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\(\bullet\) . Electric field of axons. A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0\(\mu \mathrm{m}\) in diameter, and meas- urements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e )\) enter during this process. Although the axon is a long cylinder, the charge does not all enter every- where at the same time. A plausible model would be a a series of nearly point charges moving along the axon. Let us look at a 0.10 mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10 \(\mathrm{mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 \(\mathrm{cm}\) below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0\(\mu \mathrm{N} / \mathrm{C}\) . How far from this segment of axon could a shark be and still detect its electric field?

Short Answer

Expert verified
(a) 8.96 脳 10鈦宦孤 C, (b) 3.21 脳 10鲁 N/C outward, (c) Shark detection range 鈮 283 m.

Step by step solution

01

Calculate Charge in a 0.10 mm Section

First, calculate the total number of Na鈦 ions in a 0.10 mm section of the axon. The axon contains \(5.6 \times 10^{11}\) ions per meter, so in 0.10 mm, which is 0.0001 m, the number of ions is: \(5.6 \times 10^{11} \times 0.0001 = 5.6 \times 10^7\). The charge \(q\) in coulombs is the number of ions times the charge per ion, \(e = 1.6 \times 10^{-19} \text{ C}\). Therefore, \(q = 5.6 \times 10^7 \times 1.6 \times 10^{-19} = 8.96 \times 10^{-12} \text{ C}\).
02

Calculate Electric Field at the Body Surface

Consider the point charge located 5.0 cm below the skin surface. Use Coulomb's law for the electric field due to a point charge, \(E = \frac{k|q|}{r^2}\), where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\) and \(r = 0.05 \text{ m}\) is the distance. Substituting the values, we get: \(E = \frac{8.99 \times 10^9 \times 8.96 \times 10^{-12}}{(0.05)^2} = 3.21 \times 10^3 \text{ N/C}\). The direction is radially outward from the axon.
03

Estimate Detection Distance by Shark

Given that a shark can detect electric fields as weak as \(1.0 \times 10^{-6} \text{ N/C}\), find the distance where the electric field equals this threshold. Set \(E_\text{shark} = \frac{k|q|}{r^2}\), and solve for \(r\). \(1.0 \times 10^{-6} = \frac{8.99 \times 10^9 \times 8.96 \times 10^{-12}}{r^2}\). Solving for \(r\), we find \(r = \sqrt{\frac{8.99 \times 10^9 \times 8.96 \times 10^{-12}}{1.0 \times 10^{-6}}} \approx 283 \text{ m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's law
Coulomb's law is fundamental in understanding electric fields created by charges. It describes the force between two charged particles. This force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula is given by:
\[F = \frac{k \, |q_1 q_2|}{r^2}\]where:
  • \(F\) is the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the distance between the charges.
Coulomb's law helps us calculate the electric field (\(E\)) generated by a point charge using the equation:
\[E = \frac{k \, |q|}{r^2}\]Here, \(q\) is the point charge, and \(r\) is the distance from the charge to the point where the field is being measured. This concept was crucial in the exercise as it helped determine the electric field produced by the charge inside a neuron's axon.
neuron
Neurons are the fundamental units of the brain and nervous system, responsible for transmitting information throughout the body. They process and transmit electrical and chemical signals.
A typical neuron consists of three main parts:
  • Cell Body: Contains the nucleus and other organelles.
  • Dendrites: Branch-like structures that receive messages from other neurons.
  • Axon: A long, slender projection that conducts electrical impulses away from the neuron's cell body toward other neurons or muscles.
The axon is particularly important in creating an electric field when a nerve signal is transmitted. Neurons, through rapid movements of ions like sodium \((\text{Na}^+)\), generate electrical activity that can create detectable electric fields. This electric field is an essential part of the neuron's communication process.
axon
The axon is a crucial part of a neuron. It is responsible for transmitting electrical signals from the neuron's cell body to the target cells, such as other neurons or muscles. Axons can vary in length and often range from millimeters to over a meter in long neurons.
During the transmission of nerve signals, ion channels in the axon membrane allow ions like \(\text{Na}^+\) to enter, leading to a change in electric potential along the axon. This process generates an electric potential difference that propagates as an action potential.
In the exercise, the axon was modeled as a series of point charges to simplify the calculation of the electric field produced. This simplification helps understand the effects of charges entering the axon and provides an approximation of the electric field experienced at different distances from the axon.
point charge
A point charge refers to an idealized model of a charged particle with negligible size, where the charge is concentrated at a single point in space. This concept simplifies the analysis and calculation of electric fields and forces between charges. In reality, sizes and distributions of actual charges can be complex, but the point charge model offers a good approximation for many problems.
In the context of the exercise, the axon of a neuron was treated as a series of almost point charges to calculate the electric field it produces. Despite the axon's cylindrical shape, modeling it with point charges helps in employing Coulomb's law to compute the field at varying distances, such as the detection distance by a shark. Point charges allow for straightforward application of the principles of electrostatics without the intricacies of more complex charge distributions.

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Most popular questions from this chapter

\(\bullet$$\bullet \mathrm{A}-5.00 \mathrm{nC}\) point charge is on the \(x\) axis at \(x=1.20 \mathrm{m} . \mathrm{A}\) second point charge \(Q\) is on the \(x\) axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) direction, \((\mathrm{b}) 45.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) direction?

\(\bullet\) During a violent electrical storm, a car is struck by a falling high-voltage wire that puts an excess charge of \(-850 \mu C\) on the metal car. (a) How much of this charge is on the inner sur- face of the car? (b) How much is on the outer surface?

\(\bullet$$\bullet\) A charge of \(-3.00 \mathrm{nC}\) is placed at the origin of an \(x y-\)coordi- nate system, and a charge of 2.00 \(\mathrm{nC}\) is placed on the \(y\) axis at \(y=4.00 \mathrm{cm} .\) (a) If a third charge, of \(5.00 \mathrm{nC},\) is now placed at the point \(x=3.00 \mathrm{cm}, y=4.00 \mathrm{cm},\) find the \(x\) and \(y\) com- ponents of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

\(\bullet\) Electric fields in the atom. (a) Within the nucleus. What strength of electric field does a proton produce at the distance of another proton, about \(5.0 \times 10^{-15} \mathrm{m}\) away? (b) At the elec- trons. What strength of electric field does this proton produce at the distance of the electrons, approximately \(5.0 \times 10^{-10} \mathrm{m}\) away?

\(\bullet\) \(\bullet\) (a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A\) , the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A\) , the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

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