9.3 Conservation of Electric Energy

1. AThe change in electric potential energy of the system of the particle and the field Correct
   ΔU_E = qΔV is the change in the electric potential energy of the object–field system: the particle together with the charges that produce the field.
2. BThe change in the energy stored inside the particle, which carries it from A to B
   A student who thinks potential energy is stored in the particle alone picks this. Electric potential energy belongs to the system of the particle and the field, and depends on their arrangement.
3. CThe electric potential energy of the system when the particle is at point B
   A student who uses the potential energy at one point in place of the change between two points picks this. qΔV is the change in U_E from A to B; it equals the value at B only if U_E at A happens to be zero.
4. DThe work that the electric field does on the particle as it moves from A to B
   A student who equates the work done by the field with the change in potential energy picks this. The work done by the field is −qΔV, the negative of the change in U_E.

CED 9.3.A.1 · Read this in Fix

1. A7.0 × 10⁵ m/s
   A student who thinks the electron's potential energy rises as it moves to higher potential, as a positive charge's would, subtracts 1.6 × 10⁻¹⁸ J from its kinetic energy and picks this. For an electron ΔU_E = (−e)(+10 V) < 0, so it gains kinetic energy.
2. B2.7 × 10⁶ m/s Correct
   K_i = ½(9.11 × 10⁻³¹ kg)(2.0 × 10⁶ m/s)² = 1.82 × 10⁻¹⁸ J. ΔU_E = qΔV = (−1.60 × 10⁻¹⁹ C)(+10 V) = −1.60 × 10⁻¹⁸ J, so K_f = 3.42 × 10⁻¹⁸ J and v = √(2K_f/m_e) ≈ 2.7 × 10⁶ m/s.
3. C1.9 × 10⁶ m/s
   A student who sets the final kinetic energy equal to e|ΔV| = 1.6 × 10⁻¹⁸ J, leaving out the kinetic energy the electron already had, picks this. The energy gained adds to the initial 1.82 × 10⁻¹⁸ J.
4. D3.9 × 10⁶ m/s
   A student who adds speeds instead of energies picks this: 2.0 × 10⁶ m/s plus the 1.9 × 10⁶ m/s the electron would gain from rest. Kinetic energies add; speeds combine as v = √(v₀² + 2e|ΔV|/m_e).

Working K_i = ½m_ev₀² = ½(9.11 × 10⁻³¹)(2.0 × 10⁶)² = 1.822 × 10⁻¹⁸ J. ΔK = −qΔV = −(−1.60 × 10⁻¹⁹ C)(10 V) = +1.60 × 10⁻¹⁸ J. K_f = 3.422 × 10⁻¹⁸ J; v = √(2K_f/m_e) = 2.74 × 10⁶ m/s ≈ 2.7 × 10⁶ m/s.

CED 9.3.A.2 · Read this in Fix

1. AFrom P to A
   A student who thinks the change in potential energy grows with the distance moved picks this: A is the point farthest from P. A is on the same equipotential line as P, so ΔU_E = 0.
2. BFrom P to B
   A student who treats the electron like a positive charge, whose U_E rises as it moves to higher potential, picks this: B is at +60 V, the highest potential. For an electron ΔU_E = (−e)(+20 V) = −20 eV: U_E decreases.
3. CFrom P to D Correct
   ΔU_E = qΔV with q = −e. P is at +40 V. To A: ΔV = 0; to B: ΔV = +20 V, ΔU_E = −20 eV; to C: ΔV = −40 V, ΔU_E = +40 eV; to D: ΔV = −60 V, ΔU_E = +60 eV. The largest increase is for the move to D, the lowest potential.
4. DFrom P to C
   A student who treats the potentials by their magnitudes picks this: C is at 0 V, the smallest magnitude, so the change from 40 V looks largest. With signs, D at −20 V is lower than C and gives the larger increase.

