Q#23.37
Neurons are the basic units of the nervous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium ($K^+$) ions and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller ions are able to penetrate the membrane to some degree (Fig. E23.37). This leaves an excess negative charge on the inner surface of the axon membrane and an excess positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further $K^+$ ions from leaking out. Measurements show that this potential difference is typically about 70 mV. The thickness of the axon membrane itself varies from about 5 to 10 nm, so we’ll use an average of 7.5 nm. We can model the membrane as a large sheet having equal and opposite charge densities on its faces.
(b) Which is at a higher potential: the inside surface or the outside surface of the axon membrane?
Answer:
We can model the axon membrane as a large sheet having equal but opposite charges on its opposite faces.
For two oppositely charged sheets of charge, $V_{ab}=Ed$.
The positively charged sheet is the one at higher potential.
(a) $E=\frac{V_{ab}}{d}$
$E=\frac{70 \times 10^{-3} \ V}{7.5 \times 10^{-9} \ m}$
$E=9.3 \times 10^6 \ V/m$
The electric field is directed inward, toward the interior of the axon, since the outer surface of the membrane has positive charge and $\vec{E}$ points away from positive charge and toward negative charge.
(b) The outer surface has positive charge so it is at higher potential than the inner surface.
The electric field is quite strong compared to ordinary laboratory fields in devices such as student oscilloscopes. The potential difference is only 70 mV, but it occurs over a distance of only 7.5 nm.
Q#23.38
Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.
(a) If the surface charge density for each plate has magnitude 47.0 nC/$m^2$ what is the magnitude of $\vec{E}$ in the region between the plates?
(b) What is the potential difference between the two plates?
(c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?
Answer:
For oppositely charged parallel plates, $E=\frac{\sigma}{\epsilon}$ between the plates and the potential difference between the plates is V = Ed.
(a) $E=\frac{\sigma}{\epsilon}=\frac{47.0 \times 10^{-9} \ C/m^2}{8.85 \ times 10^{-12} \ C^2/N.m^2}$
E = 5310 N/C
(b) V = Ed = (5310 N/C)(0 0220 m) = 117 V.
(c) The electric field stays the same if the separation of the plates doubles. The potential difference between the plates doubles.
The electric field of an infinite sheet of charge is uniform, independent of distance from the sheet. The force on a test charge between the two plates is constant because the electric field is constant. The potential difference is the work per unit charge on a test charge when it moves from one plate to the other. When the distance doubles, the work, which is force times distance, doubles and the potential difference doubles.
Q#23.39
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated 45.0 mm by and the potential difference between them is 360 V.
(a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
(b) What is the magnitude of the force this field exerts on a particle with charge +2.40 nC?
(c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
(d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
Answer:
The force this field exerts on the particle is given by Eq. F = |q|E. Use the equation that precedes Eq. $E=\frac{V_{ab}}{q}$ to calculate the work.
(a) $E=\frac{V_{ab}}{d}$
$E=\frac{360 \ V}{0.0450 \ m}$
$E=8000 \ V/m$
(b) F = |q|E = $(2.40 \times 10^{-9} \ C)(8000 \ V/m)$
F = $1.92 \times 10^{-5}$ N
(c) The electric field between the plates is shown in Figure 23.39.
The plate with positive charge (plate a) is at higher potential. The electric field is directed from high potential toward low potential (or, $\vec{E}$ is from + charge toward − charge), so $\vec{E}$ points from a to b. Hence the force that $\vec{E}$ exerts on the positive charge is from a to b, so it does positive work.
$\int_{a}^{b}\vec{F}.d\vec{l}=Fd$
where d is the separation between the plates.
$W=Fd=(1.92 \times 10^{-5} \ N)(0.0450 \ m)=+8.64 \times 10^{-7}$ J
We see that $W_{a→b}=-(U_b-U_a)=U_a-U_b$
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