University Physics II · Midterm Examination

A neutral solid metal sphere stands on an insulating support. A glass rod carrying positive charge is held close to the left-hand side of the sphere without touching it. While the rod is held in place, a wire connected to earth briefly touches the right-hand side of the sphere, the side farther from the rod, and is then removed. Finally the rod is taken away. What is the final charge on the sphere?

1. APositive, the same sign as the rod, spread uniformly over the surface
2. BZero, because the earth connection was broken before the rod was removed
3. CNegative, concentrated on the left-hand side where the rod had been
4. DNegative, spread uniformly over the surface

A point charge +9Q is fixed at x = 0 and a point charge −4Q is fixed at x = L on the x-axis, where Q > 0. Apart from points infinitely far away, where is the electric field zero?

1. Ax = 3L
2. Bx = 1.8L
3. Cx = 0.6L
4. DNowhere: the field is non-zero at every point a finite distance from the charges

A very long, straight, thin line of charge with uniform linear charge density λ passes perpendicularly through the centres of two opposite faces of an imaginary cube of edge length a. What is the electric flux through one of the four faces that the line does not pass through?

1. Aλa/(6ε₀)
2. Bλa/(4ε₀)
3. Cλa/(πε₀)
4. Dλa/ε₀

Four point charges sit at the corners of a square of side a centred on the origin in the xy-plane: +q at (−a/2, a/2) and at (a/2, a/2), and −q at (−a/2, −a/2) and at (a/2, −a/2), with q > 0. Take the electric potential to be zero infinitely far away and k = 1/(4πε₀). Which statement describes the electric potential V and the electric field E at the origin?

1. AV = 0 and E = 0
2. BV > 0 and E = 0
3. CV = 0, and E has magnitude 4√2kq/a² in the +y direction
4. DV = 0, and E has magnitude 4√2kq/a² in the −y direction

Two initially uncharged capacitors, of capacitance 2.0 μF and 6.0 μF, are connected in series across a 12 V battery and allowed to charge fully. What is the potential difference across the 6.0 μF capacitor?

1. A3.0 V
2. B6.0 V
3. C9.0 V
4. D12 V

Wire X and wire Y are made of the same metal, are at the same temperature, and are joined end to end so that they carry the same steady current. The diameter of X is twice the diameter of Y. What are the ratio of the drift speed in X to that in Y, v_X/v_Y, and the ratio of the electric field inside X to that inside Y, E_X/E_Y?

1. Av_X/v_Y = 1/2 and E_X/E_Y = 1/2
2. Bv_X/v_Y = 1/4 and E_X/E_Y = 1/4
3. Cv_X/v_Y = 1/4 and E_X/E_Y = 1
4. Dv_X/v_Y = 1 and E_X/E_Y = 1/4

A charged capacitor discharges through a fixed resistor. The potential difference across the capacitor falls to half its initial value in 3.0 s. How long after the discharge begins has the energy stored in the capacitor fallen to one-eighth of its initial value?

1. A1.5 s
2. B3.0 s
3. C4.5 s
4. D9.0 s

In a mass spectrometer, ions start from rest, are accelerated through the same potential difference, and then enter a uniform magnetic field perpendicular to their velocity, in which they move on circular arcs. Ion X is ⁴He⁺ (charge +e, mass 4u) and ion Y is ¹⁶O²⁺ (charge +2e, mass 16u), where u is the atomic mass unit. What is the ratio of the radius of Y's path to the radius of X's path?

1. A1/2
2. B1
3. C√2
4. D2

Three small charged beads are fixed on a horizontal insulating board, which defines the xy-plane. Bead 1 carries +4.0 nC at the origin, bead 2 carries −2.0 nC at (0.30 m, 0) and bead 3 carries +3.0 nC at (0, 0.40 m). Point P is at (0.30 m, 0.40 m). Bead 2 was initially neutral and gained its charge by being rubbed with a cloth that was also initially neutral, and no charge escaped to the surroundings. A fourth bead carrying +1.0 nC is available for part (d), and in part (e) bead 2 is replaced by a bead carrying +2.0 nC at the same position. Treat all beads as point charges and ignore the surroundings. Take k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻² and e = 1.60 × 10⁻¹⁹ C.

