁹)/(0.20)² = 6.75 × 10² N/C, pointing away from B (up). The fields are perpendicular: |E⃗| = √((4.0 × 10²)² + (6.75 × 10²)²) N/C = 7.8 × 10² N/C.
D1.8 × 10² N/C A student who divides by r instead of r² picks this: 120 N/C and 135 N/C, combined to 1.8 × 10² N/C. The field of a point charge falls off as 1/r², and only then are the units N/C.
A solid metal sphere on an insulating stand is given excess negative charge and then reaches electrostatic equilibrium. Which describes where the excess charge is and the electric field inside the metal?
Answer and reasoning
AThrough its volume, with a nonzero field inside the metal A student who pictures charge soaking into a conductor picks this. Charge spread through the volume would produce a field inside the metal and push the free electrons outward; they move until they are all on the surface and the field inside is zero.
BOn its outer surface, with a nonzero field inside the metal A student who applies the point-charge field inside the sphere picks this. The charge location is right, but a field inside the metal would move its free electrons; in equilibrium the field inside is zero.
COn its outer surface, with zero field inside the metalCorrect The excess electrons repel one another and move freely through the metal, so they spread as far apart as possible: onto the outer surface. In equilibrium no charges inside the metal move, which requires the field inside the metal to be zero.
DThrough its volume, with zero field inside the metal A student who thinks evenly spread charge gives balanced pushes everywhere inside picks this. Charge spread through a volume gives a nonzero field at off-center points; the field inside a conductor is zero because its excess charge has moved to the surface.
A solid metal sphere and a solid plastic sphere have the same radius R and the same excess positive charge. The metal sphere is in electrostatic equilibrium; the plastic sphere's charge is spread uniformly throughout its volume. Both are far from other charged objects. Emetal and Eplastic are the magnitudes of the electric fields at a point inside each sphere, a distance R/2 from its center. Which is correct?
Answer and reasoning
AEplastic = Emetal = 0 A student who applies the conductor result to every charged object picks this. The field inside a conductor is zero because its charges move until it is; the plastic's charges cannot move, and the field inside it is not zero.
BEplastic = Emetal > 0 A student who uses the point-charge model inside both spheres picks this. The model holds only outside a sphere. Inside the metal the field is zero, and inside the plastic it is smaller than the point-charge value.
CEmetal > Eplastic > 0 A student who thinks a conductor's excess charge gathers at its center picks this, expecting a strong field inside the metal. The metal's excess charge is on its surface and the field inside the metal is zero.
DEplastic > Emetal = 0Correct In the metal the free charges move to the surface and the field inside the metal is zero. The plastic's charges cannot move, so charge remains throughout its interior, and at R/2 more of it pushes outward than inward: the field there is nonzero.
Working Metal (conductor, equilibrium): excess charge on the surface, Emetal = 0 inside. Plastic (insulator): charge stays throughout the volume; at r = R/2 the charge nearer the center than the point is not balanced by the rest, so Eplastic > 0 (directed outward). Hence Eplastic > Emetal = 0.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.3.A.1 Electric field Fix
Electric field
A vector quantity defined at every point in the space around charged objects. A charge placed at the point experiences an electric force; the field is there whether or not a charge is placed at the point.
Source charge
A charged object that produces an electric field. The field at a point depends on the source charges and their positions, not on any charge placed at the point to detect it.
Students often think An electric field exists at a point only while a charge is there to experience it. In fact No. A charged object produces a field at every point around it. A charge placed at a point only reveals the field there by experiencing a force.
Students often think An electric field needs air or another material around a charged object to carry it. In fact No. Electric fields exist in a vacuum as well as in materials; electric forces act across empty space, for example between the charged particles in an atom.
8.3.A.2 Electric field (definition) Fix
Electric field (definition)
E⃗ = F⃗E/q: the electric force exerted on a test charge at a point divided by the charge of the test charge. SI unit: N/C, equivalently V/m.
Force on a charge in a field
F⃗E = qE⃗. A particle of charge q and mass m in a field E⃗ (with no other forces) has acceleration a⃗ = qE⃗/m, so particles with different charge-to-mass ratios accelerate differently in the same field.
Test charge
A point charge of small enough magnitude that its presence does not significantly change the electric field being measured, for example by moving the charges on a nearby conductor.
Direction of the field of a point charge
The electric field points away from an isolated positive charge and toward an isolated negative charge, along the line through the charge.
Direction of the electric force
The force on a positive charge is in the direction of the field; the force on a negative charge is opposite to the field.
