10.3 Electric Fields

1. AIt comes into being at a point only when a second charge is placed at that point.
   A student who thinks a field needs a charge to be present picks this. The sphere's field exists around it all the time; a second charge placed at a point experiences a force because the field is already there.
2. BIt exists along the field lines drawn from the sphere but not in between them.
   A student who reads field lines as the only places where the field exists picks this. Field lines are a model: only a few are drawn, and the field exists at every point around the sphere, including between the lines.
3. CIt needs air or some other material around the sphere to pass it along.
   A student who thinks a field needs a material to travel through picks this. Electric fields exist in a vacuum; electric forces act across empty space, for example between the particles of an atom.
4. DIt exists at points around the sphere even where no other charge is present. Correct
   A charged object produces an electric field in the space around it. The field has a magnitude and direction at every point, whether or not a charge is there; a charge placed at a point only reveals the field by experiencing a force.

CED 10.3.A.1 · Read this in Fix

1. AEast, away from the sphere, since a field points outward
   A student who thinks every field points away from its source picks this. That is true for a positive charge. A positive test charge at P would be pulled toward the negative sphere, so the field points toward it, west.
2. BWest, pointing toward the sphere, as the sphere is negative Correct
   The field at a point has the direction of the force on a positive test charge there. A positive test charge at P would be attracted toward the negative sphere, so the field at P points west, toward the sphere.
3. CEither way, depending on the sign of a charge put at P
   A student who mixes up the field with the force on a charge picks this. The force on a charge at P depends on its sign, but the field is set by the sphere alone and points toward it, west, whatever is placed at P.
4. DNeither way: no field exists at P until a charge is put there
   A student who thinks a field needs a charge to be present picks this. The sphere produces a field at P whether or not anything is placed there; the field points west, toward the negative sphere.

CED 10.3.A.2.ii · Read this in Fix

1. A5.6 × 10² N/C
   A student who adds the fields' magnitudes, ignoring their directions, picks this: 2.4 × 10² + 3.2 × 10² = 5.6 × 10² N/C. That is correct only for fields in the same direction; perpendicular fields combine by the Pythagorean theorem.
2. B3.2 × 10² N/C
   A student who thinks the stronger source alone sets the field picks this. Both objects contribute: the net field is the vector sum of the two fields, √(2.4² + 3.2²) × 10² N/C = 4.0 × 10² N/C.
3. C1.6 × 10⁵ N/C
   A student who adds the squares of the fields but forgets the square root picks this: (2.4 × 10²)² + (3.2 × 10²)² = 1.6 × 10⁵. The net field of two perpendicular fields must be larger than either one but smaller than their sum; taking the square root gives 4.0 × 10² N/C.
4. D4.0 × 10² N/C Correct
   The fields are vectors at right angles, so they combine by the Pythagorean theorem: |E⃗| = √((2.4 × 10² N/C)² + (3.2 × 10² N/C)²) = 4.0 × 10² N/C, pointing north of east.

Working Perpendicular fields: |E⃗| = √(E_A² + E_B²) = √((2.4 × 10² N/C)² + (3.2 × 10² N/C)²) = √(1.6 × 10⁵) N/C = 4.0 × 10² N/C.

CED 10.3.A.3 · Read this in Fix

1. ASpread evenly through the whole of the sphere's volume
   A student who pictures charge soaking into a conductor picks this. Charges spread through the volume would produce a field inside the metal and push one another outward; they move until they are all on the surface.
2. BGathered together at the very center of the sphere
   A student who takes the point-charge model literally picks this. The field outside the sphere is calculated as if the charge were at the center, but the excess electrons repel one another and end up on the surface, as far apart as possible.
3. CStill at the point where the rod touched the sphere
   A student who thinks charge stays where it is put picks this. That happens on an insulator. In a metal the excess electrons move freely and repel one another, so they spread over the whole surface.
4. DSpread over the outer surface of the sphere Correct
   The excess electrons repel one another and move freely through the metal, so they spread out as far as they can. In electrostatic equilibrium all the excess charge is on the sphere's outer surface, and the field inside the metal is zero.

