General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 18, Problem 5RQ
To determine
The Nemst equation relates the potential difference across a membrane to the equilibrium ratio of
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potential difference VB - VA,
What is the magnitude of the electric field in unite of N/C across an axon membrane (1.00x10^0)x10-8 m thick if the resting potential is -(7.400x10^1) mV?
Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed x pulse duration
equal and opposite charge of negative organic ions, as shown in the figure below. Model the axon as a parallel-plate capacitor and take C = K² A/d and Q
radius r = 1.6 × 10¹ μm, and cell-wall dielectric constant x = 2.9.
=
Positive
charge
layer
Negative
charge
layer
+
External fluid
Axon wall membrane
No
Axon radius = r
+
Internal fluid
+
How many sodium ions (Na+) is this?
Na+ ions
d
+
(a) Calculate the positive charge on the outside of a 0.10-m piece of axon when it is not conducting an electric pulse. (Assume an initial potential difference of 7.0 × 10-² V.)
-2
9.03E-10
C
How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 × 10-² V?
5.639E9
K+ ions
= 50.0 m/s X 0.0020 s = 0.10 m). In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall
CAV to investigate the…
Chapter 18 Solutions
General Physics, 2nd Edition
Ch. 18 - Prob. 1RQCh. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - Prob. 4RQCh. 18 - Prob. 5RQCh. 18 - Prob. 6RQCh. 18 - Prob. 7RQCh. 18 - Prob. 8RQCh. 18 - Prob. 9RQCh. 18 - Prob. 10RQ
Ch. 18 - Prob. 1ECh. 18 - Prob. 2ECh. 18 - Prob. 3ECh. 18 - Prob. 4ECh. 18 - Prob. 5ECh. 18 - Prob. 6ECh. 18 - Prob. 7ECh. 18 - Prob. 8ECh. 18 - Prob. 9ECh. 18 - Prob. 10ECh. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26E
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- Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed pulse duration = 50.0 m/s 2.0 103 s = 0.10 m). In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and opposite charge of negative organic ions, as shown in Figure P18.43. Model the axon as a parallel-plate capacitor and take C = 0A/d and Q = C V to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 1.0 108 m, axon radius r = 1.0 101 m, and cell-wall dielectric constant = 3.0. (a) Calculate the positive charge on the outside of a 0.10-m piece of axon when it is not conducting an electric pulse. How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 102 V? Is this a large charge per unit area? Hint: Calculate the charge per unit area in terms of electronic charge e per squared (2). An atom has a cross section of about 1 2 (1 = 1010 m). (b) How much positive charge must flow through the cell membrane to reach the excited state of + 3.0 102 V from the resting state of 7.0 102 V? How many sodium ions (Na+) is this? (c) If it takes 2.0 ms for the Na+ ions to enter the axon, what is the average current in the axon wall in this process? (d) How much energy does it take to raise the potential of the inner axon wall to + 3.0 102 V, starting from the resting potential of 7.0 102 V? Figure P18.43 Problem 43 and 44.arrow_forwardVoltage across the resting membrane potential…A) Can be described by the Nernst equation.B) Is established by a difference in charges across the cell membrane, with the outside more negative than the inside.C) Defines the driving force for flow of a particular ion across the cell membrane, given its equilibrium potentialD) Is equally dependent on the flow of potassium and sodium ions through leak channels across the cell membrane.E) Stays the same during an action potential.arrow_forwardThe giant axon of a squid is 0.5 mm in diameter, 10 cm long, and not myelinated. Unmyelinated cell membranes behave as capacitors with 1 μF of capacitance per square centimeter of membrane area. When the axon is charged to the -70 mV resting potential, what is the energy stored in this capacitance?arrow_forward
- Calculate the axoplasm resistance for a neuron of length 0.06 m and a radius of 5 um. The axoplasm resistivity is 2.0 Ohm.m. Give your answer in MOhmsarrow_forwardA potential difference of 81 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na+) from the interior of the cell?_______________________Jarrow_forwardAssume the length of an axon membrane of about 0.10 cm is excited by an action potential (length excited = nerve speed ✕ pulse duration = 50 m/s ✕ 2.0 ms = 10 cm). In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and opposite charge of negative organic ions, as shown in the figure below. Model the axon as a parallel-plate capacitor and take C = ??oA/d and Q = CΔV to investigate the charge as follows. Use typical values for a cylindrical axon of cell thickness d = 1.6 ✕ 10−8 m, axon radius r = 1.2 ✕ 101 ?m, and cell-wall dielectric constant ? = 2.3. A diagram shows a collection of positive and negative charges in and around an axon. The diagram is divided into three sections, one on top of the other. The top section is labeled "External fluid". A row of positive charges labeled "Positive charge layer" lies along the bottom side of this section. Above the row of positive charges, there is an even mixture of…arrow_forward
- 6. The giant axon of a squid is 0.5 mm in diameter, 10 cm long, and not myelinated. Unmycli- nated cell membranes behave as capacitors with 1 μF of capacitance per square centimeter of membrane area. When the axon is charged to the -70 mV resting potential, what is the energy stored in this capacitance?arrow_forwardWhat is the energy stored in such a cell membrane if the potential difference across it is 7.05×10−2 VV ? Express your answer using two significant figures.arrow_forwardA capacitor has an initial charge of 10 µC. It is observed that after discharging for 2ms through a resistor of 8.7 kn, the charge of the capacitor is now only 8.1 μC. What is the capacitance of the capacitor? Express your answer in micro Farad, and keep three significant digits. Answer:arrow_forward
- A 10^-8 F capacitor (10 nanofarads) is charged to 100 V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is 5 x 10^13 ohms; on a humid day, the resistance is 8 x 10^6 ohms. How long will it take the capacitor voltage to dissipate to half its original value on each day?arrow_forwardPlease help. The nonpolar core of the membrane of a muscle cell has a thickness d = 4nm and a dielectric constancy k = 20. (a) What is the surface capacity of the membrane? Express your response in microcoulombs per square centimeter. (b) During muscle contraction, the internal potential of the cell increases by about 100 mV. For 1 cm2 of membrane, how many Na+ ions must have entered the cell for this purpose? (c) The medium outside the cell initially contains 0.150 mol / L of Na + ions in solution. Considering that it extends to 1 µm from the cell, what is the decrease in Na+ concentration immediately after the change in potential? (d) Estimate the number of times the process can be repeated before the Na+ ion reserve is exhausted. Please show formulas ans steps for my own understandingarrow_forwardA 10⁹-F capacitor (1 nanofarad) is charged to 50 V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is 6x10¹3 2; on a humid day, the resistance is 7 × 10 2. How long will it take the capacitor voltage to dissipate to half its original value on each day? On the dry day, it will take seconds. (Type your answer in scientific notation, using the appropriate symbol from the math palette for any multiplication. Round to three decimal places as needed.)arrow_forward
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