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Physics 9 Fall 2011
Homework 5 - Solutions
Friday September 23, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Friday, September 30th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. (a) Show that the equivalent capacitance of two capacitors in series can be written
Ceqv =
C1 C2
.
C1 + C2
(b) Using only this formula and some algebra, show that Ceqv must always be less
than C1 and C2 , and hence must be less than the smaller of the two values.
(c) Show that the equivalent capacitance of three capacitors in series can be written
Ceqv =
C1 C2 C3
.
C1 C2 + C2 C3 + C1 C3
(d) Using only this formula and some algebra, show that Ceqv must always be less
than each of C1 , C2 , and C3 , and hence must be less than the least of the three
values.
————————————————————————————————————
Solution
(a) Two capacitors in series add to give an equivalent capacitance as
1
1
1
=
+
.
Ceqv
C1 C2
Putting the two terms over the same denominator gives
1
C2
C1
C1 + C2
=
+
=
.
Ceqv
C1 C2 C1 C2
C1 C2
Taking the reciprocal gives
Ceqv =
C1 C2
.
C1 + C2
(b) Rewriting our expression for Ceqv gives
1
Ceqv =
C1 .
1 + C1 /C2
1
Now, the term in parenthesis is smaller than one, meaning that Ceqv < C1 , so
the equivalent capacitance must be smaller than C1 . We can do exactly the same
thing with C2 , writing
1
C2 ,
Ceqv =
1 + C2 /C1
which is again less than C2 . Hence, the equivalent resistance must be smaller
than either C1 or C2 .
(c) Now, for three capacitors we have
1
1
1
1
=
+
+
.
Ceqv
C1 C2 C3
Again, putting all the terms under the same common denominator gives
1
C2 C3
C1 C3
C1 C2
C2 C3 + C1 C3 + C1 C2
=
+
+
=
.
Ceqv
C1 C2 C3 C1 C2 C3 C1 C2 C3
C 1 C2 C3
Once again, taking the reciprocal gives
Ceqv =
C1 C2 C3
.
C2 C3 + C1 C3 + C1 C3
(d) Once again, we can rewrite our expression for the equivalent capacitance, factoring
out a C2 C3 from the top and bottom,
1
Ceqv =
C1 ,
1 + C1 /C2 + C1 /C3
and again the term in parenthesis is less than one, such that Ceqv < C1 . We can
also write
1
1
Ceqv =
C2 =
C3 .
1 + C2 /C1 + C3 /C1
1 + C3 /C1 + C3 /C2
In each case the terms in parenthesis are less than one, and so Ceqv < C2 and
Ceqv < C3 . Thus, the equivalent capacitance must be less than any of the individual capacitances.
2
2. Recall that the energy density of an electric field is
uE =
0 2
E ,
2
where E is the electric field. What is the maximum electric energy density in a region
containing dry air at standard conditions?
————————————————————————————————————
Solution
The air breaks down and sparks at an electric field of about 3 million volts per meter.
Thus, the maximum energy density is
2
uE =
0 2 8.85 × 10−12 × (3 × 106 )
E =
= 40 J/m3 .
2
2
3
3. The membrane of the axon of a nerve cell can be modeled as a thin cylindrical shell
of radius 1.00 × 10−5 m, having a length of 10.0 cm and a thickness of 10.0 nm. The
membrane has a positive charge on one side and a negative charge on the other, and
the membrane acts as a parallel-plate capacitor of area 2πrL and separation d. Assume
the membrane is filled with a material whose dielectric constant is 3.00.
(a) Find the capacitance of the membrane.
(b) If the potential difference across the membrane is 70.0 mV, find the charge on the
positively charged side of the membrane.
————————————————————————————————————
Solution
(a) The capacitance of a parallel-plate capacitor, filled with a dielectric of constant
κ is
2πκ0 rL
κrL
κ0 A
=
=
,
C=
d
d
2kd
where we have plugged in the Coulomb constant, k, for convenience. Plugging in
the values gives
κrL
3 × (10−5 ) × 0.1
C=
=
= 1.67 × 10−8 F,
9
−8
2kd
2 × 9 × 10 × 10
or 16.7 nF.
(b) The charge is given in terms of the capacitance and voltage as Q = CV , and so
Q = CV = 1.67 × 10−8 × 70 × 10−3 = 1.17 × 10−9 C,
or 1.17 nC.