1. (a)

   State State the charge on the cloth after the rubbing.

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   1. 1

      +2.0 nC: the cloth loses exactly the electrons the bead gains, so the charge of cloth and bead together stays zero

2. (b)

   Calculate Calculate the number of electrons transferred during the rubbing, and state whether they moved from the cloth to the bead or from the bead to the cloth.

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   1. 1

      N = (2.0 × 10⁻⁹ C)/(1.60 × 10⁻¹⁹ C) = 1.25 × 10¹⁰ electrons, moving from the cloth to the bead

3. (c)

   Calculate Calculate the magnitude and direction of the net electrostatic force on bead 1.

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   1. 1

      Force from bead 2: (8.99 × 10⁹)(4.0 × 10⁻⁹)(2.0 × 10⁻⁹)/(0.30)² = 7.99 × 10⁻⁷ N, attractive, so along +x towards bead 2

   2. 2

      Force from bead 3: (8.99 × 10⁹)(4.0 × 10⁻⁹)(3.0 × 10⁻⁹)/(0.40)² = 6.74 × 10⁻⁷ N, repulsive, so along −y away from bead 3

   3. 3

      Combines the perpendicular components: |F| = √[(7.99 × 10⁻⁷)² + (6.74 × 10⁻⁷)²] = 1.05 × 10⁻⁶ N

   4. 4

      Direction: tan⁻¹(6.74/7.99) = 40° from the +x axis, rotated towards the −y axis (components +7.99 × 10⁻⁷ N along x and −6.74 × 10⁻⁷ N along y)

4. (d)

   Determine Determine the electric potential at P, and the work an external agent must do to bring the +1.0 nC bead slowly from a very large distance to rest at P.

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   1. 1

      Distances from P are 0.50 m, 0.40 m and 0.30 m, so V_P = (8.99 × 10⁹)[(4.0/0.50) − (2.0/0.40) + (3.0/0.30)] × 10⁻⁹ V = 117 V

   2. 2

      W = qV_P = (1.0 × 10⁻⁹ C)(117 V) = 1.17 × 10⁻⁷ J, positive work done by the agent

5. (e)

   Compare Bead 2 is replaced by the +2.0 nC bead. Compare the net force on bead 1 in the original and the new arrangements, and state one measurement at P that would distinguish the two arrangements.

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   1. 1

      The force from bead 2 reverses to −x with the same size, so the net force on bead 1 keeps its magnitude of 1.05 × 10⁻⁶ N but now points at 40° from the −x axis towards −y: same magnitude, different direction

   2. 2

      The potential at P distinguishes them: 117 V originally against 207 V in the new arrangement (equivalently 1.17 × 10⁻⁷ J against 2.07 × 10⁻⁷ J of work to bring the +1.0 nC bead to P); accept instead the electric field at P, about 386 N C⁻¹ almost exactly along +x originally against 448 N C⁻¹ at 31° above the +x axis in the new arrangement

A length of high-voltage coaxial cable has a solid inner conductor of radius a = 1.0 mm and a thin outer conducting sheath of radius b = 4.0 mm on the same axis. The space between them is filled with polyethylene of dielectric constant κ = 2.3, which breaks down when the electric field in it reaches E_bd = 2.0 × 10⁷ V m⁻¹. The inner conductor carries free charge +λ per unit length and the sheath carries −λ per unit length. The cable is long enough for end effects to be neglected. A redesign with a thinner inner conductor, of radius 0.50 mm and the same sheath, is under consideration. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹.

1. (a)

   Derive Using Gauss's law, derive an expression for the magnitude of the electric field in the polyethylene at a distance r from the axis, where a < r < b. Justify your choice of Gaussian surface.