Coulomb constant, k
k = 1/(4πε₀) = 9.0 × 10⁹ N·m²/C². Combining E⃗ = F⃗E/q with Coulomb's law gives the magnitude of the field of a point charge Q at distance r: |E⃗| = k|Q|/r².
Students often think The electric field at a point is the force exerted there, and it is the same whatever charge is placed at the point. In fact No. The field is the force per unit charge, E⃗ = F⃗E/q, in N/C. The force on a charge at the point is F⃗E = qE⃗, which depends on the charge placed there; the field does not.
Students often think Rearranging E = FE/q gives the force as the field divided by the charge, FE = E/q. In fact No. Rearranging E = FE/q gives FE = qE: the force is the field multiplied by the charge.
8.3.A.3 Vector nature of the field Fix
Vector nature of the field
The electric field has a magnitude and a direction at every point, and fields are combined by vector addition, component by component.
Superposition of electric fields
The net electric field at a point is the vector sum of the fields produced there by each charged object: E⃗net = E⃗₁ + E⃗₂ + …, each found as if the other objects were absent.
Vector field map
A representation that draws an arrow at each of many points: the arrow starts at the point, points in the direction of the field there, and has a length proportional to the field's magnitude.
Electric field line diagram
A simplified model of a field map: lines are drawn so that the field at any point is tangent to the line through it (arrows show the direction), and the field is stronger where the lines are closer together. The field also exists between the drawn lines.
Students often think The net electric field at a point is the sum of the magnitudes of the individual fields, whatever their directions. In fact Only if all the fields point the same way. Fields are vectors, so fields in opposite directions subtract and fields at an angle combine by vector addition.
Students often think In the field of each source charge, the distance r is the distance between the source charges rather than the distance from the source to the point. In fact No. r is the distance from each source charge to the point where the field is being found.
8.3.B.1 Electrostatic equilibrium Fix
Electrostatic equilibrium
The state of a conductor in which none of its charges is moving on average. In this state the electric field inside the conducting material is zero and any excess charge is on the conductor's surface.
Field at the surface of a conductor
In electrostatic equilibrium the field just outside a charged conductor is perpendicular to its surface; a component along the surface would move the surface charges.
Field outside a spherically symmetric charge distribution
Outside an isolated sphere whose charge is spherically symmetric, the field is the same as that of a point charge with the sphere's net charge at the sphere's center: |E⃗| = k|Q|/r², with r measured from the center.
Students often think The excess charge on a conductor spreads evenly through its whole volume. In fact No. In electrostatic equilibrium the excess charges of a conductor, which repel one another and can move freely, end up on its surface.
Students often think The field inside a charged conductor is nonzero, as the field of a point charge would be at that distance from the center. In fact No. In electrostatic equilibrium the field inside the conducting material is zero; a nonzero field would move the free charges.
8.3.B.2 Charge and field in an insulator Fix
Charge and field in an insulator
In an insulator the charges cannot move freely, so excess charge can remain throughout its interior as well as on its surface, and the field inside the insulator can be nonzero.
Students often think Like a conductor, a charged insulator in equilibrium has its excess charge on its surface and zero field inside. In fact No. The field inside a conductor is zero because its free charges move until it is. An insulator's charges cannot move freely, so charge can remain throughout its interior and the field inside can be nonzero.
Students often think The field at a point inside a charged sphere is that of a point charge at the center, the same as the formula gives outside. In fact No. The field of a spherically symmetric charge is that of a point charge at the center only at points outside the sphere. Inside, the point-charge formula does not apply: for a conductor the field inside is zero, and for a uniformly charged insulator it is smaller than the formula gives.
18 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 18
A particle with charge −3.0 nC is placed at point P, where it experiences an electric force of magnitude 6.0 × 10⁻⁵ N directed upward. What are the direction and magnitude of the electric field at P?
Answer and reasoning
AUpward, 2.0 × 10⁴ N/C A student who takes the field's direction to be the direction of the force on whatever charge is at P picks this. The field is defined by the force on a positive charge; the force on this negative charge is opposite to the field, so the field points downward.
BDownward, 2.0 × 10⁴ N/CCorrect E⃗ = F⃗E/q. Magnitude: (6.0 × 10⁻⁵ N)/(3.0 × 10⁻⁹ C) = 2.0 × 10⁴ N/C. Dividing the upward force by a negative charge reverses the direction, so the field points downward; the force on a negative charge is opposite to the field.
CDownward, 1.8 × 10⁻¹³ N/C A student who rearranges E = FE/q as FE = E/q, and so computes E = qFE, picks this: (3.0 × 10⁻⁹ C)(6.0 × 10⁻⁵ N). The field is the force divided by the charge, and the units check: N/C = N ÷ C.
