1. No category

PHYS3511-Biological Physics
Winter 2015, Assignment #8
Assigned on Monday 23 March 2015. Due on Monday 30 March 2015.
BioPhysics Reading: Chapter 17: All. Chapter 18, section 18.1 to 18.4
Question 1 Do Multiple-Choice MC 17.7 and 17.12 of chapter 17 of page 438-439.
MC17.7 ANSWER E
MC17.12 ANSWER A
Question 2 Do Multiple-Choice MC 18.12 and 18.13 of chapter 18 of page 462.
MC18.12 ANSWER D
MC18.13 ANSWER E
Question 3 Problem P18.1 of Chapter 18 textbook (page 463).
A large number of energetic cosmic–ray particles reach Earth's atmosphere continuously
and knock electrons out of the molecules in the air. Once an electron is released, it
responds to an electrostatic force, which is due to an electric field E produced in the
atmosphere by other point charges. Near the surface of Earth this electric field has a
magnitude of |E| = 150 N/C and is directed downward, as shown in Fig. 13.48. Calculate
the change in electric potential energy of a released electron when it moves vertically
upward through a distance d = 650 m.
The electric energy of a test charge between two uniformly charged parallel plates is
given in Eq. 18.2:
σ
Eel = qmobile y
ε0
Can we apply this formula to the charge distribution causing the processes shown in Fig.
13.48? Yes, in an approximate sense: Eqn 18.2 holds if we identify the origin of the
motion of charges in Figure above. This treats the cloud cover as one plate and Earth’s
surface as the other plate of a parallel plate capacitor. The surface of Earth is charged
negative and the cloud is charged positive since the electric field is pointing down. As
shown in Figure, a free electron accelerates toward the cloud since negative charges free
to move accelerate always toward the region of positive charges. The motion of the
electron is accompanied by a reduction in its electric potential energy. The change in the
electric potential energy is
N⎞
⎛
ΔEel = −e E Δy = − 1.6 × 10 −19 C ⎜ 150 ⎟ ( 650m ) = −1.56 × 10 −19 J , in which the
⎝
C⎠
negative sign on the right hand side of the first formula is the result of a decreasing
potential energy in the direction of increasing height y.
Question 4 Problem P18.19 of Chapter 18 textbook (page 464).
Fig. 14.33 shows an electron at the origin that is released with initial speed
v0 = 5.6 × 10 6 m / s at an angle θ 0 = 45! between the plates of a parallel plate capacitor
of plate separation D = 2.0 mm. If the potential difference between the plates is
ΔV = 100V , calculate the closest proximity of the electron to the bottom plate, d.
(
)
This problem becomes possible to solve when you realize that the electric field is uniform
between the plates of the capacitor and therefore the electron feels a constant force and
thus experiences a constant acceleration. This becomes a projectile motion problem
except that the electric force provides the acceleration in the y–direction instead of
gravity. The electric field will point down from + side to negative, and the force on the
electron is F = eE = eV / D = 1.6 × 10 −19 C × 100V / 2 × 10 −3 m = 8.0 × 10 −15 N . The force
will point up since a negative charge will always accelerate toward the direction of the
electric field. The acceleration of the electron is
ay = F / m = 8 × 10 −15 N / 9.1 × 10 −31 kg = 8.8 × 1015 m / s 2 , which is much larger than the
acceleration due to gravity of 9.8m/s2, justifying why we ignore gravitational force.
The electron has initial speed of v0 = 5.6 × 10 6 m / s , 450 below the horizontal as shown in
figure 14.33. The y-component (vertical) is v0 y = −v0 sin 45 0 = −4.0 × 10 6 m / s , negative
since it is down. Use vy2 = v02y + 2ay y . The minimum position corresponds to vy = 0 ,
(
) (
)
giving y = −v02y / 2ay = − −4.0 × 10 6 m / s / 2 8.8 × 1015 m / s 2 = −9 × 10 −4 m . So it travels
0.9 mm below the center before reversing direction. It will reach d = 0.1 mm above the
bottom plate.
