10.5 Electric Potential

1. AA particle of charge q placed at P adds U_E = q(50 V) to the system. Correct
   Potential is electric potential energy per unit charge. With a charge q at P, the system's electric potential energy is U_E = qV = q(50 V), which is 50 J for every coulomb of charge placed there.
2. BAny charged particle placed at P has 50 J of electric potential energy.
   A student who treats the potential as the energy of whatever is placed at P picks this. 50 V means 50 J per coulomb: a particle of charge q gives U_E = q(50 V), which depends on q.
3. CThe electric field produced at P by the objects is 50 V/m.
   A student who takes the potential's value as the field's magnitude picks this. The field depends on how the potential changes near P, |ΔV/Δr|, which the value 50 V at one point does not give.
4. DP has no electric potential until a charge is placed there.
   A student who thinks a potential needs a charge at the point picks this. The fixed charged objects produce the potential at P whether or not anything is there; a charge placed at P only reveals it.

Working V = U_E/q, so U_E = qV = q(+50 V) = (50 J/C)q.

CED 10.5.A.1 · Read this in Fix

1. A4.24kq/s
   A student who ignores the sign of the −q particle picks this: 3 × √2 kq/s ≈ 4.24kq/s. The negative particle contributes a negative potential, which cancels one of the positive contributions.
2. B1.41kq/s Correct
   Each particle is s/√2 from the center, so each contributes kq_i/(s/√2) = √2 kq_i/s. Adding the three as signed scalars: √2 k(q + q − q)/s = √2 kq/s ≈ 1.41kq/s.
3. C3.16kq/s
   A student who adds the three potentials as vectors picks this: the contributions of the +q and −q particles on one diagonal would line up (2√2 kq/s) and the other +q's would be perpendicular (√2 kq/s), giving √10 kq/s ≈ 3.16kq/s. Potential is a scalar; add the three terms as numbers.
4. D1.00kq/s
   A student who uses the side of the square, the distance between neighbouring particles, as r picks this: k(q + q − q)/s. In V = kq/r, r is the distance from each particle to P, which is half the diagonal, s/√2.

Working Distance from each corner to the center: r = s/√2. V = Σ kq_i/r_i = k(+q + q − q)/(s/√2) = √2 kq/s ≈ 1.41kq/s.

CED 10.5.A.2 · Read this in Fix

1. AChemical reactions make new charge, which then collects at the two terminals.
   A student who thinks a battery produces charge picks this. Charge is conserved: the reactions separate existing charge; they do not create it.
2. BA store of charge put in when it was made keeps its terminals charged until used up.
   A student who pictures a battery as a container of charge picks this. The battery keeps separating charge chemically; what runs down is the chemical energy available, not a supply of charge.
3. CIt pushes out a fixed current, and the potential difference follows from that current.
   A student who treats a battery as a constant-current source picks this. A battery maintains a potential difference; the current it delivers depends on what is connected to it.
4. DChemical reactions move positive and negative charge apart, to opposite terminals. Correct
   The chemical processes in the battery separate charge that is already present, building up positive charge at one terminal and negative charge at the other. That separation of charge is what produces the potential difference.

CED 10.5.A.3.i · Read this in Fix

1. AThe electric potential at the surface of each sphere Correct
   Electrons move through the wire as long as there is a potential difference between the spheres. They stop when the surfaces are at the same potential, which is the equilibrium condition for conductors in contact.
2. BThe amount of excess charge on each of the two spheres
   A student who thinks connected conductors share charge equally picks this. Charge moves until the potentials are equal; spheres of different radii then carry different charges, the larger sphere the larger charge.
3. CThe excess charge per unit area of each sphere's surface
   A student who thinks the charge spreads evenly over the combined surface picks this. Equal potentials, not equal charge per unit area, set the final state; the smaller sphere ends with the greater charge per unit area.
4. DThe magnitude of the electric field just outside each sphere
   A student who expects charge flow to stop when the fields balance picks this. Charge moves between the spheres while there is a potential difference between them, so the potentials become equal; the fields outside the two spheres differ.