DUpward, 6.0 × 10⁻⁵ N/C A student who takes the field to be the force at P picks this, giving it the force's value and direction. The field is the force per unit charge, 2.0 × 10⁴ N/C, and for a negative charge it is opposite to the force.
Working E⃗ = F⃗E/q with q = −3.0 × 10⁻⁹ C: |E| = 6.0 × 10⁻⁵ N / 3.0 × 10⁻⁹ C = 2.0 × 10⁴ N/C. Negative q, so E⃗ is opposite to F⃗E: downward. Distractors: E taken along F (upward); E = qF = 1.8 × 10⁻¹³; E = F = 6.0 × 10⁻⁵ upward.
The diagram shows a small charged ball hanging in equilibrium from a light insulating thread in a uniform electric field, with the ball's charge and mass and the thread's angle labeled. g is the magnitude of the gravitational field. Which expression gives the magnitude of the electric field?
Answer and reasoning
Amg tanθ/qCorrect The ball is in equilibrium under its weight mg (down), the electric force qE (horizontal) and the tension T (along the thread). Vertical: T cosθ = mg. Horizontal: T sinθ = qE. Dividing, tanθ = qE/(mg), so E = mg tanθ/q.
Bqmg tanθ A student who rearranges E = FE/q as FE = E/q picks this: the electric force mg tanθ is then set equal to E/q, giving E = qmg tanθ. The force is qE, so E = FE/q = mg tanθ/q.
Cmg/(q tanθ) A student who resolves the forces as if θ were measured from the horizontal picks this, writing tanθ = mg/(qE). The angle is measured from the vertical, so the horizontal electric force is opposite θ: tanθ = qE/(mg).
Dmg/q A student who thinks the electric force must equal the weight picks this, from qE = mg. The electric force is horizontal and balances the horizontal part of the tension; the vertical part of the tension balances the weight, so qE = mg tanθ.
Working Forces: weight mg down, qE horizontal, tension T along thread at θ to vertical. T cosθ = mg; T sinθ = qE ⇒ tanθ = qE/(mg) ⇒ E = mg tanθ/q. Distractors: F = E/q ⇒ E = qF = qmg tanθ; θ treated as from the horizontal ⇒ tanθ = mg/(qE) ⇒ E = mg/(q tanθ); qE = mg ⇒ E = mg/q.
A metal sphere with a large positive charge is on an insulating stand. A tiny positive test charge placed at point P, a distance from the sphere's center equal to three times the sphere's radius, experiences a force; the force divided by the test charge is E₀. The test charge is removed, and a particle with a positive charge 1000 times as large, comparable to the sphere's own charge, is placed at P. How does FE/q for this particle compare with E₀?
Answer and reasoning
AEqual to E₀, as the ratio FE/q is the same for any charge A student who applies 'the field does not depend on the test charge' without its condition picks this. That holds only for a charge too small to move the source charges. This particle's charge is comparable to the sphere's and moves the sphere's charges away from P.
BAbout 1000E₀, as the field grows with the charge placed at P A student who thinks the field is proportional to the charge placed at P picks this. The force grows with the charge, but FE/q does not; the only reason FE/q changes is that the sphere's charges move.
CSmaller than E₀: the sphere’s charges shift away from PCorrect A test charge must be too small to disturb the source charges. The large positive particle repels the sphere's positive charge, which is free to move on the metal, toward the far side of the sphere. The sphere's charge is then farther from P on average, so the field it produces at P, and so FE/q, is smaller than E₀.
DAbout E₀/1000, as the force on any charge placed at P is the same A student who treats the force at P as fixed picks this, dividing the same force by a charge 1000 times as large. The force on a charge is proportional to the charge; FE/q falls only because the sphere's charges are pushed away, not by a factor of 1000.
A proton released from rest in a uniform electric field has an initial acceleration of magnitude ap. The proton is replaced by an alpha particle at the same place. Treat the alpha particle as having charge +2e and mass 4mp, where e and mp are the charge and mass of the proton. Gravitational effects are negligible. What is the magnitude of the alpha particle's initial acceleration?
Answer and reasoning
A0.50apCorrect FE = qE, so the force on the alpha particle is twice that on the proton; a = FE/m, and its mass is four times as large. aα = (2eE)/(4mp) = (1/2)(eE/mp) = 0.50ap.
B1.00ap A student who expects every particle in a field to accelerate equally, as all objects in a gravitational field fall with g, picks this. The electric force depends on charge, not mass, so a = qE/m depends on the charge-to-mass ratio, which is halved.