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   1. 1

      The charge distribution is unchanged by translation along the axis and by rotation about it, so E is radial and depends only on r; a coaxial cylinder of radius r and length ℓ therefore has E normal to and constant over its curved surface, and zero flux through its flat ends

   2. 2

      Gauss's law in the dielectric in terms of free charge, ∮D⋅dA = q_free with D = κε₀E (or with the bound charge included, which reduces the enclosed charge by the factor κ): κε₀E(2πrℓ) = λℓ

   3. 3

      E = λ/(2πκε₀r), directed radially outward

2. (b)

   Sketch Sketch a graph of the magnitude of the electric field against r from r = 0 to r = 2b, marking on the r-axis the radii at which the graph changes form.

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   1. 1

      E = 0 for r < a (inside the conducting inner conductor) and E = 0 for r > b (a Gaussian cylinder there encloses zero net charge), with a and b marked on the r-axis

   2. 2

      At r = a the field jumps from zero to its maximum, then falls along a 1/r curve (concave up, not a straight line) to one-quarter of that maximum at r = b, where it drops to zero

3. (c)

   Show that Show that the capacitance per unit length of the cable is C/L = 2πκε₀/ln(b/a).

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   1. 1

      Potential difference between the conductors: V = ∫ₐᵇ E dr = (λ/(2πκε₀)) ∫ₐᵇ dr/r = (λ/(2πκε₀)) ln(b/a)

   2. 2

      Capacitance per unit length is charge per unit length over potential difference: C/L = λ/V = 2πκε₀/ln(b/a)

4. (d)

   Determine Determine the largest potential difference that can be applied between the conductors of the present design without breakdown of the polyethylene.

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   1. 1

      The field varies as 1/r, so it is greatest at r = a, the surface of the inner conductor, and breakdown begins there

   2. 2

      Eliminating λ between the results of (a) and (c): E(a) = V/(a ln(b/a)), so V_max = E_bd a ln(b/a) = (2.0 × 10⁷ V m⁻¹)(1.0 × 10⁻³ m) ln 4

   3. 3

      V_max = 2.8 × 10⁴ V (27.7 kV)

5. (e)

   Evaluate The proposed redesign thins the inner conductor to 0.50 mm on the grounds that a wider gap needs a larger potential difference to break down. Evaluate this proposal, and determine the inner radius that gives the highest voltage rating for b = 4.0 mm.

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   1. 1

      With a = 0.50 mm, V_max = (2.0 × 10⁷)(0.50 × 10⁻³) ln 8 = 2.1 × 10⁴ V, lower than 2.8 × 10⁴ V: thinning the conductor raises the field at its surface faster than ln(b/a) grows, so the proposal lowers the rating

   2. 2

      Maximising V_max = E_bd a ln(b/a) with respect to a: dV_max/da = E_bd[ln(b/a) − 1] = 0, so a = b/e = 1.47 mm

   3. 3

      The optimum rating is E_bd b/e = 2.9 × 10⁴ V, about 6% above the present design; the judgement is to reject the proposal and make the inner conductor slightly thicker, not thinner

A 12.0 V supply of internal resistance r₁ = 0.50 Ω and a rechargeable 9.0 V cell of internal resistance r₂ = 1.5 Ω are connected in parallel across a load of resistance R_L = 4.0 Ω. Both positive terminals are joined to node X, both negative terminals are joined to node Y, and the load connects X to Y. The connecting wires have negligible resistance, and the emfs and internal resistances stay constant. In parts (d) and (e) an initially uncharged 0.47 F capacitor, with a manufacturing tolerance of ±20%, is also connected directly between X and Y, and the design specification requires the load voltage to stay above 5.0 V for at least 1.0 s after both sources are disconnected.

1. (a)

   Determine Using Kirchhoff's rules, determine the current in the supply, in the cell and in the load, stating the direction of each, and the potential difference V_XY between X and Y.

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   1. 1

      Defines I₁ and I₂ as the currents leaving the positive terminals of the supply and the cell and I_L as the load current from X to Y, and applies the junction rule at X: I₁ + I₂ = I_L

   2. 2

      Loop rule for each branch: 12.0 − 0.50I₁ = V_XY, 9.0 − 1.5I₂ = V_XY and V_XY = 4.0I_L (or an equivalent pair of independent loop equations)

   3. 3

      Solving (12.0 − V_XY)/0.50 + (9.0 − V_XY)/1.5 = V_XY/4.0 gives V_XY = 10.3 V (10.29 V)

   4. 4

      I₁ = 3.43 A out of the supply's positive terminal; I₂ = −0.857 A, that is 0.857 A into the cell's positive terminal; I_L = 2.57 A from X to Y through the load

2. (b)

   Show that Show that the power delivered by the emf of the 12.0 V supply, 41.1 W, equals the total rate of energy transfer in the two internal resistances, the load and the cell.