CED 10.5.A.4 · Read this in Fix

1. A8.3 × 10² V/m
   A student who divides the potential at the point by its position picks this: 250 V/0.30 m. That is the ratio of coordinates, not the slope. The field is the change in V divided by the change in x.
2. B1.5 × 10³ V/m Correct
   From x = 0.20 m to x = 0.40 m the graph is a straight line, so the field is uniform there and equals the magnitude of the slope: |ΔV/Δx| = |100 V − 400 V|/(0.40 m − 0.20 m) = 300 V/0.20 m = 1.5 × 10³ V/m.
3. C2.5 × 10² V/m
   A student who takes the potential at the point as the field's magnitude picks this, reading 250 V at x = 0.30 m. The field depends on how fast V changes with x there, not on the value of V.
4. D1.1 × 10² V/m
   A student who uses the area under the graph from x = 0 to x = 0.30 m picks this: (400 V)(0.20 m) + ½(400 V + 250 V)(0.10 m) ≈ 1.1 × 10² V·m. The field is given by the slope of a V–x graph, not by its area, which has units of V·m.

Working Between x = 0.20 m (V = 400 V) and x = 0.40 m (V = 100 V) the graph is linear. |E| = |ΔV/Δx| = 300 V/0.20 m = 1500 V/m = 1.5 × 10³ V/m at every point of that interval, including x = 0.30 m.

CED 10.5.B.1 · Read this in Fix

1. APerpendicular to the 75 V line, toward the 100 V line
   A student who thinks the field points toward higher potential picks this. The direction is right to be perpendicular to the lines, but the field points toward lower potential, the 50 V line.
2. BPerpendicular to the 75 V line, toward the 50 V line Correct
   The field is perpendicular to the equipotential lines and points toward decreasing potential. At P that is perpendicular to the 75 V line, away from the 100 V line and toward the 50 V line.
3. CAlong the 75 V line, tangent to it at the point P
   A student who treats equipotential lines as field lines picks this. The field has no component along an equipotential line; it is perpendicular to it.
4. DIt has no direction: the field at P is zero
   A student who thinks the field is zero on an equipotential picks this. The potential changes across the line, from 100 V on one side to 50 V on the other, so there is a field at P, perpendicular to the line.

Working E⃗ is perpendicular to the equipotential lines and points toward decreasing potential: at P, perpendicular to the 75 V line, from the 100 V side toward the 50 V side.

CED 10.5.B.2.ii · Read this in Fix

1. A−2.00V₀
   A student who treats the potential as the potential energy of the particle at P picks this: the energy of the new particle is (−2q)V₀, −2 times the old value qV₀, so they multiply the potential by −2. It is U_E that changes, to (−2q)V₀; the potential, U_E per unit charge, stays V₀.
2. B+0.50V₀
   A student who reads V = U_E/q as 'V is inversely proportional to q' and compares only the sizes of the charges picks this: 2q is twice q, so they halve the potential. U_E changes in proportion to q, sign included, so the ratio U_E/q stays V₀.
3. C+1.00V₀ Correct
   The potential at P is set by the fixed objects and the position of P. Replacing the particle changes the system's U_E, from +qV₀ to −2qV₀, but not the potential energy per unit charge, V₀.
4. D−1.00V₀
   A student who thinks the sign of the potential follows the sign of the charge placed at P picks this. The sign of V is set by the fixed objects; only U_E = qV changes sign when q does.

Working V = U_E/q. Before: U_E = (+q)V₀, so V = V₀. After: U_E = (−2q)V₀, so V = (−2qV₀)/(−2q) = V₀ = +1.00V₀.

CED 10.5.A.1 · Read this in Fix

1. A6.3 × 10² V
   A student who ignores the sign of the −4.0 nC charge picks this: 540 V + 90 V. A negative charge produces a negative potential, which subtracts.
2. B2.5 × 10³ V
   A student who divides by r², as for the field of a point charge, picks this: k(12.0 nC)/(0.20 m)² − k(4.0 nC)/(0.40 m)² ≈ 2.5 × 10³. The potential of a point charge is kq/r, with r to the first power.
3. C1.2 × 10² V
   A student who uses the 0.60 m separation of the two particles as r for both picks this: k(12.0 nC − 4.0 nC)/(0.60 m). Each r is the distance from that particle to the point: 0.20 m and 0.40 m.
4. D4.5 × 10² V Correct
   Add the two potentials as signed scalars, each with its own distance to the point: k(12.0 nC)/(0.20 m) = 540 V and k(−4.0 nC)/(0.40 m) = −90 V. V = 540 V − 90 V = 450 V.

