Physics: Electromagnetism
Spring 2007
Physics: Electromagnetism
Spring 2007
PROBLEM SET 6 Solutions
2. Electrostatic Energy Basics:
Wolfson and Pasachoff, Ch. 26, Problem 7, p. 679
There are six different pairs of equal charges and the separation of any pair is a. Thus,
W = ∑ pairs kqi q j / a = 6kq 2 / a.
3.
Capacitor Basics -- Cylindrical Symmetry:
Wolfson and Pasachoff, Ch. 26, Problem 35, p. 680
The capacitance of air-filled (κ = 1) cylindrical capacitor was found in Example 26-5 in your textbook
and in the class example. Utilizing that expression, we obtain:
C=
2πε 0 A
2π (8.85 pF/m)(1 m)
=
= 55.0 pF.
ln(b / a )
ln(2.2 / 0.8)
4. Capacitor Basics -- Spherical Symmetry:
Wolfson and Pasachoff, Ch. 26, Problem 36, p. 680
Let’s assume that the outer shell with radius b is carrying a charge +Q, and the inner sphere of radius b
is carrying a charge -Q. Then the field outside the two spheres is zero, and the field between the
spheres is just kQ/r2. It is clear then that the potential between the two spheres is
a
b
a
kQ JJG
kQ
kQ
kQ
⎛1 1⎞
V = − ∫ 2 r ⋅ dr = − ∫ 2 dr − ∫ 2 dr = 0 +
= kQ ⎜ − ⎟
r
r
r
r b
⎝a b⎠
∞
∞
b
a
Therefore, the capacitance of such a capacitor is
C=
Q Q ab
ab
.
=
=
V kQ ( b − a ) k ( b − a )
5. Energy Stored in Capacitors:
Wolfson and Pasachoff, Ch. 26, Problem 47, p. 681
a)
The charge on the plates remains the same, and so does the electric field ( E = σ / ε 0 ) in the gaps
between either plate and the slab. However, the separation (i.e., the thickness of the field region)
between the plates is reduced to 40% of its original value d ′ = d1 + d 2 = 0.4d , therefore the
capacitance is increased, C ′ = ε 0 A / d ′ = ε 0 A / 0.4d = 2.5 C. (The equations V = El and
C = Q / V lead to the same result.) In fact, the configuration behaves like a series combination
of two parallel plate capacitors,
1/ Ceffective = C1−1 + C2−1 = (d1 / ε 0 A) + (d 2 / ε 0 A) = (d1 + d 2 ) / ε 0 A = 0.4d / ε 0 A = 1/ 2.5 C.
b) When the charge is constant (no connections to anything isolates the system), the energy stored is
inversely proportional to the capacitance, U = Q / 2C. Thus
2
U effective = Q 2 / 2Ceffective = Q 2 / 2(2.5C ) = 0.4U ,
or the energy decreases to 40% of its original value. (With the slab inserted, there is less field
region and less energy stored. While the slab is being inserted, work is done by electrical forces to
conserve energy.)
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Physics: Electromagnetism
6.
Spring 2007
Connecting Capacitors:
Wolfson and Pasachoff, Ch. 26, Problem 52, p. 681
(i)
C1 is in series with the parallel combination of C2 and C3. Thus,
C=
(ii)
C1 (C2 + C3 )
(0.02 µ F) × (1 + 2)
=
= 0.012 µ F.
(C1 + C2 + C3 )
(2 + 1 + 2)
The net charge on the entire combination is
Q = CV = (0.012 µ F)(100 V) = 1.2 µ C .
Since C1 is in series with the capacitors in parallel,
Q = 1.2 µ C = Q1 = Q2 + Q3 .
Moreover, for the parallel capacitors,
V2 = Q2 / C2 = V3 = Q3 / C3 ,
so
Q3 / Q2 = C3 / C2 = 2.
Thus,
Q2 = (1/ 3)Q = 0.4 µ C
and
Q3 = (2 / 3)Q = 0.8 µ C.
(In general, for two capacitors in parallel, Q2 = C2Q /(C2 + C3 ) etc.)
(iii)
Equation 26-5 in your textbook and the ones derived in class, applied to each capacitor, give
V1 = Q1 / C1 = 1.2 µC/0.02 µ F = 60 V,
And
V2 = V3 = 40 V .
7.