C2.00ap A student who scales the acceleration with the charge alone picks this. Doubling the charge doubles the force, but the alpha particle's mass is four times as large, so its acceleration is (2/4)ap = 0.50ap.
D0.25ap A student who takes the field to be a force that is the same on any particle picks this, dividing one force by four times the mass. The force is qE, twice as large on the alpha particle, so the acceleration is (2/4)ap.
Working a = qE/m. Proton: ap = eE/mp. Alpha: a = (2e)E/(4mp) = (2/4)ap = 0.50ap. Distractors: same acceleration (1.00); charge only (2.00); same force, 4× mass (0.25).
An electron is in a region where the electric field is uniform, with magnitude 3.0 × 10³ N/C, and points north. Gravitational effects are negligible. Use e = 1.60 × 10⁻¹⁹ C and me = 9.11 × 10⁻³¹ kg. What are the direction and magnitude of the electron's acceleration?
Answer and reasoning
ANorth, 5.3 × 10¹⁴ m/s² A student who thinks every charge is pushed along the field picks this. The magnitude is right, but the electron's charge is negative, so the force and the acceleration are opposite to the field: south.
BSouth, 2.1 × 10⁵² m/s² A student who takes the force to be E/q picks this: (3.0 × 10³)/(1.60 × 10⁻¹⁹) divided by 9.11 × 10⁻³¹ kg. The force is qE, 4.8 × 10⁻¹⁶ N; the units of E/q, N/C², are not those of a force.
CNorth, 3.3 × 10³³ m/s² A student who takes the field to be the force on the electron picks this, using 3.0 × 10³ as the force in newtons and its direction. The force is the field times the charge, eE = 4.8 × 10⁻¹⁶ N, and it points south for a negative charge.
DSouth, 5.3 × 10¹⁴ m/s²Correct F⃗E = qE⃗ with q = −e, so the force is opposite to the field: south, with magnitude eE = (1.60 × 10⁻¹⁹ C)(3.0 × 10³ N/C) = 4.8 × 10⁻¹⁶ N. a = F/me = (4.8 × 10⁻¹⁶ N)/(9.11 × 10⁻³¹ kg) = 5.3 × 10¹⁴ m/s², south.
Working F = eE = (1.60 × 10⁻¹⁹)(3.0 × 10³) = 4.8 × 10⁻¹⁶ N, opposite to E⃗ (electron negative): south. a = F/me = 4.8 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 5.27 × 10¹⁴ ≈ 5.3 × 10¹⁴ m/s². Distractors: north (force along field); F = E/q → a = E/(e me) = 2.1 × 10⁵²; F = E → a = E/me = 3.3 × 10³³, north.
The diagram shows two small charged spheres fixed in place and point P, with the charges and distances labeled. What is the magnitude of the net electric field at P? (k = 1/(4πε₀))
Answer and reasoning
A0.36kq/a²Correct Each sphere is r = √(a² + (2a)²) = √5 a from P, so each field has magnitude kq/(5a²). The components along the line joining the spheres cancel; the components along the bisector add, each kq/(5a²) × (2a/(√5 a)). E = 2 × kq/(5a²) × 2/√5 = 4kq/(5√5 a²) ≈ 0.36kq/a², directed away from the spheres along the bisector.
B0.18kq/a² A student who keeps the components along the line joining the spheres and cancels the components along the bisector picks this: 2 × kq/(5a²) × (a/(√5 a)) ≈ 0.18kq/a². At P the fields' components parallel to the line joining the spheres point in opposite directions and cancel; the components along the bisector point the same way and add.
C0.40kq/a² A student who adds the two fields' magnitudes picks this: 2 × kq/(5a²) = 0.40kq/a². The fields point in different directions, so only their components along the bisector add.
D0.50kq/a² A student who replaces the two spheres by a single charge 2q at their midpoint picks this: k(2q)/(2a)² = 0.50kq/a². That shortcut is a good approximation only far from the spheres; at P each field must be found from its own distance and direction.
Working r² = a² + (2a)² = 5a². Each |E| = kq/(5a²). Bisector component: × (2a)/(√5 a) = 2/√5. E = 2 · kq/(5a²) · 2/√5 = 4kq/(5√5 a²) = 0.358kq/a² ≈ 0.36kq/a². Distractors: other component (×1/√5) → 0.18; magnitudes added → 0.40; single 2q at midpoint → 2kq/(4a²) = 0.50.