Working V = k(q₁/r₁ + q₂/r₂) = (9.0 × 10⁹)[(12.0 × 10⁻⁹)/(0.20) + (−4.0 × 10⁻⁹)/(0.40)] = 540 V − 90 V = 450 V = 4.5 × 10² V.

CED 10.5.A.2 · Read this in Fix

1. AIt is zero, although the electric field at M is not zero. Correct
   M is d/2 from each particle, so the potentials kQ/(d/2) and k(−Q)/(d/2) add to zero. The fields at M both point toward −Q, so they add rather than cancel: the field is not zero where the potential is.
2. BIt is not zero, because the electric field at M is not zero.
   A student who expects potential and field to be zero or nonzero together picks this. The field at M is not zero, but the potentials, which are signed scalars, cancel exactly: +2kQ/d − 2kQ/d = 0.
3. CIt is 4kQ/d, twice the potential that +Q alone produces there.
   A student who ignores the sign of −Q picks this, adding 2kQ/d + 2kQ/d. The negative particle produces a negative potential, −2kQ/d, which cancels the positive one.
4. DIt has no value until a test charge is placed at M.
   A student who thinks potential exists only where a charge is placed picks this. The two fixed particles produce a potential at M whether or not a test charge is there; its value is zero.

Working V_M = kQ/(d/2) + k(−Q)/(d/2) = 0. E⃗ from +Q at M points away from +Q (toward −Q); E⃗ from −Q points toward −Q; same direction, |E| = 2 × kQ/(d/2)² = 8kQ/d² ≠ 0.

CED 10.5.A.2 · Read this in Fix

1. A+9.60 × 10⁻¹⁸ J
   A student who thinks moving to a higher potential always raises U_E picks this. The electron's charge is negative, so ΔU_E = qΔV is negative when ΔV is positive.
2. B−1.36 × 10⁻¹⁷ J
   A student who uses the potential at the end point instead of the potential difference picks this: (−1.60 × 10⁻¹⁹ C)(85.0 V). That is U_E at C relative to the zero of potential, not the change from A to C.
3. C−9.60 × 10⁻¹⁸ J Correct
   ΔU_E = qΔV = q(V_C − V_A) = (−1.60 × 10⁻¹⁹ C)(85.0 V − 25.0 V) = −9.60 × 10⁻¹⁸ J. Only the start and end points matter; the detour through B does not change the result.
4. D+1.28 × 10⁻¹⁷ J
   A student who adds up the change along each part of the path picks this: U_E falls by 1.12 × 10⁻¹⁷ J from A to B (70.0 V) and rises by 1.60 × 10⁻¹⁸ J from B to C (10.0 V), and they add the two amounts to get 1.28 × 10⁻¹⁷ J. The change depends only on the end points: ΔU_E = q(V_C − V_A) = −9.60 × 10⁻¹⁸ J, a decrease.

Working ΔU_E = q(V_C − V_A) = (−1.60 × 10⁻¹⁹ C)(85.0 V − 25.0 V) = (−1.60 × 10⁻¹⁹ C)(60.0 V) = −9.60 × 10⁻¹⁸ J.

CED 10.5.A.3 · Read this in Fix

1. A0.50E₀
   A student who takes the field to be the potential difference itself picks this, halving the field with ΔV. The field is the potential difference per unit distance, and the distance was halved too.
2. B0.25E₀
   A student who multiplies the potential difference by the distance picks this: (½)(½) = ¼. The average field is ΔV divided by Δr, so the two halvings cancel.
3. C2.00E₀
   A student who uses the inverse-square dependence of a point charge's field picks this: (½)/(½)² = 2. The average field is |ΔV/Δr|, with Δr to the first power.
4. D1.00E₀ Correct
   |E| = |ΔV/Δr|. For C and D, (ΔV/2)/(Δr/2) = ΔV/Δr, so the average field is the same, E₀: halving both the potential difference and the distance leaves their ratio unchanged.

