PHYS 1120: General Physics II

Discussion Worksheet 4 Answers

1.  Two-capacitor circuit

1. Q = CV.
2. U = ½ QV = ½ CV².
3. Q = 360 μC, U = 2.16 mJ.
4. Equilibrating a charged capacitor with an uncharged capacitor
   1. The new voltage I'll call V1.  Then V1 = V C1/(C1 + C2)
   2. Charges Q1 = C1 V1 = V C1²/(C1+C2)
      Q2 = C2 V1 = V C1 C2/(C1+C2)
      The capacitor with the higher capacitance has the greater charge at the same voltage.
   3. Energies U1 = ½ C1 V1² (you can expand this to ½ C1 V² (C1/[C1+C2])²
      U2 = ½ C2 V1² = ½ C2 V² (C1/[C1+C2])²
      The capacitor with the higher capacitance has the higher energy at the same voltage.
5. V1 = 3.0 V, Q1 = 90 μC, Q2 = 270 μC, Q = 360 μC (same as before, as it must be), U1 = 135 μJ, U2 = 405 μJ, ΣU = 540 μJ which is less than before.

2.  Capacitor with and without a dielectric

1. Q = CV.
2. U = ½ QV = ½ CV² = ½ Q²/C.
3. Now I'll call the new capacitance C1 = κC.  The new voltage I'll call V1.  V1 = Q/C1 = CV/C1 = V/κ.
4. Energy U1 = 1/2 C1 V1² = ½ (κC)(V/κ)² = 1/2 C/κ.  The energy is less by a factor of κ for the same charge.
5. I'll call the new charge Q2 = C1 V = κC V. This is a factor of κ greater than without the dielectric.
6. Energy U2 = ½ C1 V² = ½ κCV².  This is a factor of κ greater than without the dielectric.

3.  Current in a resistor

1. I = V/R = 1.5 mA.
2. 9.36 × 10¹⁵ electrons per second.
3. The cross sectional area of the wire is 3.14 × 10⁻¹² m².
   RA/ρ = 0.187 m = 18.7 cm.
4. 1.10 × 10⁻¹³ m³.
5. V = LA, so L = V/A = 3.51 × 10⁻² m = 3.51 cm.
6. Drift speed is 3.51 cm/s.
7. Conservation of energy says qV = ½ mv². The charge q = e, the elementary charge.  This solves for v = (2 eV/m)^½ = 7.26 × 10⁵ m/s.
8. The drift speed is much less than the accelerated speed.  Electrons clearly do not flow down the copper wire unimpeded.