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   1. 1

      Dissipation: I₁²r₁ = (3.43)²(0.50) = 5.88 W, I₂²r₂ = (0.857)²(1.5) = 1.10 W and I_L²R_L = (2.57)²(4.0) = 26.4 W, totalling 33.4 W

   2. 2

      The cell stores ε₂|I₂| = (9.0 V)(0.857 A) = 7.71 W as chemical energy because current is driven into its positive terminal, and 33.4 W + 7.71 W = 41.1 W = (12.0 V)(3.429 A)

3. (c)

   Predict Without the capacitor, the load resistance is gradually reduced from 4.0 Ω. Predict the load resistance at which the current in the 9.0 V cell becomes zero, and what the cell does when the load resistance is smaller still.

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   1. 1

      Zero current in the cell means no potential drop across r₂, so V_XY = 9.0 V; then I₁ = (12.0 − 9.0)/0.50 = 6.0 A, all of which passes through the load, so R_L = 9.0 V/6.0 A = 1.5 Ω

   2. 2

      For R_L < 1.5 Ω, V_XY falls below 9.0 V, so current leaves the cell's positive terminal: the cell discharges and shares the load current with the supply (at R_L = 1.0 Ω it delivers 0.55 A)

4. (d)

   Calculate With the load at 4.0 Ω, the 0.47 F capacitor is connected between X and Y and the circuit is left for a long time. At t = 0 both the supply and the cell are disconnected, leaving only the capacitor and the load. Calculate the time at which the potential difference across the load has fallen to 5.0 V.

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   1. 1

      In the steady state no current flows into the capacitor, so the currents found in (a) are unchanged and the capacitor is charged to V₀ = V_XY = 10.3 V

   2. 2

      After disconnection it discharges through the load alone: V = V₀e^(−t/RC) with RC = (4.0 Ω)(0.47 F) = 1.88 s

   3. 3

      t = RC ln(V₀/V) = (1.88 s) ln(10.29/5.0) = 1.36 s

5. (e)

   Evaluate Evaluate whether every capacitor within the ±20% tolerance meets the specification that the load voltage stays above 5.0 V for at least 1.0 s after disconnection.

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   1. 1

      The worst case is the smallest capacitance, 0.80 × 0.47 F = 0.376 F, giving RC = (4.0 Ω)(0.376 F) = 1.504 s; the starting voltage of 10.3 V does not depend on C

   2. 2

      t = (1.504 s) ln(10.29/5.0) = 1.08 s, which exceeds 1.0 s, so every capacitor in tolerance meets the specification, but with a margin of only about 0.08 s (about 8%), so a lower load resistance or a lower starting voltage could cause failure

A thin rectangular strip of doped semiconductor lies in the xy-plane. It is 20 mm long in the x-direction, w = 5.0 mm wide in the y-direction and t = 0.50 mm thick in the z-direction. A steady current I = 4.0 mA flows through it in the +x direction. A uniform magnetic field of adjustable magnitude is applied along the z-axis, and a high-resistance voltmeter connected to contacts on the two long edges records the Hall voltage V_H = V(y = w) − V(y = 0). A positive value of B means the field points in the +z direction; the final reading was taken with the field reversed. Take e = 1.60 × 10⁻¹⁹ C. For part (e), copper has density 8.96 × 10³ kg m⁻³ and molar mass 63.5 g mol⁻¹, and N_A = 6.02 × 10²³ mol⁻¹.

1. (a)

   Justify Justify, from the sign of the readings for positive B, whether the charge carriers in the strip are positive or negative.