4
4. A parallel-plate capacitor that has no dielectric
in the space between the plates has a capacitance
C0 and a plate separation d. Two dielectric slabs
that have dielectric constants of κ1 and κ2 , respectively, are then inserted between the plates
as shown in the figure to the right. Each slap has
a thickness of d/2 and has area A, the same area
as each capacitor plate.
(a) When the charge on the positively charged capacitor plate is Q, find the electric
field in each dielectric.
(b) Find the potential difference between the plates.
(c) Show that the capacitance of the system after the slabs are inserted is given by
C=
2κ1 κ2
C0 .
(κ1 + κ2 )
(d) Show that your result from part (c) is the equivalent capacitance of a series
combination of two capacitors, each having plates of area A and a gap width
equal to d/2. The space between the plates of one is filled with a material that
has a dielectric constant equal to κ1 and the space between the plates of the other
is filled with a material that has a dielectric constant equal to κ2 .
————————————————————————————————————
Solution
(a) The dielectric of constant κ drops the electric field by the same factor,
Ediel =
E0
,
κ
where E0 is the field with no dielectric. Now, for a capacitor, E = η0 , where
η=Q
is the surface charge density on the plates. Thus, we find the electric fields
A
in each region as
Q
Q
E1 =
,
E2 =
,
κ1 0 A
κ2 0 A
and are both constant.
(b) The total difference in potential between the two plates is the sum of the potential
differences in each slab, V = V1 + V2 . Since we know the (constant) electric field
in each slab, then the potential is just the field, times the thickness of the slab,
V1 = d2 E1 , and V2 = d2 E2 . Thus,
d
1
Qd κ1 + κ2
d
Qd
1
V = E1 + E2 =
+
=
.
2
2
20 A κ1 κ2
20 A
κ1 κ2
5
(c) The capacitance is defined as C = Q/V , and so
C=
Q
=
V
Qd
20 A
Q
κ1 +κ2
κ1 κ2
=
20 A κ1 κ2
.
d κ1 + κ2
Now, capacitance of an empty capacitor is C0 = 0dA , and so our total capacitance
is
2κ1 κ2
C0 .
C=
κ1 + κ2
(d) Two capacitors in series add together to give an equivalent capacitance
1
1
1
C1 C2
=
+
⇒ Ceqv =
.
Ceqv
C1 C2
C1 + C2
Now, if we have a capacitor with area A, thickness d/2 and filled with dielectric
constant κ1 , then the capacitance is
C1 =
2κ2 0 A
,
d
and similarly for C2 . Thus, the equivalent capacitance is
Ceqv =
42 A2
C1 C2
= 02
C1 + C2
d
20 A
d
κ1 κ2
0 A 2κ1 κ2
.
=
d κ1 + κ2
(κ1 + κ2 )
This is our answer from part (c), confirming the equivalence.
6
5. A parallel-plate capacitor that has plate area A is
filled with two dielectrics of equal size, as seen in
the figure to the right.
(a) Show that this system can be modeled as two capacitors that are connected in
parallel and each have an area A/2.
(b) Show that the capacitance is given by
C=
1
(κ1 + κ2 ) C0 ,
2
where C0 is the capacitance if there were no dielectric materials in the space
between the plates.
————————————————————————————————————
Solution
(a) Both dielectrics experience the same difference in potential (since they are both
hooked up to the same plates), which is exactly the same situation as two parallel
capacitors. Since each dielectric covers only half the area, then the equivalent
parallel capacitors would have only half the area, A/2, as claimed.
(b) The capacitance of a single parallel-plate capacitor with no dielectric is C0 = 0dA ,
where A is the area of the plates, and d is the separation distance. Including
a dielectric of constant κ between the plates increases this capacitance by κ, so
C = κC0 .
For two capacitors, connected in parallel, the net capacitance is just the sum of
the capacitances,
Ceqv = C1 + C2 .
Now, recalling that the area of these individual equivalent capacitors is only A/2
we have
κ2 0 A
1
0 A
1
κ1 0 A
+
= (κ1 + κ2 )
= (κ1 + κ2 ) C0 ,
Ceqv =
d
2
d
2
2
d
2
as claimed.
7