Working E_AB = ΔV/Δr = E₀. E_CD = (ΔV/2)/(Δr/2) = ΔV/Δr = E₀ = 1.00E₀.

CED 10.5.B.1 · Read this in Fix

1. APerpendicular to the lines toward −20 V, since a negative charge is pushed toward higher potential Correct
   The field points toward decreasing potential, from the −20 V line toward the −40 V line, and perpendicular to the lines. The force on an electron is F⃗ = qE⃗ with q < 0, opposite to E⃗: perpendicular to the lines, toward the −20 V line, which is the higher potential.
2. BPerpendicular to the lines toward −40 V, since every charge is pushed toward lower potential
   A student who thinks every charge moves 'downhill' to lower potential picks this. That is true for a positive charge. The electron's charge is negative, so the force on it is opposite to the field, toward higher potential.
3. CAlong the −30 V line through P, since the force on a charge points along an equipotential line
   A student who treats equipotential lines as field lines picks this. The field, and so the force, is perpendicular to the equipotential lines and has no component along them.
4. DIt stays at rest at P, since the field is zero at every point on an equipotential line
   A student who thinks the field vanishes on an equipotential picks this. The potential changes across the line, from −40 V to −20 V, so the field at P is not zero; it is perpendicular to the line.

Working E⃗ points perpendicular to the isolines toward decreasing V (from −20 V toward −40 V). Electron: F⃗ = qE⃗, q = −e, so F⃗ is opposite to E⃗: perpendicular to the lines toward the −20 V line (increasing V). It starts from rest, so it initially moves along F⃗.

CED 10.5.B.2 · Read this in Fix

1. AP to B
   A student who thinks a longer move gives a larger change picks this, since B is the point farthest from P. B is on the +15 V line, so ΔV = −15 V; the change depends only on the potentials at the start and end points, not on the distance.
2. BP to C
   A student who compares potentials by size alone picks this: taking −30 V at D to equal +30 V at P, they find no change for the move to D and the largest change, 30 V, for the move to C on the 0 V line. Potentials are signed numbers: going from +30 V to −30 V is a 60 V drop, twice the drop to C.
3. CP to D Correct
   P is on the +30 V line and D on the −30 V line, so ΔV = −60 V, the largest potential difference of the four moves, and |ΔU_E| = q|ΔV| is largest. The distance between the points and the path taken do not matter.
4. DP to A
   A student who treats equipotential lines as field lines picks this: moving from P to A follows the +30 V line, which they take to be the direction of the field and force, while the other moves cross the lines. P and A are on the same equipotential line, so ΔV = 0 and U_E does not change at all.

Working V_P = +30 V. ΔV: to A (+30 V): 0; to B (+15 V): −15 V; to C (0 V): −30 V; to D (−30 V): −60 V. |ΔU_E| = q|ΔV| is largest for P → D.

CED 10.5.B.2.i · Read this in Fix

1. APoint A
   A student who thinks the field points toward higher potential picks this: A is where the field lines start, next to +Q. Potential decreases along the field, so A, next to +Q, has one of the highest potentials.
2. BPoint B
   A student who takes zero to be the lowest potential picks this: at the midpoint the two potentials cancel, V = 0. C has a negative potential, which is lower than zero.
3. CPoint D
   A student who thinks the potential is lowest where the field is weakest picks this: the field lines are sparse at D. D is on the +Q side, where the potential is positive; a weak field means the potential changes slowly there, not that it is low.
4. DPoint C Correct
   The field points toward decreasing potential. Following the field lines from +Q toward −Q, the potential falls all the way, so the lowest of the four is C, next to −Q, where V = kQ/r₊ − kQ/r₋ is strongly negative.

Working Field lines run from +Q to −Q and V decreases along them. With separation d (figure geometry: A and C each 0.24d from the nearer particle, D 0.63d to the left of +Q), V = kQ(1/r₊ − 1/r₋): V_A ≈ +2.9kQ/d, V_D ≈ +1.0kQ/d, V_B = 0, V_C ≈ −2.9kQ/d. Lowest: C.