The diagram shows two charged particles fixed on the x-axis, with their charges and positions labeled. At what position on the x-axis, other than at infinity, is the net electric field zero?
Answer and reasoning
Ax = 0.67d A student who thinks the fields of opposite charges oppose each other between them picks this, solving 4/x² = 1/(d − x)² for x between the particles. There the field of +4q points away from it and the field of −q points toward −q: both point in the +x direction, so they add.
Bx = 2.00dCorrect Between the particles both fields point toward −q and cannot cancel. To the left of +4q the field of +4q is always the larger, as that point is nearer the larger charge. To the right of −q the fields are opposite: 4kq/x² = kq/(x − d)² gives x = 2(x − d), so x = 2d = 2.00d.
Cx = 1.33d A student who takes the fields to fall off as 1/r picks this: 4/x = 1/(x − d) gives x = 4d/3. The fields fall off as 1/r², so the balance point is where x = 2(x − d), x = 2d.
Dx = 0.50d A student who thinks the net field is zero halfway between any two charges picks this. That is true only for two equal charges of the same sign; here, at the midpoint, both fields point toward −q and add.
Working x > d: E = 4kq/x² (+x) − kq/(x − d)² ⇒ zero when x = 2(x − d) ⇒ x = 2d. 0 < x < d: both fields +x (no zero). x < 0: |E(+4q)| > |E(−q)| always. Distractors: solving in the middle region 2(d − x) = x ⇒ 2d/3; 1/r ⇒ 4(x − d) = x ⇒ 4d/3; midpoint d/2.
The vector field map shows the electric field in a region. Point P is at the tail of one arrow, and point Q lies between the arrows. An electron is placed at Q. Which statement about the electric force on the electron is correct?
Answer and reasoning
AZero, because no field arrow is drawn at the point Q itself A student who thinks the field exists only where arrows are drawn picks this. A map shows the field at a sample of points; the field is also present between them, and here it is the same everywhere.
BRightward, with the same magnitude as it would have at P A student who thinks every charge is pushed along the field picks this. The magnitude is right, but the force on a negative charge is opposite to the field, to the left.
CRightward, and equal in magnitude to the field at Q A student who takes the field to be the force picks this. The force is FE = qE: its magnitude is eE, in newtons, and for an electron it points opposite to the field.
DLeftward, with the same magnitude as it would have at PCorrect All the arrows in the map have the same length and direction, so the field is uniform: the same at Q, between the arrows, as at P. The electron is negative, so the force on it is opposite to the field, to the left, with the same magnitude eE as at P.
The field line diagram shows the electric field in a region. Points A, B and C are marked; A lies on a field line, and B and C lie between lines. Which ranking of the magnitudes of the electric field at the three points is correct?
Answer and reasoning
AEB = EC < EA A student who thinks the field exists only on the drawn lines picks this, taking the field at B and C to be zero. The field exists between the lines too; its strength there is shown by how close the nearby lines are.
BEA = EB = EC A student who thinks the field has one magnitude throughout the diagram picks this. The spacing of the lines changes across the diagram, and the field is stronger where they are closer together.
CEA < EB < ECCorrect In a field line diagram the field is stronger where the lines are closer together. The lines are widely spaced near A, closer near B and closest near C, so EA < EB < EC. Whether a point lies on a drawn line or between lines does not matter.
DEA > EB > EC A student who thinks the field weakens along the direction of its lines picks this, as A is 'upstream'. Strength is shown by line spacing, not by position along a line; the lines are closest at C.
Working Line density increases from left to right (lines converge): EA < EB < EC. On-line vs between-line position is irrelevant.
A student sketches the electric field lines of two charged particles, and two of the lines cross each other at point P, where the net electric field is not zero. Which evaluation of the sketch is correct?
Answer and reasoning
AAcceptable: the lines come from different charges, and each charge’s lines keep their own paths A student who pictures each charge keeping its own lines picks this. Field lines show the net field of both charges together, and the net field at P has only one direction.
BWrong: field lines are the paths of charged particles, and two particles cannot both be at P A student who thinks field lines are particle paths picks this. The sketch is wrong, but not for that reason: field lines give the direction of the net field, which has only one direction at P; they are not paths.
CAcceptable: two lines through P just show that the field at P is twice as strong as nearby A student who counts lines through a point to find the field strength picks this. Strength is shown by the spacing of lines around a point, and where the field is not zero only one line passes through each point.
DWrong: the net field at P has one direction, so only one field line can pass through PCorrect A field line is drawn along the net field, the vector sum of the fields of all the charges. At a point where that sum is not zero it has one direction, so one line passes through the point; two crossing lines would give the field two directions at P.