CED 10.3.B.1 · Read this in Fix

1. AIt is zero, as it is inside a charged object in equilibrium.
   A student who applies the conductor rule to every object picks this. The field is zero inside a conductor because its charges move until it is zero. In plastic the charges cannot move, and the field inside is not zero.
2. BIt is nonzero, and it is directed away from the center of the sphere. Correct
   In an insulator the charges cannot move to cancel the field, so excess charge can remain inside and the field inside can be nonzero. At the halfway point, more of the positive charge pushes outward than inward, so the field points away from the center.
3. CIt is nonzero, and it is directed toward the center of the sphere.
   A student who takes the field to point the way an electron would be pushed picks this. The field points in the direction of the force on a positive test charge, which the positive charge of the sphere pushes outward, away from the center.
4. DIt is larger than just outside, as all charge acts from the center.
   A student who uses the point-charge model inside the sphere picks this. The model holds only outside the charge. Inside, the charge on the far side of the center pushes a positive test charge the opposite way to the nearer charge, so the field weakens toward zero at the center.

CED 10.3.B.2 · Read this in Fix

1. A1.8 × 10⁻⁴ N Correct
   The field at P is set by the source charges: E = F_E/q = (6.0 × 10⁻⁵ N)/(2.0 × 10⁻⁹ C) = 3.0 × 10⁴ N/C, and it does not change when the charge at P is changed. The force is proportional to the charge: F_E = qE = (6.0 × 10⁻⁹ C)(3.0 × 10⁴ N/C) = 1.8 × 10⁻⁴ N, three times the original force.
2. B5.4 × 10⁻⁴ N
   A student who thinks the field grows with the charge placed at P picks this: tripling the field to 9.0 × 10⁴ N/C and then multiplying by the tripled charge gives nine times the force. The field depends only on the source charges, so only the charge factor of 3 applies.
3. C6.0 × 10⁻⁵ N
   A student who takes the field at P to be the force there, the same for any charge, picks this. The field is the force per unit charge; a charge three times as large experiences three times the force, 1.8 × 10⁻⁴ N.
4. D5.0 × 10¹² N
   A student who rearranges E = F_E/q as F_E = E/q picks this: (3.0 × 10⁴ N/C)/(6.0 × 10⁻⁹ C). The force is the field multiplied by the charge, F_E = qE; the units check, (N/C) × C = N, shows the error.

Working Field at P: E = F_E/q = (6.0 × 10⁻⁵ N)/(2.0 × 10⁻⁹ C) = 3.0 × 10⁴ N/C, set by the source charges and unchanged by the new charge. Force on +6.0 nC: F_E = qE = (6.0 × 10⁻⁹ C)(3.0 × 10⁴ N/C) = 1.8 × 10⁻⁴ N (three times the original force, since F_E ∝ q).

CED 10.3.A.2 · Read this in Fix

1. AE = kQq₀/r²
   A student who takes the field to be the force on the test charge picks this. kQq₀/r² is the force, in newtons; dividing it by q₀ gives the field, kQ/r², in N/C.
2. BE = kQ/r² Correct
   Coulomb's law gives the force on the test charge, F_E = kQq₀/r². The field is the force per unit charge of the test charge: E = F_E/q₀ = kQ/r². The test charge cancels, so the field depends only on the source charge Q and the distance r.
3. CE = kQ/r
   A student who drops the square from Coulomb's law picks this. The force, and so the field, is proportional to 1/r²: at twice the distance the field is one-quarter as large.
4. DE = kq₀/r²
   A student who thinks the test charge produces the field it measures picks this. The field is produced by the source charge Q; when the force kQq₀/r² is divided by q₀, the test charge cancels and Q remains.

Working Coulomb's law: F_E = kQq₀/r². Definition of field: E = F_E/q₀ = (kQq₀/r²)/q₀ = kQ/r². Units: (N·m²/C²)(C)/m² = N/C.

CED 10.3.A.2 · Read this in Fix

1. AThe same value as before, since the ratio F_E/q does not depend on the charge used
   A student who applies 'the field does not depend on the test charge' without its condition picks this. That holds only while the charge used is too small to move the source charges. A charge 1000 times as large moves the charges on the metal plate and changes the field it is measuring.
2. BA value 1000 times the original field, since the field is proportional to the charge
   A student who thinks the field grows with the charge placed at the point picks this. The force grows in proportion to the charge, so F_E/q would stay the same if the plate's charges did not move; the field does not scale with the measuring charge.
3. CA value that differs from the original field, as the large charge moves the charges on the plate Correct
   A test charge must be small enough not to disturb the charges that produce the field. The metal plate's charges are free to move, and a large charge placed nearby pushes or pulls them into a new arrangement. F_E/q then gives the field of the rearranged plate, not the field that existed before.
4. DA value 1/1000 of the original field, since the force is fixed and the charge is larger
   A student who treats the force at a point as fixed picks this. The force on a charge is proportional to the charge, so dividing by a larger charge does not shrink the ratio by that factor; the only reason the result differs is that the plate's charges have moved.