CED 10.5.B.2.iii · Read this in Fix

1. ACorrect: a field at a point would make the potential change, so on the line it is zero.
   A student who thinks the field vanishes on an equipotential picks this. A field makes the potential change along its own direction; perpendicular to the line the potential does change, so the field there is not zero.
2. BIncorrect: the field is not zero, and it points along the line from one end to the other.
   A student who treats equipotential lines as field lines picks this. The field is indeed not zero, but it is perpendicular to the line; it has no component along it.
3. CIncorrect: the field has no component along the line but can be perpendicular to it. Correct
   An unchanging potential along the line means only that the field has no component along the line. The potential does change across the line, so the field can be, and usually is, nonzero and perpendicular to it.
4. DIncorrect: the field equals the potential on the line, which is usually not zero.
   A student who takes the potential's value as the field's magnitude picks this. The field depends on how the potential changes with position, |ΔV/Δr|, not on the value of the potential; a line at 0 V can still have a field at its points.

CED 10.5.B.2.iv · Read this in Fix

1. A0.50d
   A student who expects the potential to be zero midway between a positive and a negative particle picks this. That holds only for equal and opposite charges. At the midpoint the +3q particle's contribution, 6kq/d, is three times the size of the −q particle's, −2kq/d, so V = +4kq/d there, not zero.
2. B0.75d Correct
   The potentials of the two particles add as signed scalars: k(3q)/x − kq/(d − x) = 0 gives 3(d − x) = x, so x = 3d/4 = 0.75d. The zero lies closer to the particle with the smaller charge, so that its shorter distance makes up for its smaller charge.
3. C0.63d
   A student who uses the inverse-square dependence of the field for the potential picks this: 3/x² = 1/(d − x)² gives x = √3d/(1 + √3) ≈ 0.63d. A point charge's potential is kq/r, with r to the first power, which puts the zero at 0.75d.
4. D0.25d
   A student who takes each particle's potential to grow in proportion to the distance from it picks this: 3x = d − x gives x = 0.25d. The potential kq/r falls with distance, so the zero is far from the larger charge, at 0.75d, not near it.

Working At P (0 < x < d): V = k(3q)/x + k(−q)/(d − x) = 0, so 3/x = 1/(d − x), 3d − 3x = x, x = 3d/4 = 0.75d.

CED 10.5.A.2 · Read this in Fix

1. A+1.00kQq/d
   A student who uses the potential at the end point as if it were the potential difference picks this: qV(2d) = kQq/d. The change in U_E depends on V_final − V_initial; the potential at x = d, 4kQ/(3d), is higher than at x = 2d, so U_E decreases.
2. B−0.67kQq/d
   A student who adds the two potentials as vectors picks this: the particles lie on opposite sides of each point, so the contributions are subtracted, giving V(d) = kQ/d − kQ/(3d) = 2kQ/(3d) and V(2d) = 0, and ΔU_E = −2kQq/(3d). Potentials are scalars: both positive contributions add.
3. C−0.33kQq/d Correct
   Add the potentials as signed scalars at each point: V(d) = kQ/d + kQ/(3d) = 4kQ/(3d) and V(2d) = kQ/(2d) + kQ/(2d) = kQ/d. Then ΔU_E = q(V_final − V_initial) = q(kQ/d − 4kQ/(3d)) = −kQq/(3d) ≈ −0.33kQq/d.
4. D−1.33kQq/d
   A student who thinks the potential is zero where the field is zero picks this: the fields cancel at x = 2d, so V(2d) is taken as 0 and ΔU_E = q(0 − 4kQ/(3d)). The fields cancel there, but the two positive potentials add to kQ/d.

Working V(d) = kQ/d + kQ/(3d) = 4kQ/(3d). V(2d) = kQ/(2d) + kQ/(2d) = kQ/d. ΔV = V(2d) − V(d) = −kQ/(3d). ΔU_E = qΔV = −kQq/(3d) ≈ −0.33kQq/d.

CED 10.5.A.3 · Read this in Fix