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   1. 1

      For current in +x, positive carriers move in +x and negative carriers in −x; in both cases F = qv × B with B along +z points in −y (q(v x̂) × (B ẑ) = −qvB ŷ for positive carriers, (−e)(−v x̂) × (B ẑ) = −evB ŷ for electrons), so carriers of either sign accumulate on the y = 0 edge

   2. 2

      V_H = V(y = w) − V(y = 0) is positive, so the y = 0 edge is at the lower potential and the charge accumulated there is negative: the carriers are electrons and the material is n-type

2. (b)

   Explain The reading at B = 0 is not zero. Explain how this offset can arise, and use the readings at B = +0.50 T and B = −0.50 T to obtain the Hall voltage at 0.50 T with the offset removed.

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   1. 1

      The two edge contacts are not exactly opposite each other, so they also pick up part of the resistive potential drop along the strip; this offset keeps its sign when B is reversed, whereas the Hall voltage reverses sign

   2. 2

      Half the difference of the two readings removes the offset: V_H = [12.92 − (−12.08)]/2 = 12.50 mV (half their sum, 0.42 mV, recovers the offset and agrees with the zero-field reading)

3. (c)

   Determine Using the seven readings with B ≥ 0, determine the gradient of a graph of V_H against B.

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   1. 1

      Uses all seven readings through a single best-fit straight line of V_H against B, drawn on a graph or found by least squares, rather than one reading or two adjacent readings

   2. 2

      Reads the gradient from two widely separated points on that line, for example about (0.00 T, 0.43 mV) and (0.60 T, 15.40 mV), giving (15.40 − 0.43) mV/(0.60 − 0.00) T, or takes the least-squares slope directly

   3. 3

      Gradient = 25.0 mV T⁻¹ = 2.50 × 10⁻² V T⁻¹ (accept 24.6 to 25.4 mV T⁻¹)

4. (d)

   Calculate Calculate the number density of charge carriers in the semiconductor.

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   1. 1

      Balancing the magnetic and Hall-field forces on the carriers gives V_H = IB/(n e t), so the gradient equals I/(n e t) and n = I/(e t × gradient)

   2. 2

      n = (4.0 × 10⁻³)/[(1.60 × 10⁻¹⁹)(0.50 × 10⁻³)(2.50 × 10⁻²)] = 2.0 × 10²¹ m⁻³

5. (e)

   Estimate Estimate the Hall voltage that the same current and a field of 0.50 T would produce in a copper strip of the same dimensions, stating your assumption about the number of conduction electrons, and say what your estimate implies for the choice of material in a Hall probe.

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   1. 1

      Assuming one conduction electron per copper atom: n ≈ (8.96 × 10³ kg m⁻³/0.0635 kg mol⁻¹)(6.02 × 10²³ mol⁻¹) ≈ 8.5 × 10²⁸ m⁻³

   2. 2

      V_H = IB/(n e t) ≈ (4.0 × 10⁻³)(0.50)/[(8.5 × 10²⁸)(1.60 × 10⁻¹⁹)(0.50 × 10⁻³)] ≈ 3 × 10⁻¹⁰ V, about 4 × 10⁷ times smaller than the semiconductor's 12.5 mV and far below its 0.42 mV offset, so Hall probes are made from low-carrier-density semiconductors

A thin flat annulus, shaped like a washer, of inner radius a and outer radius b lies in the xy-plane centred on the origin and carries a uniform surface charge density σ > 0. The z-axis passes through the centre of the hole, perpendicular to the annulus. Take the electric potential to be zero infinitely far away and neglect gravity. For part (e), the particle is an electron of charge magnitude q = 1.60 × 10⁻¹⁹ C and mass m = 9.11 × 10⁻³¹ kg, with a = 2.0 cm, b = 6.0 cm, σ = 1.0 × 10⁻⁹ C m⁻² and ε₀ = 8.85 × 10⁻¹² F m⁻¹.

1. (a)

   Derive Starting from the potential of a point charge, derive an expression for the electric potential V(z) at a point on the z-axis.