Biophysics!!!
Wolfson and Pasachoff, Ch. 26, Problem 70, p. 682
If we assume that the inner and outer surfaces of the membrane act like a parallel plate capacitor, with
the space between the plates filled with material of dielectric constant κ = 3 , then the capacitance per
unit area is
C κε 0
=
.
A
d
Thus,
d=
8.
3(8.85 pF/m)
= 2.7 nm.
(1 µ F/cm 2 )
More Biophysics!!!
A neuron, or nerve cell, has a long signal-carrying axon (which, from spine to fingers can be more
than a meter in length). The axon membrane is usually positive on the outside and negative on the
inside. The dielectric constant of the membrane has been measured to be about 7.
a) Given that the membrane wall is a mere 6.0 µm thick and that the axon radius is 5 mm, determine
its capacitance per unit area. (Hint: the axon is a tube. Because the wall thickness is much
smaller than both the inner and outer diameters of the tube, you can treat it as a parallel-plate
capacitor.)
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Physics: Electromagnetism
Spring 2007
The axon can be treated as a parallel plate capacitor whose capacitance is C = ε A d , where
ε = κε 0 .
Then, the capacitance of the axon per unit area is
ε
C
8.85 × 10−12 F/m
=κ 0 =7
= 1.0 × 10−2 F/m 2
−9
6.0 × 10 m
A
d
b) Determine the surface charge density of the axon. In the resting state, when there is no signal
being transmitted, the potential difference across the membrane is about 70 mV.
Using the definition of capacitance, C = Q / V , we can find the surface charge density of the
axon to be
σ=
Q CV
=
= (1.0 × 10−2 F/m 2 )( 70 × 10−3 V ) = 7 × 10−4 C/m 2 ,
A
A
where we have used the result of part (a) for the capacitance of the axon per unit area.
c)
When a pulse is propagated down the axon, there is a shift of ions across a segment of membrane
that causes a reversal of polarity and an overall change in potential of about 100 mV. This
voltage spike (and the associated repolarization of the membrane) propagates along from one
region to the next and constitutes the signal, just as a flame propagates down a length of fuse.
After a few milliseconds, this so-called action potential pulse passes and given point of the axon,
which then returns to its resting potential (~70 mV). How much energy is required to recharge 1m length of axon in the wake of a pulse, so that it will be ready to transmit the next pulse?
For a 1-m long axon, the capacitance is
C=
C
2π rL = (1.0 ×10−2 F/m 2 ) 2π ( 5 × 10−6 m ) (1 m ) = 3.1× 10−7 F.
A
The energy needed to recharge the axon is
2
1
1
∆U = CV 2 = ( 3.1×10−7 F )(100 × 10−3 V ) = 1.6 × 10−9 J = 1.6 nJ.
2
2
9. Energy Stored in an Isolated and Connected Capacitors.
a) Wolfson and Pasachoff, Ch. 26, Problem 81, p. 683
a.
Ignoring fringing fields, the electric field between the plates is uniform,
E=
V
,
d
where V ≠ V0 since the dielectric is inserted, and E depends on x . On the left side of the
capacitor on Figure 26-36, where the slab has penetrated,
⎛ 1 ⎞⎛ σ
E = ⎜ ⎟⎜ l
⎝ κ ⎠ ⎝ ε0
⎞
⎟,
⎠
and on the right of the capacitor,
E=
where
σl
and
σ l = κε 0 E
σr
ε0
,
σr
are the charge densities on the left and right sides, respectively. Thus,
and
σ r = ε0 E,
and the charge can be written in terms of geometrical variables
superposed on Figure below as
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Physics: Electromagnetism
Spring 2007
⎛V ⎞
q = σ A wx + σ r w( L − x) = ε 0 Ew(κ x + L − x) = ε 0 ⎜ ⎟ w(κ x + L − x).
⎝d⎠
Then, the effective capacitance can be written as
q C0 (κ x + L − x)
(1)
=
,
V
L
ε A
where C0 = 0
and A = Lw . Although the question specifies x = 12 L , for which value the
d
C=
effective capacitance is
C = 12 C0 (κ + 1),
we give C as a function of x in (1), because we will need to differentiate with respect to x in part
(c).
b.
When the battery is disconnected, the capacitor is isolated and the charge on it is a
constant,
q = q0 .