A particle with charge +Q is fixed a distance r to the left of point P, where it produces an electric field of magnitude E₀. A second particle, also with charge +Q, is then fixed a distance 2r to the right of P, on the same line. What is the magnitude of the net electric field at P?
Answer and reasoning
A1.25E₀ A student who adds the magnitudes of the two fields picks this: E₀ + 0.25E₀. Both charges are positive and on opposite sides of P, so their fields at P point in opposite directions and subtract.
B0.75E₀Correct The first particle's field at P points away from it, to the right, with magnitude E₀. The second particle is twice as far away, so its field is E₀/2² = 0.25E₀, pointing away from it, to the left. The fields are opposite: 1.00E₀ − 0.25E₀ = 0.75E₀, to the right.
C0.50E₀ A student who takes the field to fall off as 1/r picks this: the second field would be E₀/2, giving E₀ − 0.50E₀. At twice the distance the field is one-quarter as large, 0.25E₀.
D1.00E₀ A student who thinks only the nearer charge produces the field at P picks this. The farther charge's field is weaker but not blocked: its 0.25E₀, in the opposite direction, reduces the net field to 0.75E₀.
Working E₁ = E₀ (→, away from left charge). E₂ = kQ/(2r)² = E₀/4 (←, away from right charge). Net = E₀ − E₀/4 = 0.75E₀. Distractors: sum 1.25E₀; 1/r ⇒ E₀ − E₀/2 = 0.50E₀; nearest only 1.00E₀.
A solid metal block with excess charge is isolated and in electrostatic equilibrium. Which reasoning correctly supports the claim that the electric field inside the metal is zero?
Answer and reasoning
AFree electrons in the metal would move in any nonzero field, and in equilibrium none are movingCorrect A metal contains free electrons that move whenever a field acts on them. Electrostatic equilibrium means no charges are moving, so there can be no field inside the metal; the surface charges arrange themselves so that the fields of all the charges cancel there.
BMetal blocks electric fields from getting into it, in the way that a wall blocks light A student who thinks of shielding as a barrier picks this. Fields are not stopped at a surface; the free charges in the metal rearrange until their field cancels all other fields inside.
CNo excess charge is inside the metal, and a field at a point comes only from charge there A student who thinks the field at a point is produced only by charge at that point picks this. The surface charges do produce fields at points inside the metal; those fields add up to zero there, and that requires the equilibrium argument.
DThe excess charge spreads evenly through the metal, so its pushes balance at every point A student who thinks evenly spread charge balances everywhere inside picks this. The excess charge of a conductor is on its surface, not through its volume, and charge spread through a volume would not give zero field at off-center points.
The diagram shows a metal block with excess negative charge, in electrostatic equilibrium, and four arrows drawn from point P on its flat top surface. Which arrow shows the direction of the electric field just outside the surface at P?
Answer and reasoning
AArrow 1 A student who thinks a field always points away from the charged object picks this. Arrow 1 is perpendicular to the surface, but the block is negative, and the field points toward negative charge.
BArrow 3Correct Just outside a charged conductor in equilibrium, the field is perpendicular to the surface; a component along the surface would move the surface charges. The block is negative, so the field points toward it, into the surface: arrow 3, straight down.
CArrow 2 A student who thinks the field runs along the surface picks this. A field component along the surface would push the free surface charges along it, so the conductor would not be in equilibrium.
DArrow 4 A student who thinks the field always lies along the line through the object's center picks this. That is true for a sphere; for this block the field at P is perpendicular to the flat top surface, not directed at the center.
Working Equilibrium: field just outside a conductor is perpendicular to the surface (no tangential component). Negative charge: field points toward the surface. At P on the flat top face that is straight down, arrow 3. Arrow 4 (toward the center) is not perpendicular to the face.
An insulating sphere of radius 0.10 m has a charge of +6.0 × 10⁻⁸ C spread uniformly throughout its volume and is far from other charged objects. A proton is released from rest at a point 0.20 m from the sphere's surface. Use k = 9.0 × 10⁹ N·m²/C², e = 1.60 × 10⁻¹⁹ C and mp = 1.67 × 10⁻²⁷ kg; gravitational effects are negligible. What is the magnitude of the proton's initial acceleration?
Answer and reasoning
A5.7 × 10¹¹ m/s²Correct Outside the sphere its field is that of a point charge at the center, with r = 0.10 m + 0.20 m = 0.30 m: E = (9.0 × 10⁹)(6.0 × 10⁻⁸)/(0.30)² = 6.0 × 10³ N/C. F = eE = 9.6 × 10⁻¹⁶ N, so a = F/mp = (9.6 × 10⁻¹⁶ N)/(1.67 × 10⁻²⁷ kg) = 5.7 × 10¹¹ m/s².