CED 10.3.A.2.i · Read this in Fix

1. AEqual in magnitude, opposite in direction Correct
   F⃗_E = qE⃗. The proton's charge is +e and the electron's is −e, so the forces have equal magnitudes, eE. The force on the positive proton points east, along the field; the force on the negative electron points west, opposite to it.
2. BLarger on the electron, which has much less mass
   A student who mixes up force and acceleration picks this. The electric force, |q||E⃗|, is the same size for both, since their charges are equal in magnitude; the electron's much smaller mass gives it a much larger acceleration, not a larger force.
3. CLarger on the proton, which has much more mass
   A student who expects the electric force to depend on mass, like gravity, picks this. The electric force depends on charge, and the proton and electron carry charges of equal magnitude.
4. DEqual in magnitude, with both directed east
   A student who thinks every charge is pushed along the field picks this. The force on a negative charge is opposite to the field, so the electron is pushed west.

CED 10.3.A.2.iii · Read this in Fix

1. A2.0 × 10² N/C
   A student who thinks the fields of opposite charges always subtract picks this: 4.0 × 10² − 2.0 × 10² N/C. Between a positive and a negative charge, the field of each points toward the negative charge, so at P the two fields are in the same direction and add.
2. B1.5 × 10² N/C
   A student who uses the separation of the spheres, 0.60 m, as the distance picks this: k(6.0 × 10⁻⁹ C)/(0.60 m)². The field of each sphere at P depends on its distance from P, which is 0.30 m.
3. C6.0 × 10² N/C Correct
   Each sphere is 0.30 m from P. A's field at P points away from A, toward B: k(4.0 × 10⁻⁹ C)/(0.30 m)² = 4.0 × 10² N/C. B's field points toward B, the negative charge: k(2.0 × 10⁻⁹ C)/(0.30 m)² = 2.0 × 10² N/C. Both point toward B, so they add: 6.0 × 10² N/C.
4. D1.8 × 10² N/C
   A student who divides by r instead of r² picks this: k(6.0 × 10⁻⁹ C)/(0.30 m). The field of a point charge falls off as 1/r²; with r² = 0.090 m² the net field is 6.0 × 10² N/C.

Working Distance from each sphere to P: 0.60 m/2 = 0.30 m. E_A = kq_A/r² = (9.0 × 10⁹)(4.0 × 10⁻⁹)/(0.30)² = 4.0 × 10² N/C, directed away from A (toward B). E_B = (9.0 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 2.0 × 10² N/C, directed toward B. Same direction: E = 4.0 × 10² + 2.0 × 10² = 6.0 × 10² N/C, toward B.

CED 10.3.A.3.i · Read this in Fix

1. AVertical, pointing directly away from both charges
   A student who thinks every field points away from its source picks this. The field of −q points toward −q, so the vertical parts of the two fields cancel and the horizontal parts, both toward −q, add.
2. BHorizontal, pointing toward the side where −q is Correct
   At P, the field of +q points away from +q, up and toward the −q side; the field of −q points toward −q, down and toward the −q side. The two fields have equal magnitudes, so their vertical parts cancel and their horizontal parts add: the net field is horizontal, toward the −q side.
3. CHorizontal, pointing toward the side where +q is placed
   A student who takes the field's direction to be the push on an electron picks this. The field is defined by the force on a positive test charge, which is repelled by +q and attracted by −q: the net field points toward the −q side.
4. DThere is none, because the two opposite fields cancel out
   A student who thinks the fields of opposite charges always cancel picks this. At P the fields point in different directions: their vertical parts cancel, but their horizontal parts both point toward −q and add.

CED 10.3.A.3.i · Read this in Fix

1. AReverse every arrow so that each one points toward the sphere
   A student who takes the field's direction to be the push on an electron picks this. The field points in the direction of the force on a positive test charge, which +Q repels, so the arrows correctly point away from the sphere.
2. BNo change: the arrows in a field map show direction, not strength
   A student who thinks arrow length means nothing in a field map picks this. The length of each arrow represents the field's magnitude at its starting point, and the field is weaker farther from the sphere, so the outer arrows must be shorter.
3. CRemove every arrow, as no test charge has been placed near the sphere
   A student who thinks a field exists only where a charge is placed to experience it picks this. The field of +Q exists at every point around the sphere whether or not a charge is there; a vector field map shows it at many points. The map's error is the arrow lengths: the field is weaker farther from the sphere, so the outer arrows must be shorter.
4. DMake the arrows farther from the sphere shorter than the nearer ones Correct
   In a vector field map each arrow's length is proportional to the field's magnitude at the point where it starts. The field of +Q is k|Q|/r², so at about three times the distance it is roughly one-ninth as strong: the outer arrows must be much shorter. The directions, away from +Q, are already correct.