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   1. 1

      Divides the annulus into thin rings of radius r and width dr, each carrying dq = σ(2πr dr), all of it at the same distance √(r² + z²) from the field point

   2. 2

      dV = dq/(4πε₀√(r² + z²)) = (σ/(2ε₀)) r dr/√(r² + z²), added as scalars with no components to resolve

   3. 3

      Integrates from r = a to r = b: ∫ₐᵇ r dr/√(r² + z²) = √(b² + z²) − √(a² + z²)

   4. 4

      V(z) = (σ/(2ε₀))[√(b² + z²) − √(a² + z²)]

2. (b)

   Determine Determine the z-component of the electric field on the axis from your answer to (a), and explain why the potential on the axis alone is enough to give the complete field there.

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   1. 1

      E_z = −dV/dz = (σz/(2ε₀))[1/√(a² + z²) − 1/√(b² + z²)], positive for z > 0, so the field on the axis points away from the annulus

   2. 2

      V known only along the axis gives only ∂V/∂z, but rotational symmetry about the z-axis forbids any transverse component on the axis, so E_x = E_y = 0 there and E_z is the whole field

3. (c)

   Show that Show that for |z| ≫ b the field on the axis reduces to that of a point charge Q = πσ(b² − a²), and state the field obtained in the limit a → 0, b → ∞, naming the charge distribution this limit describes.

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   1. 1

      For |z| ≫ b, 1/√(a² + z²) ≈ (1/|z|)(1 − a²/(2z²)) and likewise for b, so the bracket in E_z is approximately (b² − a²)/(2|z|³)

   2. 2

      E_z ≈ σ(b² − a²)/(4ε₀z²) for z > 0, which equals Q/(4πε₀z²) with Q = πσ(b² − a²), the total charge of the annulus

   3. 3

      With a → 0 and b → ∞ the bracket tends to 1/|z|, so E_z = σ/(2ε₀) directed away from the plane on each side and independent of distance: the field of an infinite uniformly charged plane

4. (d)

   Derive A particle of charge −q and mass m is released from rest on the axis at z = z₀, where 0 < z₀ ≪ a. Using Newton's second law, derive the equation of motion for small z, show that the motion along the axis is simple harmonic, and derive its angular frequency ω.

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   1. 1

      For |z| ≪ a, 1/√(a² + z²) ≈ 1/a and 1/√(b² + z²) ≈ 1/b, so E_z ≈ (σz/(2ε₀))(1/a − 1/b) = σ(b − a)z/(2ε₀ab)

   2. 2

      Newton's second law with force −qE_z: m d²z/dt² = −[qσ(b − a)/(2ε₀ab)]z, a restoring force proportional to displacement, so the motion is simple harmonic about the centre

   3. 3

      ω = √[qσ(b − a)/(2ε₀mab)]

5. (e)

   Calculate Calculate the period of the oscillation for the electron described in the stem.

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   1. 1

      ω² = (1.60 × 10⁻¹⁹)(1.0 × 10⁻⁹)(0.040)/[2(8.85 × 10⁻¹²)(9.11 × 10⁻³¹)(0.020)(0.060)] = 3.3 × 10¹⁴ s⁻², so ω = 1.82 × 10⁷ rad s⁻¹

   2. 2

      T = 2π/ω = 3.5 × 10⁻⁷ s (3.45 × 10⁻⁷ s)

6. (f)

   Predict In practice the electron cannot be released exactly on the axis. By applying Gauss's law to a small closed cylinder of radius ρ and length 2h, centred on the origin with its axis along z and with ρ and h both much smaller than a, predict the radial component of the electric field at a small distance ρ from the axis near the centre, and hence predict whether the oscillation in (d) survives a small sideways displacement.

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   1. 1

      The cylinder lies inside the hole and encloses no charge; with E_z ≈ Kz, where K = σ(b − a)/(2ε₀ab), the outward flux through the two end caps is 2(Kh)(πρ²)

   2. 2

      Zero total flux requires an equal inward flux through the curved surface: E_ρ(2πρ)(2h) = −2πKhρ², so E_ρ = −Kρ/2, directed towards the axis

   3. 3

      The force on the electron, −qE_ρ, points away from the axis and grows with ρ, so the centre is unstable sideways: a small transverse displacement grows, the electron drifts off the axis towards the annulus and the axial oscillation is not sustained, as Earnshaw's theorem requires of any electrostatic equilibrium