Recitation Week 4
Chapter 24
Problem 14. For the system of capacitors shown in Fig. 24.24, find the equivalent capacitance (a) between b and c, and (b)
between a and c.
a
|
=== 15pF
|
b
|
+--+--+
|9.0pF|
===
=== 11pF
|
|
+--+--+
|
c
(a) The 9 and 11 pF capacitors are in parallel, so the equivalent capacitance between b and c is
Cbc = CL + CR = (9.0 + 11) pF = 20 pF
(1)
(b) The 15 pF capacitor is in series with the equivalent capacitance Cbc , so the equivalent capacitance between a and c is
Cac =
1
1
+
CT
Cbc
−1
=
1
1
+
15 20
−1
pF =
4
3
+
12 12
−1
· 5 pF =
60
pF = 8.6 pF
7
(2)
Problem 18. In Fig. 24.26, C1 = 6.00 µF, C2 = 3.00 µF, and C3 = 5.00 µF. The capacitor network is connected to an
applied potential Vab . After the charges on the capacitors have reached their final values, the charge on C2 is 40.0 µC. (a)
What are the charges on capacitors C1 and C3 ? (b) What is the applied voltage Vab ?
1
+-||-+
|
|
a--+
+-+
| 2 | |
+-||-+ |
d
3
|
b----||---+
(a) Because C1 and C2 are in parallel, they must have the same voltage differences. Using this and the definition of capacitance
Q = CV , we have
Q1
Q2
= V1 = V2 =
C1
C2
C1
6.00
Q1 =
Q2 =
40.0 µC = 80.0 µC .
C2
3.00
(3)
(4)
Since the wire d was initially uncharged, all the charge on C1 and C2 had to come from C3 . Assuming Va > Vb (if Vb > Va ,
just flip all the charge signs in the following solution), the left hand sides of C1 and C2 will be at a higher potential than the
right, so 0 < Q1L = Q1 = −Q1R and 0 < Q2L = Q2 = −Q2R . Balancing the charge on the wire d
0 = Q1R + Q2R + Q3R
(5)
Q3R = −Q1R − Q2R = Q1 + Q2 = 120 µC > 0
(6)
Q3L = −Q3R = −120 µC < 0 .
(7)
(b)
Vab = Vad + Vdb
(8)
We can find Vdb easily enough from Q3 = C3 V3 . Because C1 and C2 are in parallel, V1 = V2 = Vad (as we pointed out in
(a)), so
Q2
120 µC
40.0 µC
Q3
+
=
+
= 24.0 V + 13.3 V = 37.3 V
(9)
Vab =
C3
C2
5.00 µF 3.00 µF
Problem 32. For the capacitor network shown in Fig. 24.28, the potential difference across ab is 36 V. Find (a) the total
charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energy
stored in each capacitor; (e) the potential differences across each capacitor.
a--||--||--b
150 120
nF nF
(a) The equivalent capacitance is
1
1
+
150 120
−1
nF = 66.7 nF .
(10)
Q = CV = 66.7 nF · 36 V = 2.40 µC .
(11)
Cab =
The total charge on the capacitors is then
(b) Each capacitor must also have Q = 2.40 µC, since the wire connecting the capacitors must remain uncharged as a whole.
(c) The total capacitative energy is
EC =
1
1
CV 2 = · 66.7 nF · (36 V)2 = 43.2 µJ
2
2
(12)
(d) We don’t know the voltage of each capacitor seperately, but we could find it using Q = CV , since we know Q and C. We
could also just plug the formula in directly for V in our energy formula
2
1
1
Q
1 Q2
2
EC = CV = C
(13)
=
2
2
C
2 C
The energy stored in each capacitor is then
1 Q2L
1 (2.40 µC)2
= 19.2 µJ
=
2 CL
2 150 nF
1 Q2R
1 (2.40 µC)2
=
=
= 24.0 µJ .
2 CR
2 120 nF
ECL =
(14)
ERL
(15)
Note that ECL + ECR = 43.2 µJ = EC , as it should be, since the equivalent capacitor situation is equivalent (i.e. interchangable) with the two-capacitor situation.
(e) Here we find the potential differences the way we mentioned in (d).
QL
2.40 µC
=
= 16.0 V
CL
150 nF
QR
2.40 µC
VR =
=
= 20.0 V .
CR
120 nF
VL =
(16)
(17)
Note that VL + VR = 36.0 V = Vab .
Problem 52. Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially
permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside
and outside faces of such a membrane,and these charges prevent additional charges from passing throughthe cell wall. We
can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic
material to give the membrane a dielectric constant of about 10. (See Fig. 24.30.) (a) What is the capacitance per square
centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane.
What is the electric field inside this membrane?
Outside axon
++++++++++++++++++++++++
________________________
Axon membrane
|- 7.5nm
________________________
-----------------------Inside Axon
(a) For parallel-plate capacitors (from Gauss’ law)
C=
A
A
Q
= ε = kε0 .
V
d
d
(18)
So the capacitance per square centimeter will be
kε
10 · 8.85 · 10−12 F/m
C
=
=
= 1.2 · 10−2 F/m2 ·
A
d
7.5 · 10−9 m
1m
100 cm
2
= 1.2 µF/cm2
(19)
(b) Since the electric field generated by parallel plates is constant and perpendicular to the plates, we can easily integrate
across the membrane from the bottom to the top
Z t
Z t
∆V =
E · dx = E ·
dx = E · d .
(20)
b
b
Notice that for a perpendicular path (cutting straight across the membrane), E and d are in the same direction, so ∆V =
E · d = Ed. Flipping this around to give E, we have
E=
V
85 mV
=
= 11 MV/m
d
7.5 nm
(21)