CED 10.3.A.3.ii · Read this in Fix

1. AIt is greater at X, where the field lines are closer together. Correct
   In a field line diagram the field is stronger where the lines are closer together. X lies in the crowded region between the two charges, where both fields point toward −q and add; Y lies where the lines are spread far apart. So the field is stronger at X.
2. BIt is greater at Y, which is on a field line while X is not.
   A student who thinks the field exists only on the drawn lines picks this. Only a sample of lines is drawn; the field exists between them too, and its strength is shown by how close together the lines are, closest around X.
3. CIt is equal, since X and Y are the same distance from +q.
   A student who considers only the +q sphere picks this. The −q sphere contributes too: at X its field points the same way as the field of +q and adds to it, while at Y it is weaker and points partly across. The lines are much closer together at X.
4. DIt is smaller at X, where the fields of +q and −q cancel out.
   A student who thinks the fields of opposite charges always cancel picks this. Between +q and −q both fields point toward −q, so at X they add; the crowded lines there show a strong field.

CED 10.3.A.3.iii · Read this in Fix

1. AYes: a field line is, by definition, the path that a charged particle follows
   A student who thinks field lines are paths picks this. A field line is defined by the direction of the force on a positive charge at each point, and a particle's velocity need not be along its force; on a curved line the proton does not stay on the line.
2. BYes: the force on it is along the line, and an object moves in the direction of its force
   A student who thinks an object always moves in the direction of the force on it picks this. The force sets the direction of the acceleration, not of the velocity; once the proton is moving, its velocity lags behind the turning force, and it leaves the curved line.
3. CNo: the force is tangent to the line, but the proton's inertia carries it off the curve Correct
   At every point the force on the proton is along the field line through that point. Once the proton is moving, its velocity changes gradually, so it cannot turn as sharply as the curved line: it drifts to the outside of the curve. A field line gives the direction of the force, not the path.
4. DNo: a proton is pushed in the direction opposite to the arrows on the field lines
   A student who thinks the field shows the push on an electron picks this. The arrows show the direction of the force on a positive charge, so the proton is pushed along them; the claim is wrong for a different reason, that the proton does not stay on a curved line.

CED 10.3.A.3.iii · Read this in Fix

1. AE_A > E_B > E_C
   A student who thinks the charge gathers at the center picks this, expecting the field to be strongest nearest the center. The excess charge is on the surface, and the field inside the metal is zero, at the center as elsewhere.
2. BE_C > E_A = E_B Correct
   A and B are inside the metal, where the field is zero in electrostatic equilibrium: E_A = E_B = 0. C is just outside the surface, where the sphere's excess charge produces a nonzero field. So E_C is the largest, and E_A and E_B are equal.
3. CE_C > E_B > E_A
   A student who thinks the excess charge is spread through the volume picks this: the field would then be zero at the center by symmetry but not at B. On a conductor the charge is all on the surface, and the field is zero everywhere inside the metal, at B as well as at A.
4. DE_A = E_B = E_C
   A student who thinks the field is zero everywhere around a conductor in equilibrium picks this. The field is zero only inside the metal; just outside, at C, the excess charge produces a field perpendicular to the surface.

CED 10.3.B.1 · Read this in Fix

1. AArrow 2
   A student who thinks the field always points directly away from the object's center picks this. That holds for a sphere, but on an oval at P the direction away from the center is not perpendicular to the surface; the field is perpendicular to the surface.
2. BArrow 3
   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.
3. CArrow 1 Correct
   Just outside a charged conductor in equilibrium the field is perpendicular to the surface; if it had a component along the surface, the free surface charges would move. The conductor is positive, so the field points outward. Arrow 1 is perpendicular to the surface and points out of the conductor.
4. DArrow 4
   A student who takes the field to point the way an electron would be pushed picks this. The field points in the direction of the force on a positive test charge, which the positive conductor repels: outward, not into the conductor.

Working In electrostatic equilibrium the field just outside a conductor is perpendicular to its surface (no component along it), and for a positive conductor it points outward, away from the surface. At P on an oval the outward perpendicular (arrow 1) differs from the direction away from the center (arrow 2).

