/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 103 An axon has the outer part of it... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 103

An axon has the outer part of its membrane positively charged and the inner part negatively charged. The membrane has a thickness of \(4.4 \mathrm{nm}\) and a dielectric constant \(\kappa=5 .\) If we model the axon as a parallel plate capacitor whose area is \(5 \mu \mathrm{m}^{2},\) what is its capacitance?

Short Answer

Expert verified
Answer: The capacitance of the axon modeled as a parallel plate capacitor is approximately \(5.01 \times 10^{-14} \mathrm{F}\).

Step by step solution

01

Write down the given values

Given the problem, we have the following values: - Thickness of the membrane (distance between the plates), d = \(4.4 \times 10^{-9} m\) - Dielectric constant, \(\kappa = 5\) - Area of the capacitor, A = \(5 \times 10^{-12} m^2\)
02

Use the vacuum permittivity constant

In order to determine the capacitance, we need to use the vacuum permittivity constant, \(\epsilon_0\). The value of vacuum permittivity is \(8.854 \times 10^{-12} \mathrm{F/m}\).
03

Calculate the capacitance using the formula

Now, we can use the formula for the capacitance of a parallel plate capacitor with a dielectric: $$ C=\kappa \epsilon_0\frac{A}{d} $$ Substituting the known values into the formula: $$ C= 5 \times (8.854 \times 10^{-12} \mathrm{F/m}) \times \frac{5 \times 10^{-12} m^2}{4.4 \times 10^{-9} m} $$
04

Solve for capacitance

Performing the multiplication and division, we get: $$ C \approx 5.01 \times 10^{-14} \mathrm{F} $$ The capacitance of the axon modeled as a parallel plate capacitor is approximately \(5.01 \times 10^{-14} \mathrm{F}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two metal spheres are separated by a distance of \(1.0 \mathrm{cm}\) and a power supply maintains a constant potential difference of \(900 \mathrm{V}\) between them. The spheres are brought closer to one another until a spark flies between them. If the dielectric strength of dry air is $3.0 \times 10^{6} \mathrm{V} / \mathrm{m},$ what is the distance between the spheres at this time?
A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)
A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Electric Charge Field and Potential

Read Explanation

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Quantum Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.