Working ΔU_E = (−e)(V_f − 40 V): A 0; B −20 eV; C +40 eV; D +60 eV (1 eV = 1.60 × 10⁻¹⁹ J). Largest increase: D.

CED 9.3.A.1 · Read this in Fix

1. A+6.0 × 10⁻⁴ J
   A student who treats the negative particle as if it were positive, so that moving to higher potential raises U_E, picks this: (3.0 × 10⁻⁶ C)(200 V). With q = −3.0 μC, ΔU_E = qΔV is negative.
2. B−2.4 × 10⁻⁴ J
   A student who uses the potential at B alone in place of the potential difference picks this: (−3.0 × 10⁻⁶ C)(80 V). The difference is V_B − V_A = +80 V − (−120 V) = +200 V.
3. C+1.2 × 10⁻⁴ J
   A student who treats −120 V as 120 V picks this: ΔV = 80 V − 120 V = −40 V, and (−3.0 × 10⁻⁶ C)(−40 V) = +1.2 × 10⁻⁴ J. The potential at A is 120 V below zero, so ΔV = +200 V.
4. D−6.0 × 10⁻⁴ J Correct
   ΔV = V_B − V_A = +80 V − (−120 V) = +200 V. ΔU_E = qΔV = (−3.0 × 10⁻⁶ C)(+200 V) = −6.0 × 10⁻⁴ J: a negative charge moving to higher potential lowers the system's potential energy.

Working ΔV = (+80 V) − (−120 V) = +200 V. ΔU_E = qΔV = (−3.0 × 10⁻⁶ C)(200 V) = −6.0 × 10⁻⁴ J.

CED 9.3.A.1 · Read this in Fix

1. A(2/3)kQe/a Correct
   Energy is conserved: K = −ΔU_E = U_i − U_f = kQe/a − kQe/(3a) = (2/3)kQe/a.
2. BkQe/a
   A student who takes the potential energy at the start, kQe/a, as the energy converted picks this. That would be the kinetic energy only when the proton is infinitely far away; at 3a the system still has U_E = kQe/(3a).
3. C2kQe/a
   A student who multiplies the force at the release point, kQe/a², by the distance moved, 2a, picks this. The force falls off as the proton moves away, so the work is less; it equals −ΔU_E.
4. D(8/9)kQe/a²
   A student who uses U_E = kQe/r² picks this: kQe(1/a² − 1/(9a²)). That has units of force, not energy; U_E = kQe/r.

Working U_i = kQe/a, U_f = kQe/(3a). K_f = U_i − U_f = (2/3)kQe/a. Distractors: all of U_i → kQe/a; F(a) × 2a = 2kQe/a; 1/r² → (8/9)kQe/a².

CED 9.3.A.2 · Read this in Fix

1. AK_α/K_p = 1
   A student who thinks every particle gains the same kinetic energy through the same potential difference picks this. The energy gained is |q||ΔV|, and the alpha particle has twice the charge.
2. BK_α/K_p = 2 Correct
   With only the electric force acting, K = −qΔV = |q||ΔV|. The alpha particle's charge is twice the proton's, so it gains twice the kinetic energy; the mass does not enter.
3. CK_α/K_p = 4
   A student who thinks both particles reach the same speed, as objects dropped from the same height do, picks this: at equal speeds K ∝ m, giving 4. The kinetic energy is set by the charge: |q||ΔV|.
4. DK_α/K_p = ½
   A student who thinks the kinetic energy gained is proportional to q/m picks this: (2e/4m)/(e/m) = ½. The charge-to-mass ratio sets the acceleration and final speed, not the energy.

Working K = |q||ΔV|: K_α = 2e|ΔV|, K_p = e|ΔV|, ratio 2.

