Physics 132
Week 7
Problem 1: Capacitance
Two large circular plates made of metal have area A = 2 m2. They are arranged parallel to
each other and are separated by a distance d = 1.5 cm. The gap between the plates is filled
with air. Together, the plates form a parallel-plate capacitor.
(a) What is the capacitance of this system?
(b) Is the capacitance a property of the geometry and the material between the plates, or does
it depend on the voltage and charge applied to the capacitor?
(c) Design a parallel-plate capacitor with capacitance 100 pF (pF = “picoFarad” = 10-12
Farads).
(d) The capacitor has been built according to the original design specifications (A = 2 m2,
d = 1.5 cm), but your troublesome client now insists that its capacitance be increased by a
factor of 2. Without changing the dimensions of the capacitor, what can be done?
(e) Finally, consider again the original capacitor geometry (A = 2 m2, d = 1.5 cm). This
time, try filling the left half of the gap between the plates with
Aluminum Oxide, a common material that has dielectric constant 7 (i.e.
with  = 70). The figure shows this partially-filled capacitor. What is
its capacitance C? Hint: this device could be treated as two separate


capacitors, connected in series or in parallel (which one?).
area A
Problem 2: Capacitors
A battery of voltage E is hooked up to a network of 5 capacitors.
Calculate the Q and V for all the capacitors below.
C1
A
C2
C4
C3
B
E C5
E = 12 V
C1 = 5 F
C2 = 8 F
C3 = 15 F
C4 = 3 F
C5 = 10 F
d
Problem 3: Circuits
This week, we will analyze circuits involving resistors. As you know, the formulas for
combining resitors in series and in parallel are “opposite” to those for capacitors. We also
need an expanded set of rules for “breaking up” combinations of devices:
 For devices connected in parallel: the voltage across them is always the same.
 For devices connected in series: the charge and current is always the same.
But otherwise, the procedure is the same!
(a) A light bulb is a familiar example of a resistor at work: when current passes through the
filament, resistive heating causes it to glow. Higher current makes the bulb glow more
brightly. Consider the three simple circuits shown below, in which a battery is connected to
one or two bulbs. Given that the three batteries are the five bulbs are identical, rank the
bulbs B1 through B5 in order of brightness.
B2
B4
B1
B5
B3
(b) How would the brightnesses B1 through B3 compare in the circuit below? (c) How would
the voltages V1, V2, V3 across the bulbs compare?
B1
B2
B3
(d) Have a look at the circuit below. Suppose we remove bulb #3, leaving an open break.
Will the brightnesses of bulbs B1, B2, and B4 increase, decrease, or stay the same? (e) Instead,
suppose we had removed bulb #3 and replaced it with a piece of wire. How would B1, B2,
and B4 have changed in this case?
B1
B2
B4
B3
Physics 132
Week 7 Practical: Cheat Sheet
Answers to Problems
1.)
2.)
3.)
a. 1.18 nF b. geometry c. up to you d. dielectric with  = 2 e. 400 pF
a. Q1 = 20.9 C, Q2 = 33.5 C, Q3 = 54 C, Q4 = 12.6 C, Q5 = 41.9 C
V1 = 4.2 V, V 2 = 4.2 V, V 3 = 3.6 V, V 4 = 4.2 V, V 5 = 4.2 V
a. B1 = B4 = B5 > B2 = B3
b. B1 > B2 = B3,
c. V1 > V2 = V3
d. B1 and B4 decrease, B2 increase
e. B1 and B4 increase, B2 decrease