CED 10.3.B.1.i · Read this in Fix

1. A4.0 × 10³ N/C Correct
   Outside the sphere its field is that of a point charge at the center, so r is measured from the center: r = 0.10 m + 0.20 m = 0.30 m. E = kQ/r² = (9.0 × 10⁹)(4.0 × 10⁻⁸)/(0.30)² = 4.0 × 10³ N/C.
2. B9.0 × 10³ N/C
   A student who measures r from the surface picks this: (9.0 × 10⁹)(4.0 × 10⁻⁸)/(0.20)². The sphere acts as a point charge at its center, so r = 0.30 m.
3. C1.2 × 10³ N/C
   A student who divides by r instead of r² picks this: (9.0 × 10⁹)(4.0 × 10⁻⁸)/(0.30). The field of a point charge falls off as 1/r², and the units come out right only with r².
4. D7.2 × 10³ N/C
   A student who writes (0.10 + 0.20)² as 0.10² + 0.20² = 0.050 picks this. Add the distances first: r = 0.30 m, so r² = 0.090 m².

Working The sphere acts as a point charge at its center, so r = R + d = 0.10 m + 0.20 m = 0.30 m. E = kQ/r² = (9.0 × 10⁹ N·m²/C²)(4.0 × 10⁻⁸ C)/(0.30 m)² = 3.6 × 10²/0.090 N/C = 4.0 × 10³ N/C.

CED 10.3.B.1.ii · Read this in Fix

1. A2kq/(25d²)
   A student who uses the separation of the two particles, 5d, as the distance for both fields picks this: k(3q)/(5d)² − kq/(5d)² = 2kq/(25d²). The field of each particle at P depends on its own distance from P: 2d for +q and 3d for +3q.
2. Bkq/(4d²)
   A student who counts only the nearer charge picks this. The +3q charge is farther away, but its larger charge gives it the larger field at P, kq/(3d²), opposite to that of +q; the net field is kq/(12d²).
3. C(7/12)kq/d²
   A student who adds the two fields' magnitudes, ignoring their directions, picks this: kq/(4d²) + kq/(3d²) = (7/12)kq/d². Both charges are positive, so each field points away from its source: the field of +q points right and the field of +3q points left. Opposite fields subtract: kq/(3d²) − kq/(4d²) = kq/(12d²).
4. Dkq/(12d²) Correct
   The field of +q at P points away from it, to the right, with magnitude kq/(2d)² = kq/(4d²). The field of +3q points away from it, to the left, with magnitude k(3q)/(3d)² = kq/(3d²). They are opposite, so the net field is kq/(3d²) − kq/(4d²) = kq/(12d²), to the left: the farther charge contributes the larger field.

Working E₁ (from +q, distance 2d) = kq/(2d)² = kq/(4d²), directed away from +q (to the right). E₂ (from +3q, distance 3d) = k(3q)/(3d)² = kq/(3d²), directed away from +3q (to the left). Net: kq/(3d²) − kq/(4d²) = (4 − 3)kq/(12d²) = kq/(12d²), to the left.

CED 10.3.A.3.i · Read this in Fix

1. A√(2md/(qE)) Correct
   From E = F_E/q, the force on the particle is F_E = qE, constant and in the direction of the field, so its acceleration is a = qE/m. Starting from rest with constant acceleration, d = (1/2)at² = qEt²/(2m), so t² = 2md/(qE) and t = √(2md/(qE)).
2. B√(md/(qE))
   A student who takes the distance as the final speed times the time picks this: d = (at)t = at² gives t = √(md/(qE)). The particle starts from rest, so its average speed is only half its final speed: d = (1/2)at², and t = √(2md/(qE)).
3. C√(2qmd/E)
   A student who rearranges E = F_E/q as F_E = E/q picks this: a = E/(qm), and d = (1/2)at² gives t = √(2qmd/E). Multiplying both sides of E = F_E/q by q gives F_E = qE, so a = qE/m and t = √(2md/(qE)).
4. Dmd/(qE)
   A student who treats the acceleration as if it were a speed picks this: t = d/a = d/(qE/m) = md/(qE). The acceleration is the rate at which the speed changes, not the speed; starting from rest, d = (1/2)at², so t = √(2md/(qE)). The units show the slip: md/(qE) is in s², not s.

Working E = F_E/q, so F_E = qE, constant and along the field. a = F_E/m = qE/m. From rest with constant acceleration, d = (1/2)at² = qEt²/(2m), so t = √(2md/(qE)). Errors: distance taken as final speed × time, d = (at)t = at² → √(md/(qE)); F_E = E/q → a = E/(qm) → √(2qmd/E); acceleration used as a speed, t = d/a → md/(qE).

CED 10.3.A.2 · Read this in Fix