CED 9.3.A.2 · Read this in Fix

1. Av₁ > v₃ > v₂
   A student who thinks a charge moves fastest where the potential is highest picks this: V₁ = 40 V, V₃ = −10 V, V₂ = −40 V. For a proton K + eV is constant, so it is fastest where V is lowest.
2. Bv₃ > v₁ > v₂
   A student who thinks the proton is fastest where the force on it is largest picks this: the graph is steepest at point 3, less steep at point 1 and level at point 2. The slope gives the force, which sets the acceleration; the speed follows from V.
3. Cv₁ > v₂ > v₃
   A student who thinks the proton uses up energy as it travels, so that it is slower the farther it has gone, picks this. With only the electric force acting, K + U_E is constant; the proton is fastest where U_E = eV is lowest.
4. Dv₂ > v₃ > v₁ Correct
   Energy conservation: K = K₀ + e(V₀ − V) with V₀ = 60 V at x = 0. The proton's kinetic energy is largest where V is lowest: V₂ = −40 V, then V₃ = −10 V, then V₁ = 40 V. So v₂ > v₃ > v₁.

Working K(x) = K₀ + e[V(0) − V(x)] = K₀ + e(60 V − V(x)). V₁ = 40 V (x = 0.10 m), V₂ = −40 V (x = 0.35 m), V₃ = −10 V (x = 0.45 m): K₂ = K₀ + 100 eV > K₃ = K₀ + 70 eV > K₁ = K₀ + 20 eV. Field magnitudes |dV/dx|: point 1, 200 V/m; point 2, 0; point 3, 600 V/m.

CED 9.3.A.2 · Read this in Fix

1. ATheir speeds are equal, and so the proton has more kinetic energy
   A student who thinks both reach the same speed, as in free fall, picks this; at equal speeds the heavier proton would have more kinetic energy. In fact each gains e(100 V), so the kinetic energies are equal and the lighter electron is faster.
2. BThe electron has more kinetic energy, and it is also moving faster
   A student who thinks the energy gained depends on the charge-to-mass ratio picks this. Each gains e(100 V) = 100 eV; the smaller mass gives the electron a higher speed, not more energy.
3. CTheir kinetic energies are equal, and the electron is moving faster Correct
   Each particle has charge of magnitude e and crosses 100 V, so each gains K = e(100 V) = 100 eV = 1.60 × 10⁻¹⁷ J. With equal kinetic energies, the particle with the smaller mass, the electron, has the greater speed.
4. DThe proton has more kinetic energy, and it is moving faster than the electron
   A student who thinks the electron’s U_E rises as it moves to higher potential, as a positive charge’s would, expects it to gain little kinetic energy and so picks this. The electron is negative: moving to higher potential lowers U_E, and it gains 100 eV, the same as the proton.

Working Proton X → Y must be to lower potential (it starts from rest), so V_X − V_Y = +100 V. Proton: ΔK = −eΔV = +100 eV. Electron Y → X (to higher potential): ΔK = −(−e)(+100 V) = +100 eV. Equal K; v = √(2K/m), so v_e/v_p = √(m_p/m_e) ≈ 43.

CED 9.3.A.2 · Read this in Fix

1. A(kQq/d) ln 2 Correct
   With x the distance from the particle, λ = Q/d and the rod runs from x = d to x = 2d: V = kλ∫dx/x = (kQ/d) ln 2. Energy conservation from rest to infinity: K = qV = (kQq/d) ln 2.
2. BkQq/d
   A student who treats all the rod's charge as if it were at the end nearest the particle picks this: kQq/d. Most of the rod is farther than d from the particle, so the potential energy is smaller.
3. CkQq/(2d²)
   A student who integrates dq/x² picks this: kλ(1/d − 1/(2d))q = kQq/(2d²). That has units of force, not energy; the potential integrand is dq/x.
4. D(2/3)kQq/d
   A student who replaces the rod by a point charge Q at its center, 1.5d from the particle, picks this: kQq/(1.5d). The nearer half of the rod contributes more than the farther half, so the true value, (ln 2)kQq/d ≈ 0.69kQq/d, is larger.

