Discussion Questions:

18. The energy goes into polarizing the individual atoms of the dielectric. That is, the energy goes into distorting the shapes of the individual atoms in the dielectric.

20. After you close the switches, charge will flow as the capacitors try to neutralize each other. Charge flows until the potentials are equalized which happens when each capacitor has a charge Q = 1/2 [Q1(original)-Q2(original)].

Problems

25. C = Q/V ===> Q = CV = 100 pico Farad x 1.5 Volts = 1.5 x 10**(-10) C

28. C = Q/V = Q/[kQ/R(Earth)] = R(Earth)/k = 7.1 x 10**(-4) Farad

32. C = Q/V:

• Area, A = 100 square centimeters and separation, d = 1 mm of air
• Q = charge density x A
• V = charge density/epsilon(air) x d

40. The 2 capacitors on the left are in series and are parallel to the 1 capacitor on the right. We then have

• 1/C(left) = 1/(8 pico F) + 1/(24 pico F) = 1/(6 pico F) ===> C(left) = 6 pico F
• C(total) = 6 pico F + 20 pico F = 26 pico F

Chapter 20

Discussion Questions:

5. When the switch is closed, the capacitor is uncharged and so is at nearly zero potential. Since the lamp and the capacitor are in parallel, the lamp is also near zero potential and it has a huge resistance. After the switch is closed, the capacitor starts to charge up. As the capacitor charges, its potential and the potential at the lamp increases. When the lamp's breakown voltage is reached (before steady-state is reached), its resistance goes down and the capacitor discharges causing the current through the lamp to go up and it flashes. The process then repeats itself.

7.

• Shape of pulse across the capacitor: When the pulse first appears, the charge Q on the capacitor (and therefore the V(C) on the capacitor) start to increase. When the pulse disappears and the driving V drops to zero, the charge leaks off of the capacitor and Q on the capacitor (and therefor V(C) on the capacitor) decays. In the case presented, the charge leaks off before the next pulse occurs ==> RC is less than the length of each pulse
• Output at the resistor: When the pulse first appears, the potential at the resistor is V. As the charge moves onto the capacitor (charge builds up) the potential at the resistor goes down. When the pulse finishes and V drops to zero, the charge flows off of the capacitor (decays). The current is in the opposite sense and the sense of the potential changes; it changes from positive to negative. As the capacitor slowly discharges, the potential at the resistor approaches zero.

  Problems

  27. According to Figure 20.19, it takes the system a time RC to charge to 63 % of its maximum charge. RC = 800 x 10**(-6) F x 5 x 10**3 ohm = 4.0 s.

  28. According to Figure 20.19, it takes the system a time RC to decay to 37 % of its value. RC = 600 x 10**(-6) F x 4 x 10**3 ohm = 2.4 s.

  59. In P59, the 4.0 ohm resistor and the capacitor are connected in parallel to each other. In steady state, the current flows through the resistor's branch and we effectively have the battery and the 8 ohm and 4 ohm resitor in a series circuit. The current through the circuit is then,

  0 = 12 Volts - I x 12 ohm ===> I = 1 A
  Now, since the capacitor and 4 ohm resistor are in parallel, they have the same potential drop,
  • For the resistor, V = I R = 1 A x 4 ohm = 4 Volts
  • For the capacitor, C = Q/V ===> Q = CV = 2 x 10**(-6) F x 4 Volts = 8 micro C

  65.

  • With only switch S(1) closed, the current can only flow along the outer perimeter of the circuit, i.e., through the 12 V battery, the 3 micro F capacitor, the 6 ohm resistor and the 20 ohm resistor in series. So, what is the current through this circuit? The potential drops are,
    0 = 12 Volts - Q/C - I x 6 ohm - I x 20 ohm
    In steady-state, the charge on the capacitor adjusts itself so that the current approaches zero. The potential drop across the resistor is then zero.
  • What is the charge on the 3 micro F capacitor? As noted above, we essentially have 12 Volts - Q/C = 0 in steady state ===> Q = 12 Volts x C = 12 Volts x 3 x 10**(-6)F = 3.6 x 10**(-5) C.