Question 5 DNA in vivo. As discussed in class, DNA in cells are negatively charge, and
are often modeled as a line charge with electric field strength E = κ / 2π rε , where r is the
distance from the line charge, ε = 80ε 0 is the dielectric constant in water, and κ is the
charge per unit length (C/m). Consider a single (+) counterion that is moved from a
distance of a to R from the line charge (DNA). The change in internal energy ΔU equals
R
R
eκ R dx
eκ
R
the work done ΔU = Wa→ R = ∫ Fe dx = ∫ eE dx =
=
ln . Using the idea
∫
a
a
2πε a x 2πε a
2
that the entropy is S = kB ln Ω ≈ kB ln π r l , where Ω is the volume accessible to the
counterion located at a distance r away from the DNA, and l is the length of the DNA. As
such the change in entropy in moving the counterion from distance a to R is
ΔS = kB ln π R 2l / π a 2l = 2kB ln ( R / a ) . A) Using the discussed information, calculate the
change in free energy ΔF = ΔU − T ΔS . B) If ΔF < 0 then the (+) counterion would
spontaneously move far away from the (−) DNA. Determine the condition for
counterions to distribute near the DNA. C) Assume a DNA basepair has two negatively
charge phosphate, and that the length of a basepair is 0.34 nm. Using the criterion of part
B, do (+) counterions distribute about the DNA?
eκ
R
R ⎛ eκ
⎞ R
ln − 2kBT ln = ⎜
− 2kBT ⎟ ln
A) ΔF = ΔU − T ΔS =
⎠ a
2πε a
a ⎝ 2πε
eκ
B) The condition for counterion distribution near DNA ΔF > 0 →
− 2k BT > 0 . Note
2πε
that since R > a, ln R / a > 0 . Simplifying, the condition is κ > 4πε kBT / e .
C) Using the data on DNA κ = 2e / 0.34 × 10 −9 m = 9.4 × 10 −10 C / m . Also
4πε k BT / e = 4π 80 × 8.85 × 10 −12 C 2 m −2 N −1 1.381 × 10 −23 J / K ( 300K ) / 1.6 × 10 −19 C
(
)
(
)(
)
−10
= 2.3 × 10 C / m
Hence κ > 4πε kBT / e , and counterions will distribute about the DNA.
Question 6 Born electric-self-energy of a charge spherical membrane. Suppose that a
spherical bacteria is made up of lipids with positively charged head groups, with +e
charge per head, and each head group occupying an area of 2.5 × 10 −19 m 2 . A) Calculate
the total charge on a spherical membrane of r = 1µ m and r = 10 µ m . B) Calculate the
Born electric self-energy for r = 1µ m and r = 10 µ m . C) Using the result of part B)
calculate the self-energy per kg for r = 1µ m and r = 10 µ m . Assume the bacteria are
made up of mainly water.
q2 1
As discussed the Born electric self energy of a charge spherical shell is E el =
,
8πε r
where ε = 80ε 0 = 80 × 8.85 × 10 −12 C 2 m −2 N −1 .
r = 1µ m Since the surface area of the bacteria is
(
A = 4π r 2 = 4π 1 × 10 −6 m
)
2
= 1.26 × 10 −11 m 2 and the area per lipid head is 2.5 × 10 −19 m 2 ,
the number of lipid head if N head = 1.26 × 10 −11 m 2 ÷ 2.5 × 10 −19 m 2 = 5 × 10 7 . Assume +e
per lipid, the total charge is q = N head e = 8 × 10 −12 C . This gives
E el =
(
( 8 × 10
−12
8π 80 × 8.85 × 10
)
C
−12
2
2
−2
C m N
−1
)
1
= 3.6 × 10 −9 J .
1.0 × 10 −6 m
Assume that the cell is mostly water ρ == 1000
(
kg
, the mass of the bacteria is
m3
)
3
4
kg 4
M = ρ π r 3 = 1000 3 π 1 × 10 −6 m = 4.19 × 10 −15 kg , giving
3
m 3
−9
E el
3.6 × 10 J
J
=
= 8.6 × 10 5
−15
M
4.19 × 10 kg
kg
r = 10 µ m Since the surface area of the bacteria is
(
A = 4π r 2 = 4π 1 × 10 −5 m
)
2
= 1.26 × 10 −9 m 2 and the area per lipid head is 2.5 × 10 −19 m 2 ,
the number of lipid head if N head = 1.26 × 10 −9 m 2 ÷ 2.5 × 10 −19 m 2 = 5 × 10 9 . Assume +e
per lipid, the total charge is q = N head e = 8 × 10 −10 C . This gives
E el =
(
( 8 × 10
−10
C
)
2
8π 80 × 8.85 × 10 −12 C 2 m −2 N −1
(
)
1
= 3.6 × 10 −6 J .
−5
1.0 × 10 m
)
3
4
kg 4
M = ρ π r 3 = 1000 3 π 10 × 10 −6 m = 4.19 × 10 −12 kg , giving
3
m 3
−6
E el
3.6 × 10 J
J
=
= 8.6 × 10 5
, so the electric energy per Kg are the same.
−12
M
4.19 × 10 kg
kg