Working λ = Q/d; element dq = λ dx at distance x ∈ [d, 2d]. V = k∫_d^2d λ dx/x = (kQ/d) ln 2. K_∞ = U_i − 0 = qV = (kQq/d) ln 2 ≈ 0.693kQq/d. Distractors: nearest end → kQq/d; center → (2/3)kQq/d ≈ 0.667kQq/d; 1/x² → kQq/(2d²).

CED 9.3.A.2 · Read this in Fix

1. AThe field does positive work on the electron, and that work is stored as potential energy
   A student who thinks the work done by the field is stored as U_E picks this. The work done by the field is W = −ΔU_E. Here the force on the electron is opposite its displacement, so the field does negative work, and that is why U_E increases.
2. BThe value of U_E = qV at its final position is positive, so U_E has increased
   A student who uses the potential energy at one point in place of the change picks this. The value of U_E at a single point depends on where V = 0 is chosen and says nothing about whether U_E rose; what matters is ΔU_E = qΔV. Moving along the field, ΔV < 0, and with q = −e, ΔU_E > 0.
3. CThe electron carries more stored energy the farther it has moved into the field
   A student who thinks potential energy is stored in the electron alone and grows with the distance it travels into the field picks this. U_E belongs to the electron–field system and depends on the change in potential, ΔU_E = qΔV, not on distance traveled.
4. DIt moves to lower potential, and ΔU_E = qΔV is positive when q and ΔV are both negative Correct
   The field points toward lower potential, so the electron moves to lower potential: ΔV < 0. With q = −e, ΔU_E = qΔV = (−e)(negative) > 0, so U_E increases.

Working E⃗ points toward lower V, so moving along E⃗: ΔV < 0. q = −e < 0, so ΔU_E = qΔV > 0. Also W_field = qE⃗·Δr⃗ < 0 = −ΔU_E.

CED 9.3.A.1 · Read this in Fix

1. A1.41 q√(k/(md))
   A student who gives all of the potential energy kq²/d to particle 1, as if particle 2 were held fixed, picks this: ½mv₁² = kq²/d. Particle 2 also moves off, and it carries a third of the energy.
2. B1.15 q√(k/(md)) Correct
   The electric forces are internal, so the total momentum stays zero: mv₁ = 2mv₂. The potential energy kq²/d becomes the kinetic energy of both particles: kq²/d = ½mv₁² + ½(2m)(v₁/2)² = ¾mv₁², so v₁ = √(4kq²/(3md)) ≈ 1.15 q√(k/(md)).
3. C1.00 q√(k/(md))
   A student who splits the potential energy equally between the particles picks this: ½mv₁² = kq²/(2d). Momentum conservation makes the lighter particle move twice as fast, so it takes two thirds of the energy.
4. D0.82 q√(k/(md))
   A student who gives both particles the same speed, because the forces on them are equal and opposite, picks this: ½(3m)v² = kq²/d. Equal forces acting for equal times give momenta of equal magnitude, so particle 1, with half the mass, moves twice as fast as particle 2.

Working Take both particles as the system. The electric forces are internal, so momentum stays zero: mv₁ = 2mv₂, so v₂ = v₁/2. Energy: the potential energy kq²/d becomes the kinetic energy of both particles: kq²/d = ½mv₁² + ½(2m)(v₁/2)² = ¾mv₁², so v₁ = √(4kq²/(3md)) = (2/√3)q√(k/(md)) ≈ 1.15 q√(k/(md)). Distractors (sympy-checked): particle 2 treated as fixed, ½mv₁² = kq²/d, v₁ = √2 q√(k/(md)) ≈ 1.41 q√(k/(md)); energy shared equally, ½mv₁² = kq²/(2d), v₁ = 1.00 q√(k/(md)); equal speeds, ½(3m)v² = kq²/d, v = √(2/3) q√(k/(md)) ≈ 0.82 q√(k/(md)).

CED 9.3.A.2 · Read this in Fix