The stored energy then is
U=
q02 L
U0L
q2
=
=
,
2C 2C0 (κ x + L − x) (κ x + L − x)
q02 1
C0V 02
2
1
where U 0 =
= 2 C0V 0 . For x = 2 L , the energy is U =
.
C0
(κ + 1)
1
2
c.
The force on a part of an isolated system is related to the potential energy of the system
by
JG
JG
F = −∇U .
(2)
The force due to the capacitor on the slab acts in one dimension along x-axis and therefore is
Fx = −
FG
H
IJ
K
U0 L
U 0 L(κ − 1)
dU
d
,
=−
=
dx
dx κ x + L − x
(κ x + L − x ) 2
in the direction of increasing x (so as to pull the slab into the capacitor). For x =
1
2
L , the
magnitude of the force is
2C0V 02 (κ − 1)
.
L(κ + 1) 2
It turns out that if we rewrite the force, for any value of x , in terms of the voltage for that x ,
F=
using
q0 = C0V0 = CV =
C0V (κ x + L − x)
,
L
the expression can be used in the succeeding problem. Thus,
C V 2 L(κ − 1) C0 ⎛ V ⎞
C0V 2 (κ − 1)
Fx = 0 0
=
L
−
=
.
(
κ
1)
⎜ ⎟
2(κx + L − x) 2
2 ⎝ L⎠
2L
2
b) Wolfson and Pasachoff, Ch. 26, Problem 82, p. 683
The capacitance depends on the configuration and electrical properties of the plates and insulating
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Physics: Electromagnetism
Spring 2007
materials, not on the external connections, so
C=
C0 (κ x + L − x)
L
as in the preceding problem.
(i)
If the capacitor remains connected to a battery, the voltage is constant, V = V0 , and
U = 12 CV 02 =
For x =
1
2
1
2
C0V 02 (κ x + L − x)
.
L
L,
U = 14 C0V 02 (κ + 1) .
(This is different from the preceding problem, because the battery does work.)
i.
When the capacitor is connected to a battery, Equation (2) above for the force does not
apply since the charge on plates is a function of both potential energy and continuously
changing charge. (The force, in this case, is derived in more advanced texts.) However,
for particular values of charge and voltage on the capacitor, the force on the slab considered
here is the same, regardless of the external connections. In the preceding problem we
found that
Fx =
1
2
C0V 2 (κ − 1)
,
L
where V was the particular voltage (and, because of the special form of the capacitance,
C ( x) , the particular charge q did not appear -- necessary condition for this formula to be
applicable in this problem.) Since V = V0 in this problem,
Fx =
1
2
C0V 02 (κ − 1)
.
L
y
L
σA
σr
w
x
x
10.
An Engineering Problem
A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric, one with
constant Κ 1 and one with constant Κ 2 , as shown in the figure below. Each slab has thickness d / 2 ,
where d is the plate separation. Show that the capacitance is
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Physics: Electromagnetism
Spring 2007
C=
2ε 0 A ⎛ κ1κ 2 ⎞
⎜
⎟.
d ⎝ κ1 + κ 2 ⎠
K1
d/2
K2
d/2
The effective capacitance of a capacitor filled with two slabs of dielectrics of dielectric constants κ1
and κ2 is equivalent to that of two capacitors in series with the above two dielectric constants,
respectively:
−1
−1
Ceffective
−1
⎛ ⎛ d ⎞ −1 ⎛ d ⎞ −1 ⎞
⎛ ⎛ ε A ⎞ −1 ⎛ ε A ⎞−1 ⎞
⎛ d ⎛ 1 1 ⎞⎞
1
2
= ⎜⎜
+⎜
⎟ +⎜
⎟ ⎟⎟ = ⎜
⎜ + ⎟⎟ .
⎟ ⎟ = ⎜⎜ ⎜
⎜ ⎝ d 2 ⎟⎠
d 2⎠ ⎟
2
ε
A
2
ε
A
2
ε
A
⎝
⎝ 2 ⎠ ⎠
⎝ 0 ⎝ κ1 κ1 ⎠ ⎠
⎝
⎠
⎝⎝ 1 ⎠
Then, simplifying we obtain
Ceffective =
2ε 0 A ⎛ κ 1κ 2 ⎞
⎜
⎟
d ⎝ κ 1+ κ 2 ⎠
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