B. Conductors Homework Part II

**100 points**

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**16.3 Capacitance**

**Remember this: **

**Potential Energy =****DU _{c} = 1/2CV^{2}
= 1/2QV = 1/2Q^{2}/C = Work = Area under a Voltage vs. Charge (Q)
graph. It has units of N/C, J, N*m**

**62-66, 69-71**

71: Power has units
of Watts (N*m/sec). Energy has units of
Joules (N*m). Therefore, Power x time = Energy. Energy can be either kinetic (KE) or
potential (U). We know a formula for U
right? U = 1/2CV^{2} (where V = voltage)

**16.4 Dielectrics **

**72, 73, 76, 79, 80**

73: Remember that a dielectric increases the capacitance. Since C = Q/V and V is a constant from the battery, as C increases, so must Q.

76: K = C/C_{o}

79: K = C/C_{o}. Remember that C = Q/V and Q remains
constant, but V changes from 12 Volts to 5 Volts)

80: U_{c} =
1/2CV^{2} solve for C knowing the voltage and the U_{c}. Then, find K by comparing C to C_{o}.

**16.5 Capacitors in Series and Parallel**

**81-87, 90, 91**

85: U_{total}
= 1/2C_{s}V^{2} solve for Cs.
Since you are given the capacitance of one, you can figure out the other
with 1/C_{s} = 1/C_{1 }+ 1/C_{2}).

86: They are all parallel. This is easy!

90: Realize that each capacitor holds a charge in ratio since the voltage is the same for each capacitor in parallel.

91: C_{1}
and C_{2} are parallel. C_{3}
and C_{4} are parallel. The
total C_{p} for 1 and 2 is in series with the total C_{p} for 3
and 4. So, find the charge (Q) on C_{s1234}. Once you find the total charge (SQ)
you can find the voltage on each plate and then the charges (Q) on each plate.

**Answers to Even Problems:**

62: d

64: decreases

66: 2.2 x 10^{-9}
F

70: a) same
b) doubles to 5.0 x 10^{-8}J

72: b

76: 3.0

80: 2.50

82: a

84: a) 0.24mF b) 1.0mF

86: 1.2mF

90: C_{1} =
0.60 mC C_{2} = 1.2mC C_{3} = 1.8 mC