B1.3 × 10¹² m/s² A student who measures r from the sphere's surface picks this: E = kQ/(0.20 m)² = 1.35 × 10⁴ N/C. The sphere acts as a point charge at its center, so r = 0.30 m.
C1.7 × 10¹¹ m/s² A student who divides by r instead of r² picks this: kQ/(0.30 m) = 1.8 × 10³, used as the field. The field falls off as 1/r², giving 6.0 × 10³ N/C.
D3.6 × 10³⁰ m/s² A student who takes the field to be the force on the proton picks this, dividing 6.0 × 10³ directly by mp. The force is the field times the charge, eE = 9.6 × 10⁻¹⁶ N.
Working r = 0.10 + 0.20 = 0.30 m. E = kQ/r² = (9.0 × 10⁹)(6.0 × 10⁻⁸)/0.090 = 6.0 × 10³ N/C. F = eE = 9.6 × 10⁻¹⁶ N. a = F/mp = 5.75 × 10¹¹ ≈ 5.7 × 10¹¹ m/s². Distractors: r = 0.20 m → 1.29 × 10¹²; 1/r → 1.72 × 10¹¹; F = E → 3.59 × 10³⁰.
An insulating sphere of radius R has charge +2q spread uniformly throughout its volume. A particle with charge −q is fixed a distance 3R from the sphere's center. Point P lies on the line joining them, between the sphere and the particle, a distance 2R from the sphere's center. What is the magnitude of the net electric field at P? (k = 1/(4πε₀))
Answer and reasoning
A3.00kq/R² A student who measures the sphere's distance from its surface, R, picks this: k(2q)/R² + kq/R² = 3.00kq/R². The sphere acts as a point charge at its center, 2R from P.
B0.50kq/R² A student who thinks the fields of opposite charges oppose each other between them picks this: |0.50 − 1.00|kq/R². Between a positive and a negative charge both fields point toward the negative charge, so they add.
C1.50kq/R²Correct Outside the sphere, its field is that of a point charge +2q at its center: k(2q)/(2R)² = 0.50kq/R², pointing away from the sphere, toward the particle. The particle is R from P; its field, kq/R², points toward the negative particle, the same way. The fields add: 0.50kq/R² + 1.00kq/R² = 1.50kq/R².
D0.25kq/R² A student who combines the charges into one, +2q − q = +q, at the sphere's center picks this: kq/(2R)² = 0.25kq/R². Only the sphere can be replaced by a point charge at its center; the particle's field must be found from its own distance to P.
Working Sphere (outside, point-charge model): E₁ = k(2q)/(2R)² = 0.50kq/R², away from sphere (toward −q). Particle −q at distance R: E₂ = kq/R², toward −q. Same direction: E = 1.50kq/R². Distractors: r from surface → 2kq/R² + kq/R² = 3.00; opposite → |0.50 − 1.00| = 0.50; lumped +q at center → 0.25.
The diagram shows four particles with charges +q, +q, −q and −q at the corners of a square of side s, and point P at the center of the square. What is the magnitude of the net electric field at P?
Answer and reasoning
A2q/(πε₀s²) A student who adds the magnitudes of all four fields picks this: 4 × 2kq/s² = 8kq/s² = 2q/(πε₀s²). The two diagonal sums are perpendicular, so they combine to √2 × 4kq/s².
B√2q/(πε₀s²)Correct Each corner is s/√2 from P, so each field has magnitude kq/(s²/2) = 2kq/s². Along each diagonal, the field of +q (away from it) and the field of the opposite −q (toward it) point the same way, giving 4kq/s² along each diagonal. The two diagonals are perpendicular: E = √2 × 4kq/s² = 4√2kq/s² = √2q/(πε₀s²), toward the side with the negative charges.
C√2q/(2πε₀s²) A student who uses the side s as each particle's distance from P picks this: each field kq/s², giving 2√2kq/s². The distance from a corner to the center is half the diagonal, s/√2.
Dq/(πε₀s) A student who takes each field to fall off as 1/r picks this: kq/(s/√2) = √2kq/s for each, combined to 4kq/s = q/(πε₀s). The field of a point charge is proportional to 1/r², and the units then come out in N/C.
Working r = s/√2 ⇒ each |E| = kq/(s²/2) = 2kq/s². Diagonal TL→BR: +q(TL) field toward BR, −q(BR) field toward BR ⇒ 4kq/s². Diagonal TR→BL similarly 4kq/s². Perpendicular ⇒ |E| = 4√2 kq/s² = √2q/(πε₀s²) (k = 1/(4πε₀)), pointing from the +q side toward the −q side. Checked with sympy. Distractors: scalar 8kq/s² = 2q/(πε₀s²); r = s ⇒ 2√2kq/s² = √2q/(2πε₀s²); 1/r ⇒ 4kq/s = q/(πε₀s).
A solid metal sphere of radius R is isolated and in electrostatic equilibrium. Its excess charge is spread uniformly over its surface, with charge per unit area σ. What is the magnitude of the electric field at a point a distance 2R from the sphere's center?
Answer and reasoning
Aσ/(4ε₀)Correct The sphere's total charge is Q = 4πR²σ. Outside the sphere its field is that of a point charge Q at the center, so at r = 2R, E = 4πR²σ/(4πε₀(2R)²) = σ/(4ε₀).
Bσ/ε₀ A student who measures r from the sphere's surface picks this: r = R gives E = 4πR²σ/(4πε₀R²) = σ/ε₀. The point-charge model places all the charge at the center, so r is measured from the center: r = 2R.
CσR/(2ε₀) A student who takes the field of a point charge to fall off as 1/r picks this: Q/(4πε₀·2R) = σR/(2ε₀). Its units, N·m/C, show it is not a field; the field falls off as 1/r², giving σ/(4ε₀).
Dπσ/ε₀ A student who writes the point-charge field as Q/(ε₀r²), taking k as 1/ε₀, picks this: 4πR²σ/(ε₀·4R²) = πσ/ε₀. Since k = 1/(4πε₀), the field is Q/(4πε₀r²), and the 4π cancels to leave σ/(4ε₀).
Working Total charge Q = σ(4πR²). Outside a sphere whose charge is spherically symmetric, the field is that of a point charge Q at the center: E = Q/(4πε₀r²) = σ(4πR²)/(4πε₀(2R)²) = σ/(4ε₀). Distractors (sympy-checked): r measured from the surface, r = R ⇒ σ/ε₀; field taken ∝ 1/r, Q/(4πε₀·2R) ⇒ σR/(2ε₀), in N·m/C rather than N/C; 4π dropped, Q/(ε₀r²) ⇒ πσ/ε₀. Units: σ/ε₀ is (C/m²)/(C²/(N·m²)) = N/C, so σ/(4ε₀), σ/ε₀ and πσ/ε₀ all have the units of a field.
Particles with charges +q, +q and +2q are fixed at the corners of an equilateral triangle of side a. Point P is the midpoint of the side joining the two +q particles. What is the magnitude of the net electric field at P? (k = 1/(4πε₀))
Answer and reasoning
A10.7kq/a² A student who adds the magnitudes of the three fields picks this: 4kq/a² + 4kq/a² + 8kq/(3a²) ≈ 10.7kq/a². The fields of the two +q particles point in opposite directions at P and cancel; only the +2q particle's field remains.
B2.00kq/a² A student who uses the distance between the source charges, a, in each field picks this: the +q fields still cancel, and the +2q particle gives 2kq/a². In each field, r is the distance from that particle to P; the +2q particle is (√3/2)a from P, giving 8kq/(3a²).
C2.67kq/a²Correct The two +q particles are each a/2 from P, and their fields there point in opposite directions, so they cancel. The +2q particle is (√3/2)a from P, so E = k(2q)/(3a²/4) = 8kq/(3a²) ≈ 2.67kq/a².
D48.0kq/a² A student who replaces the three particles by one charge 4q at the triangle's center picks this: the center is a/(2√3) from P, so 4kq/(a²/12) = 48.0kq/a². Point charges near P cannot be lumped at their center; each particle's field must be found at P and the vectors added.
Working The two +q particles are each a/2 from P; their fields there are equal in magnitude and opposite in direction, so they cancel. The +2q particle is the triangle's height, (√3/2)a, from P: E = k(2q)/((√3/2)a)² = 8kq/(3a²) ≈ 2.67kq/a², directed away from the +2q particle. Distractors (sympy-checked): magnitudes added, 4kq/a² + 4kq/a² + 8kq/(3a²) ≈ 10.7kq/a²; r taken as the distance a between the source charges, so the +q fields still cancel and the +2q particle gives 2kq/a²; all the charge lumped as 4q at the triangle's center, a/(2√3) from P, 4kq/(a²/12) = 48.0kq/a².
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
